🔬 Model Lab

## Step 1: Problem 1 - Restate the problem
We need to find how many whole numbers from 1 through 46 are divisible by either 3 or 5 or both.

## Step 2: Problem 1 - Calculate numbers divisible by 3
First, let's find the numbers divisible by 3. The largest number less than or equal to 46 that is divisible by 3 is 45, and the number of multiples of 3 from 1 to 46 is 45 / 3 = 15.

## Step 3: Problem 1 - Calculate numbers divisible by 5
Next, find the numbers divisible by 5. The largest number less than or equal to 46 that is divisible by 5 is 45, and the number of multiples of 5 from 1 to 46 is 45 / 5 = 9.

## Step 4: Problem 1 - Calculate numbers divisible by both 3 and 5
Numbers divisible by both 3 and 5 are divisible by 15. The largest number less than or equal to 46 that is divisible by 15 is 45, and the number of multiples of 15 from 1 to 46 is 45 / 15 = 3.

## Step 5: Problem 1 - Apply the principle of inclusion-exclusion
To find the total number of numbers divisible by either 3 or 5 or both, we use the principle of inclusion-exclusion: (numbers divisible by 3) + (numbers divisible by 5) - (numbers divisible by both) = 15 + 9 - 3 = 21.

## Step 6: Problem 1 - Provide the answer
ANSWER 1: B

## Step 7: Problem 2 - Restate the problem
Jamal has 6 green socks, 18 purple socks, and 12 orange socks. After adding more purple socks, the probability of picking a purple sock is 60%. We need to find how many purple socks Jamal added.

## Step 8: Problem 2 - Set up the initial conditions
Initially, Jamal has 6 + 18 + 12 = 36 socks.

## Step 9: Problem 2 - Set up the equation for the probability of picking a purple sock after adding x purple socks
After adding x purple socks, the total number of socks is 36 + x, and the number of purple socks is 18 + x. The probability of picking a purple sock is (18 + x) / (36 + x) = 0.6.

## Step 10: Problem 2 - Solve the equation for x
(18 + x) / (36 + x) = 0.6. Multiplying both sides by (36 + x) gives 18 + x = 0.6(36 + x). Expanding, we get 18 + x = 21.6 + 0.6x. Rearranging, x - 0.6x = 21.6 - 18, so 0.4x = 3.6. Dividing both sides by 0.4, x = 9.

## Step 11: Problem 2 - Provide the answer
ANSWER 2: B

## Step 12: Problem 3 - Restate the problem
We need to find which year is NOT a lucky year, where a lucky year has at least one date where the month times the day equals the last two digits of the year.

## Step 13: Problem 3 - Analyze the condition for a lucky year
For a year to be lucky, there must exist a month (1-12) and a day (1-31) such that month * day = last two digits of the year.

## Step 14: Problem 3 - Check each given year
Checking each year from 1990 to 1994 to see if any date satisfies the condition.

## Step 15: Problem 3 - Analyze 1990 to 1994
- For 1990, the last two digits are 90. It's difficult to get 90 by multiplying two numbers between 1 and 12 and 1 and 31, respectively, because the factors of 90 that could work are (1,90), (2,45), (3,30), (5,18), (6,15), (9,10). Only (9,10) is plausible, and it works as 9/10/90.
- For 1991, 91 can be achieved with 7/13/91, so it's lucky.
- For 1992, 92 can be achieved, for example, with 4/23/92, so it's lucky.
- For 1993, 93 can be achieved with 3/31/93, so it's lucky.
- For 1994, 94 is not easily achieved because its factors are (1,94), (2,47). No month-day combination gives 94.

## Step 16: Problem 3 - Provide the answer
Given the analysis, 1994 is not lucky because there's no month-day product equal to 94.

ANSWER 3: E

## Step 17: Problem 4 - Restate the problem
We need to find the smallest difference possible by placing the digits 2, 4, 5, 6, 9 in the boxes of a subtraction problem.

## Step 18: Problem 4 - Understand the structure of the subtraction problem
To minimize the difference, we should minimize the minuend and maximize the subtrahend.

## Step 19: Problem 4 - Arrange the digits to minimize the difference
To achieve the smallest difference, the hundreds place of the minuend should be as small as possible, and the hundreds place of the subtrahend should be as large as possible. So, the minuend could be 2** and the subtrahend 9**. The remaining digits are 4, 5, 6.

## Step 20: Problem 4 - Try different combinations
Let's try to make the minuend as small as possible and the subtrahend as large as possible: 245 - 986 is not valid because it results in a negative number. A valid attempt is 246 - 985 or similar combinations.

## Step 21: Problem 4 - Calculate the difference for a valid combination
For 964 - 852 or similar, the difference is not minimal. Let's directly check: 245 - 196 = 49 (not an option), but trying to minimize, we can get 246 - 195 = 51 (still not an option). A plausible arrangement is 254 - 196 = 58.

## Step 22: Problem 4 - Provide the answer
The smallest difference is achieved with an arrangement like 254 - 196.

ANSWER 4: A

## Step 23: Problem 5 - Restate the problem
Given W, X, Y, Z are different integers from 1 to 4, and WX - YZ = 1, we need to find W + Y.

## Step 24: Problem 5 - Understand the equation WX - YZ = 1
Since W, X, Y, Z are different integers from 1 to 4, we need to consider the possible values of WX and YZ.

## Step 25: Problem 5 - Analyze possible values for WX and YZ
The possible products are: 1*2=2, 1*3=3, 1*4=4, 2*1=2, 2*3=6, 2*4=8, 3*1=3, 3*2=6, 3*4=12, 4*1=4, 4*2=8, 4*3=12. We need a difference of 1 between two of these products.

## Step 26: Problem 5 - Check for a difference of 1
The products are 2, 3, 4, 6, 8, 12. The pairs that differ by 1 are (2,3), (3,4). So, either WX = 3 and YZ = 2 or WX = 4 and YZ = 3.

## Step 27: Problem 5 - Solve for W + Y
For WX = 3 and YZ = 2, W = 3, X = 1, Y = 2, Z = 1 (not valid because Z = X), or W = 1, X = 3, Y = 1, Z = 2 (not valid because W = Y). A valid case is W = 3, X = 1, Y = 1, Z = 2 (not valid), but if we consider W = 1, X = 3, Y = 2, Z = 1, it's valid and gives W + Y = 1 + 2 = 3.

## Step 28: Problem 5 - Provide the answer
For the valid case where W = 1, X = 3, Y = 2, Z = 1 doesn't work due to repetition, but

Problem 1:

We need count the whole numbers from 1 through 46 that are divisible by 3 or 5 or both.

Count multiples of 3:

\[
\left\lfloor \frac{46}{3} \right\rfloor = 15
\]

Count multiples of 5:

\[
\left\lfloor \frac{46}{5} \right\rfloor = 9
\]

But multiples of both 3 and 5, meaning multiples of 15, were counted twice, so subtract them once:

\[
\left\lfloor \frac{46}{15} \right\rfloor = 3
\]

Therefore the total is

\[
15+9-3=21
\]

So choices like 24 or 27 overcount because they do not remove the overlap.

ANSWER 1: B

Problem 2:

Initially Jamal has:

\[
6+18+12=36
\]

socks total, of which 18 are purple.

Suppose he adds \(x\) purple socks. Then the number of purple socks is

\[
18+x
\]

and the total number of socks is

\[
36+x
\]

The probability of choosing a purple sock is now 60%, or \(\frac35\). So

\[
\frac{18+x}{36+x}=\frac35
\]

Cross multiply:

\[
5(18+x)=3(36+x)
\]

\[
90+5x=108+3x
\]

\[
2x=18
\]

\[
x=9
\]

So Jamal added 9 purple socks.

ANSWER 2: B

Problem 3:

A lucky year has some valid date month/day/year where

\[
\text{month} \times \text{day} = \text{last two digits of the year}
\]

Check each year:

For 1990:

\[
9 \times 10 = 90
\]

So 9/10/90 works.

For 1991:

\[
7 \times 13 = 91
\]

So 7/13/91 works.

For 1992:

\[
4 \times 23 = 92
\]

So 4/23/92 works.

For 1993:

\[
3 \times 31 = 93
\]

So 3/31/93 works.

For 1994:

\[
94=2 \times 47
\]

But day 47 is not valid. The only factor pairs are \(1 \times 94\) and \(2 \times 47\), and neither gives a valid date.

Thus 1994 is not lucky.

ANSWER 3: E

Problem 4:

The subtraction problem uses the digits 2, 4, 5, 6, 9 exactly once in a 3-digit number minus a 2-digit number.

To make the difference as small as possible, the 3-digit number should be as small as possible, and the 2-digit number should be as large as possible.

The hundreds digit of the 3-digit number should be the smallest digit, 2.

Now we want the rest of the top number small and the bottom number large. Using the remaining digits 4, 5, 6, 9, the smallest possible two digits for the top number are 4 and 5, making 245, while the largest possible 2-digit number from the remaining digits is 96.

So try:

\[
245-96=149
\]

If we tried to get below 149, we would need either a smaller 3-digit number than 245 or a larger 2-digit number than 96, but neither is possible with these digits.

Therefore the smallest difference is 149.

ANSWER 4: C

Problem 5:

The equation is

\[
\frac{W}{X}-\frac{Y}{Z}=1
\]

where \(W, X, Y, Z\) are the numbers 1, 2, 3, 4 in some order.

We need find an arrangement that works.

Try to make the first fraction bigger than 1. One successful choice is

\[
\frac{3}{1}-\frac{4}{2}=1
\]

because

\[
3-2=1
\]

So

\[
W=3,\quad X=1,\quad Y=4,\quad Z=2
\]

Then

\[
W+Y=3+4=7
\]

The other choices are smaller sums, but the only working arrangement gives 7.

ANSWER 5: E

Problem 6:

There are four-digit integers from 1000 through 9999. The total number is

\[
9999-1000+1=9000
\]

We want odd integers whose digits are all distinct.

For a number to be odd, the units digit must be one of

\[
1,3,5,7,9
\]

So there are 5 choices for the units digit.

The thousands digit cannot be 0 and cannot equal the units digit. Since there are 9 possible nonzero thousands digits, and one is already used, there are

\[
8
\]

choices.

The hundreds digit can be any digit except the two already used:

\[
8
\]

choices.

The tens digit can be any digit except the three already used:

\[
7
\]

choices.

So the number of favorable integers is

\[
5 \cdot 8 \cdot 8 \cdot 7=2240
\]

The probability is

\[
\frac{2240}{9000}
\]

Reduce:

\[
\frac{2240}{9000}=\frac{224}{900}=\frac{56}{225}
\]

ANSWER 6: B

Problem 7:

Let \(n\) be the number of employees, and let \(F\) be the amount of money in the fund.

The manager planned to give each employee $50, but the fund was $5 short. So

\[
F=50n-5
\]

Instead, the manager gave each employee $45 and had $95 left. So

\[
F=45n+95
\]

Set these equal:

\[
50n-5=45n+95
\]

\[
5n=100
\]

\[
n=20
\]

Now find the fund:

\[
F=50(20)-5=1000-5=995
\]

So there were $995 in the fund.

ANSWER 7: E

Problem 8:

We are given

\[
ABA \times CD = CDCD
\]

The number \(ABA\) is

\[
101A+10B
\]

The number \(CDCD\) is a repeated two-digit number, so

\[
CDCD=101 \times CD
\]

Thus the equation becomes

\[
ABA \times CD = 101 \times CD
\]

Since \(CD\) is a two-digit number, it is not 0, so we can divide both sides by \(CD\):

\[
ABA=101
\]

Therefore

\[
A=1,\quad B=0
\]

So

\[
A+B=1+0=1
\]

ANSWER 8: A

Problem 9:

We need choose 3 numbers from

\[
\{89,95,99,132,166,173\}
\]

so that their sum is even.

Classify by parity.

Odd numbers:

\[
89,95,99,173
\]

There are 4 odds.

Even numbers:

\[
132,166
\]

There are 2 evens.

A sum of three numbers is even if it contains either:

- 0 odd numbers and 3 even numbers, or
- 2 odd numbers and 1 even number.

We cannot choose 3 even numbers because there are only 2 even numbers.

So we must choose 2 odds and 1 even.

Number of ways:

\[
\binom{4}{2}\binom{2}{1}=6 \cdot 2=12
\]

ANSWER 9: D

Problem 10:

The set has 5 positive integers with mean 5, so the total sum is

\[
5 \cdot 5=25
\]

The median is 5, so when the numbers are ordered, the middle number is 5.

Also, 8 is the only mode, meaning 8 appears more often than any other number. Since there are only five numbers and the median is 5, the two largest numbers must be 8 and 8:

\[
\_,\_,5,8,8
\]

Their sum so far is

\[
5+8+8=21
\]

So the first two numbers must sum to

\[
25-21=4
\]

They are positive integers less than or equal to 5. The possible pairs are:

\[
1,3 \quad \text{or} \quad 2,2
\]

But \(2,2\) would make 2 appear twice, tying 8 as a mode. Since 8 must be the only mode, the pair must be

\[
1,3
\]

So the set is

\[
1,3,5,8,8
\]

The difference between the largest and smallest integers is

\[
8-1=7
\]

ANSWER 10: D

Problem 11:

We need count three-digit whole numbers between 99 and 999 that contain exactly one 0.

A t

**Problem 1:**
How many whole numbers from 1 through 46 are divisible by either 3 or 5 or both?

**Step-by-step solution:**
To find the number of integers divisible by 3, 5, or both, we can use the Principle of Inclusion-Exclusion. 
1. First, we count the numbers from 1 to 46 that are divisible by 3:
   $\lfloor 46 \div 3 \rfloor = 15$
2. Next, we count the numbers from 1 to 46 that are divisible by 5:
   $\lfloor 46 \div 5 \rfloor = 9$
3. Then, we count the numbers that are divisible by BOTH 3 and 5 (which means they are divisible by their least common multiple, 15):
   $\lfloor 46 \div 15 \rfloor = 3$
4. Finally, we add the multiples of 3 and 5, and subtract the multiples of 15 so they aren't counted twice:
   $15 + 9 - 3 = 21$

ANSWER 1: B

---

**Problem 2:**
Jamal has a drawer containing 6 green socks, 18 purple socks, and 12 orange socks. After adding more purple socks, Jamal noticed that there is now a 60% chance that a sock randomly selected from the drawer is purple. How many purple socks did Jamal add?

**Step-by-step solution:**
1. Initially, Jamal has $6 + 18 + 12 = 36$ socks in total.
2. Let $x$ be the number of purple socks Jamal added. 
3. The new number of purple socks is $18 + x$.
4. The new total number of socks is $36 + x$.
5. We are told the new probability of selecting a purple sock is 60%, which simplifies to $\frac{60}{100} = \frac{3}{5}$.
6. We set up the equation:
   $\frac{18 + x}{36 + x} = \frac{3}{5}$
7. Cross-multiply to solve for $x$:
   $5(18 + x) = 3(36 + x)$
   $90 + 5x = 108 + 3x$
   $2x = 18$
   $x = 9$

Jamal added 9 purple socks.

ANSWER 2: B

---

**Problem 3:**
A lucky year is one in which at least one date, written as month/day/year, has the property that the month times the day equals the last two digits of the year. For example, 1956 is lucky because 7/8/56 has 7 × 8 = 56. Which of the following is NOT a lucky year?

**Step-by-step solution:**
We need to find if the last two digits of each given year can be factored into a valid month (1 through 12) and a valid day (1 through 31).
- **A. 1990:** $90 = 9 \times 10$. This corresponds to September 10th (9/10/90). Lucky.
- **B. 1991:** $91 = 7 \times 13$. This corresponds to July 13th (7/13/91). Lucky.
- **C. 1992:** $92 = 4 \times 23$. This corresponds to April 23rd (4/23/92). Lucky.
- **D. 1993:** $93 = 3 \times 31$. This corresponds to March 31st (3/31/93). Lucky.
- **E. 1994:** $94 = 2 \times 47$. Since a month cannot have 47 days, and 94 has no other integer pairs that fit the calendar constraints (1×94 is also invalid), 1994 is NOT a lucky year.

ANSWER 3: E

---

**Problem 4:**
When placing each of the digits 2, 4, 5, 6, 9 in exactly one of the boxes of this subtraction problem, what is the smallest difference that is possible?       −

**Step-by-step solution:**
We are given 5 digits to place in a subtraction problem. To yield a positive difference using 5 digits, the setup must be a 3-digit number minus a 2-digit number.
To get the smallest possible difference, we need the 3-digit number to be as small as possible and the 2-digit number to be as large as possible.
1. The smallest possible hundreds digit for the 3-digit number is 2. So the number is $2\_ \_$.
2. We are left with the digits 4, 5, 6, and 9. 
3. To minimize the 3-digit number, the remaining two digits should be the smallest available: 4 and 5, making the number 245.
4. To maximize the 2-digit number being subtracted, we use the largest remaining digits: 9 and 6, making the number 96.
5. Subtracting the two gives: $245 - 96 = 149$.
(Testing other close combinations like $254 - 96 = 158$ or $246 - 95 = 151$ confirms 149 is the absolute minimum difference.)

ANSWER 4: C

---

**Problem 5:**
Each of the letters W , X , Y , and Z represents a different integer in the set {1, 2, 3, 4}, but not necessarily in that order. If W X − Y Z = 1, then the sum of W and Y is

**Step-by-step solution:**
The equation given is meant to be read as fractions (a common formatting artifact drops the fraction bar): $\frac{W}{X} - \frac{Y}{Z} = 1$.
Let's test the values {1, 2, 3, 4} to satisfy this equation:
- If we set $W = 3$, $X = 1$, $Y = 4$, and $Z = 2$:
  $\frac{3}{1} - \frac{4}{2} = 3 - 2 = 1$
This perfectly uses all four distinct digits exactly once.
The question asks for the sum of $W$ and $Y$:
$W + Y = 3 + 4 = 7$

*(Note: Checking $W \times X - Y \times Z = 1$ yields no solutions where all 4 digits are distinct, confirming the fractional interpretation).*

ANSWER 5: E

---

**Problem 6:**
An integer between 1000 and 9999, inclusive, is chosen at random. What is the probability that it is an odd integer whose digits are all distinct?

**Step-by-step solution:**
1. The total number of integers from 1000 to 9999 inclusive is $9999 - 1000 + 1 = 9000$.
2. We want to form a 4-digit odd number (let's call its digits A B C D) with all distinct digits.
3. Because it must be odd, the units digit (D) must be 1, 3, 5, 7, or 9. There are **5** choices for D.
4. The thousands digit (A) cannot be 0, and it cannot be the digit used for D. Thus, out of 10 possible digits, 2 are restricted, leaving **8** choices for A.
5. The hundreds digit (B) can be 0, but cannot be the digits used for A or D. This leaves **8** choices for B.
6. The tens digit (C) cannot be the digits used for A, B, or D. This leaves **7** choices for C.
7. The total number of such integers is $5 \times 8 \times 8 \times 7 = 2240$.
8. The probability is $\frac{2240}{9000} = \frac{224}{900} = \frac{56}{225}$.

ANSWER 6: B

---

**Problem 7:**
The manager of a company planned to give a $50 bonus to each employee from the company fund, but the fund was $5 short of what was needed. Instead the manager gave each employee a $45 bonus and kept the remaining $95 in the fund. How much money was in the company fund before any bonuses were paid?

**Step-by-step solution:**
Let $n$ be the number of employees.
1. If the manager gave $50 to each employee, the fund would be $5 short. So, the total fund is $5

## Step 1: Problem 1 - Restate the problem
We need to find how many whole numbers from 1 through 46 are divisible by either 3 or 5 or both.

## Step 2: Problem 1 - Calculate numbers divisible by 3
First, let's find the numbers divisible by 3. The largest number less than or equal to 46 that is divisible by 3 is 45, and the number of multiples of 3 from 1 to 46 is 45 / 3 = 15.

## Step 3: Problem 1 - Calculate numbers divisible by 5
Next, find the numbers divisible by 5. The largest number less than or equal to 46 that is divisible by 5 is 45, and the number of multiples of 5 from 1 to 46 is 45 / 5 = 9.

## Step 4: Problem 1 - Calculate numbers divisible by both 3 and 5
Numbers divisible by both 3 and 5 are divisible by 15. The largest number less than or equal to 46 that is divisible by 15 is 45, and the number of multiples of 15 from 1 to 46 is 45 / 15 = 3.

## Step 5: Problem 1 - Apply the principle of inclusion-exclusion
To find the total number of numbers divisible by either 3 or 5 or both, we use the principle of inclusion-exclusion: (numbers divisible by 3) + (numbers divisible by 5) - (numbers divisible by both) = 15 + 9 - 3 = 21.

## Step 6: Problem 1 - Provide the answer
ANSWER 1: B

## Step 7: Problem 2 - Restate the problem
Jamal has 6 green socks, 18 purple socks, and 12 orange socks. After adding more purple socks, the probability of picking a purple sock is 60%. We need to find how many purple socks Jamal added.

## Step 8: Problem 2 - Set up the initial conditions
Initially, Jamal has 6 + 18 + 12 = 36 socks.

## Step 9: Problem 2 - Set up the equation for the probability of picking a purple sock after adding x purple socks
After adding x purple socks, the total number of socks is 36 + x, and the number of purple socks is 18 + x. The probability of picking a purple sock is (18 + x) / (36 + x) = 0.6.

## Step 10: Problem 2 - Solve the equation for x
(18 + x) / (36 + x) = 0.6. Multiplying both sides by (36 + x) gives 18 + x = 0.6(36 + x). Expanding, we get 18 + x = 21.6 + 0.6x. Rearranging, x - 0.6x = 21.6 - 18, so 0.4x = 3.6. Dividing both sides by 0.4, x = 9.

## Step 11: Problem 2 - Provide the answer
ANSWER 2: B

## Step 12: Problem 3 - Restate the problem
We need to find which year is NOT a lucky year, where a lucky year has at least one date where the month times the day equals the last two digits of the year.

## Step 13: Problem 3 - Analyze the condition for a lucky year
For a year to be lucky, there must exist a month (1-12) and a day (1-31) such that month * day = last two digits of the year.

## Step 14: Problem 3 - Check each given year
Checking each year from 1990 to 1994 to see if any date satisfies the condition.

## Step 15: Problem 3 - Analyze 1990 to 1994
- For 1990, the last two digits are 90. It's difficult to get 90 by multiplying two numbers between 1 and 12 and 1 and 31, respectively, because the factors of 90 that could work are (1,90), (2,45), (3,30), (5,18), (6,15), (9,10). Only (9,10) is plausible, and it works as 9/10/90.
- For 1991, 91 can be achieved with 7/13/91, so it's lucky.
- For 1992, 92 can be achieved, for example, with 4/23/92, so it's lucky.
- For 1993, 93 can be achieved with 3/31/93, so it's lucky.
- For 1994, 94 is not easily achieved because its factors are (1,94), (2,47). No month-day combination gives 94.

## Step 16: Problem 3 - Provide the answer
Given the analysis, 1994 is not lucky because there's no month-day product equal to 94.

ANSWER 3: E

## Step 17: Problem 4 - Restate the problem
We need to find the smallest difference possible by placing the digits 2, 4, 5, 6, 9 in the boxes of a subtraction problem.

## Step 18: Problem 4 - Understand the structure of the subtraction problem
To minimize the difference, we should minimize the minuend and maximize the subtrahend.

## Step 19: Problem 4 - Arrange the digits to minimize the difference
To achieve the smallest difference, the hundreds place of the minuend should be as small as possible, and the hundreds place of the subtrahend should be as large as possible. So, the minuend could be 2** and the subtrahend 9**. The remaining digits are 4, 5, 6.

## Step 20: Problem 4 - Try different combinations
Let's try to make the minuend as small as possible and the subtrahend as large as possible: 245 - 986 is not valid because it results in a negative number. A valid attempt is 246 - 985 or similar combinations.

## Step 21: Problem 4 - Calculate the difference for a valid combination
For 964 - 852 or similar, the difference is not minimal. Let's directly check: 245 - 196 = 49 (not an option), but trying to minimize, we can get 246 - 195 = 51 (still not an option). A plausible arrangement is 254 - 196 = 58.

## Step 22: Problem 4 - Provide the answer
The smallest difference is achieved with an arrangement like 254 - 196.

ANSWER 4: A

## Step 23: Problem 5 - Restate the problem
Given W, X, Y, Z are different integers from 1 to 4, and WX - YZ = 1, we need to find W + Y.

## Step 24: Problem 5 - Understand the equation WX - YZ = 1
Since W, X, Y, Z are different integers from 1 to 4, we need to consider the possible values of WX and YZ.

## Step 25: Problem 5 - Analyze possible values for WX and YZ
The possible products are: 1*2=2, 1*3=3, 1*4=4, 2*1=2, 2*3=6, 2*4=8, 3*1=3, 3*2=6, 3*4=12, 4*1=4, 4*2=8, 4*3=12. We need a difference of 1 between two of these products.

## Step 26: Problem 5 - Check for a difference of 1
The products are 2, 3, 4, 6, 8, 12. The pairs that differ by 1 are (2,3), (3,4). So, either WX = 3 and YZ = 2 or WX = 4 and YZ = 3.

## Step 27: Problem 5 - Solve for W + Y
For WX = 3 and YZ = 2, W = 3, X = 1, Y = 2, Z = 1 (not valid because Z = X), or W = 1, X = 3, Y = 1, Z = 2 (not valid because W = Y). A valid case is W = 3, X = 1, Y = 1, Z = 2 (not valid), but if we consider W = 1, X = 3, Y = 2, Z = 1, it's valid and gives W + Y = 1 + 2 = 3.

## Step 28: Problem 5 - Provide the answer
For the valid case where W = 1, X = 3, Y = 2, Z = 1 doesn't work due to repetition, but

Problem 1:

We need count the whole numbers from 1 through 46 that are divisible by 3 or 5 or both.

Count multiples of 3:

\[
\left\lfloor \frac{46}{3} \right\rfloor = 15
\]

Count multiples of 5:

\[
\left\lfloor \frac{46}{5} \right\rfloor = 9
\]

But multiples of both 3 and 5, meaning multiples of 15, were counted twice, so subtract them once:

\[
\left\lfloor \frac{46}{15} \right\rfloor = 3
\]

Therefore the total is

\[
15+9-3=21
\]

So choices like 24 or 27 overcount because they do not remove the overlap.

ANSWER 1: B

Problem 2:

Initially Jamal has:

\[
6+18+12=36
\]

socks total, of which 18 are purple.

Suppose he adds \(x\) purple socks. Then the number of purple socks is

\[
18+x
\]

and the total number of socks is

\[
36+x
\]

The probability of choosing a purple sock is now 60%, or \(\frac35\). So

\[
\frac{18+x}{36+x}=\frac35
\]

Cross multiply:

\[
5(18+x)=3(36+x)
\]

\[
90+5x=108+3x
\]

\[
2x=18
\]

\[
x=9
\]

So Jamal added 9 purple socks.

ANSWER 2: B

Problem 3:

A lucky year has some valid date month/day/year where

\[
\text{month} \times \text{day} = \text{last two digits of the year}
\]

Check each year:

For 1990:

\[
9 \times 10 = 90
\]

So 9/10/90 works.

For 1991:

\[
7 \times 13 = 91
\]

So 7/13/91 works.

For 1992:

\[
4 \times 23 = 92
\]

So 4/23/92 works.

For 1993:

\[
3 \times 31 = 93
\]

So 3/31/93 works.

For 1994:

\[
94=2 \times 47
\]

But day 47 is not valid. The only factor pairs are \(1 \times 94\) and \(2 \times 47\), and neither gives a valid date.

Thus 1994 is not lucky.

ANSWER 3: E

Problem 4:

The subtraction problem uses the digits 2, 4, 5, 6, 9 exactly once in a 3-digit number minus a 2-digit number.

To make the difference as small as possible, the 3-digit number should be as small as possible, and the 2-digit number should be as large as possible.

The hundreds digit of the 3-digit number should be the smallest digit, 2.

Now we want the rest of the top number small and the bottom number large. Using the remaining digits 4, 5, 6, 9, the smallest possible two digits for the top number are 4 and 5, making 245, while the largest possible 2-digit number from the remaining digits is 96.

So try:

\[
245-96=149
\]

If we tried to get below 149, we would need either a smaller 3-digit number than 245 or a larger 2-digit number than 96, but neither is possible with these digits.

Therefore the smallest difference is 149.

ANSWER 4: C

Problem 5:

The equation is

\[
\frac{W}{X}-\frac{Y}{Z}=1
\]

where \(W, X, Y, Z\) are the numbers 1, 2, 3, 4 in some order.

We need find an arrangement that works.

Try to make the first fraction bigger than 1. One successful choice is

\[
\frac{3}{1}-\frac{4}{2}=1
\]

because

\[
3-2=1
\]

So

\[
W=3,\quad X=1,\quad Y=4,\quad Z=2
\]

Then

\[
W+Y=3+4=7
\]

The other choices are smaller sums, but the only working arrangement gives 7.

ANSWER 5: E

Problem 6:

There are four-digit integers from 1000 through 9999. The total number is

\[
9999-1000+1=9000
\]

We want odd integers whose digits are all distinct.

For a number to be odd, the units digit must be one of

\[
1,3,5,7,9
\]

So there are 5 choices for the units digit.

The thousands digit cannot be 0 and cannot equal the units digit. Since there are 9 possible nonzero thousands digits, and one is already used, there are

\[
8
\]

choices.

The hundreds digit can be any digit except the two already used:

\[
8
\]

choices.

The tens digit can be any digit except the three already used:

\[
7
\]

choices.

So the number of favorable integers is

\[
5 \cdot 8 \cdot 8 \cdot 7=2240
\]

The probability is

\[
\frac{2240}{9000}
\]

Reduce:

\[
\frac{2240}{9000}=\frac{224}{900}=\frac{56}{225}
\]

ANSWER 6: B

Problem 7:

Let \(n\) be the number of employees, and let \(F\) be the amount of money in the fund.

The manager planned to give each employee $50, but the fund was $5 short. So

\[
F=50n-5
\]

Instead, the manager gave each employee $45 and had $95 left. So

\[
F=45n+95
\]

Set these equal:

\[
50n-5=45n+95
\]

\[
5n=100
\]

\[
n=20
\]

Now find the fund:

\[
F=50(20)-5=1000-5=995
\]

So there were $995 in the fund.

ANSWER 7: E

Problem 8:

We are given

\[
ABA \times CD = CDCD
\]

The number \(ABA\) is

\[
101A+10B
\]

The number \(CDCD\) is a repeated two-digit number, so

\[
CDCD=101 \times CD
\]

Thus the equation becomes

\[
ABA \times CD = 101 \times CD
\]

Since \(CD\) is a two-digit number, it is not 0, so we can divide both sides by \(CD\):

\[
ABA=101
\]

Therefore

\[
A=1,\quad B=0
\]

So

\[
A+B=1+0=1
\]

ANSWER 8: A

Problem 9:

We need choose 3 numbers from

\[
\{89,95,99,132,166,173\}
\]

so that their sum is even.

Classify by parity.

Odd numbers:

\[
89,95,99,173
\]

There are 4 odds.

Even numbers:

\[
132,166
\]

There are 2 evens.

A sum of three numbers is even if it contains either:

- 0 odd numbers and 3 even numbers, or
- 2 odd numbers and 1 even number.

We cannot choose 3 even numbers because there are only 2 even numbers.

So we must choose 2 odds and 1 even.

Number of ways:

\[
\binom{4}{2}\binom{2}{1}=6 \cdot 2=12
\]

ANSWER 9: D

Problem 10:

The set has 5 positive integers with mean 5, so the total sum is

\[
5 \cdot 5=25
\]

The median is 5, so when the numbers are ordered, the middle number is 5.

Also, 8 is the only mode, meaning 8 appears more often than any other number. Since there are only five numbers and the median is 5, the two largest numbers must be 8 and 8:

\[
\_,\_,5,8,8
\]

Their sum so far is

\[
5+8+8=21
\]

So the first two numbers must sum to

\[
25-21=4
\]

They are positive integers less than or equal to 5. The possible pairs are:

\[
1,3 \quad \text{or} \quad 2,2
\]

But \(2,2\) would make 2 appear twice, tying 8 as a mode. Since 8 must be the only mode, the pair must be

\[
1,3
\]

So the set is

\[
1,3,5,8,8
\]

The difference between the largest and smallest integers is

\[
8-1=7
\]

ANSWER 10: D

Problem 11:

We need count three-digit whole numbers between 99 and 999 that contain exactly one 0.

A t

**Problem 1:**
How many whole numbers from 1 through 46 are divisible by either 3 or 5 or both?

**Step-by-step solution:**
To find the number of integers divisible by 3, 5, or both, we can use the Principle of Inclusion-Exclusion. 
1. First, we count the numbers from 1 to 46 that are divisible by 3:
   $\lfloor 46 \div 3 \rfloor = 15$
2. Next, we count the numbers from 1 to 46 that are divisible by 5:
   $\lfloor 46 \div 5 \rfloor = 9$
3. Then, we count the numbers that are divisible by BOTH 3 and 5 (which means they are divisible by their least common multiple, 15):
   $\lfloor 46 \div 15 \rfloor = 3$
4. Finally, we add the multiples of 3 and 5, and subtract the multiples of 15 so they aren't counted twice:
   $15 + 9 - 3 = 21$

ANSWER 1: B

---

**Problem 2:**
Jamal has a drawer containing 6 green socks, 18 purple socks, and 12 orange socks. After adding more purple socks, Jamal noticed that there is now a 60% chance that a sock randomly selected from the drawer is purple. How many purple socks did Jamal add?

**Step-by-step solution:**
1. Initially, Jamal has $6 + 18 + 12 = 36$ socks in total.
2. Let $x$ be the number of purple socks Jamal added. 
3. The new number of purple socks is $18 + x$.
4. The new total number of socks is $36 + x$.
5. We are told the new probability of selecting a purple sock is 60%, which simplifies to $\frac{60}{100} = \frac{3}{5}$.
6. We set up the equation:
   $\frac{18 + x}{36 + x} = \frac{3}{5}$
7. Cross-multiply to solve for $x$:
   $5(18 + x) = 3(36 + x)$
   $90 + 5x = 108 + 3x$
   $2x = 18$
   $x = 9$

Jamal added 9 purple socks.

ANSWER 2: B

---

**Problem 3:**
A lucky year is one in which at least one date, written as month/day/year, has the property that the month times the day equals the last two digits of the year. For example, 1956 is lucky because 7/8/56 has 7 × 8 = 56. Which of the following is NOT a lucky year?

**Step-by-step solution:**
We need to find if the last two digits of each given year can be factored into a valid month (1 through 12) and a valid day (1 through 31).
- **A. 1990:** $90 = 9 \times 10$. This corresponds to September 10th (9/10/90). Lucky.
- **B. 1991:** $91 = 7 \times 13$. This corresponds to July 13th (7/13/91). Lucky.
- **C. 1992:** $92 = 4 \times 23$. This corresponds to April 23rd (4/23/92). Lucky.
- **D. 1993:** $93 = 3 \times 31$. This corresponds to March 31st (3/31/93). Lucky.
- **E. 1994:** $94 = 2 \times 47$. Since a month cannot have 47 days, and 94 has no other integer pairs that fit the calendar constraints (1×94 is also invalid), 1994 is NOT a lucky year.

ANSWER 3: E

---

**Problem 4:**
When placing each of the digits 2, 4, 5, 6, 9 in exactly one of the boxes of this subtraction problem, what is the smallest difference that is possible?       −

**Step-by-step solution:**
We are given 5 digits to place in a subtraction problem. To yield a positive difference using 5 digits, the setup must be a 3-digit number minus a 2-digit number.
To get the smallest possible difference, we need the 3-digit number to be as small as possible and the 2-digit number to be as large as possible.
1. The smallest possible hundreds digit for the 3-digit number is 2. So the number is $2\_ \_$.
2. We are left with the digits 4, 5, 6, and 9. 
3. To minimize the 3-digit number, the remaining two digits should be the smallest available: 4 and 5, making the number 245.
4. To maximize the 2-digit number being subtracted, we use the largest remaining digits: 9 and 6, making the number 96.
5. Subtracting the two gives: $245 - 96 = 149$.
(Testing other close combinations like $254 - 96 = 158$ or $246 - 95 = 151$ confirms 149 is the absolute minimum difference.)

ANSWER 4: C

---

**Problem 5:**
Each of the letters W , X , Y , and Z represents a different integer in the set {1, 2, 3, 4}, but not necessarily in that order. If W X − Y Z = 1, then the sum of W and Y is

**Step-by-step solution:**
The equation given is meant to be read as fractions (a common formatting artifact drops the fraction bar): $\frac{W}{X} - \frac{Y}{Z} = 1$.
Let's test the values {1, 2, 3, 4} to satisfy this equation:
- If we set $W = 3$, $X = 1$, $Y = 4$, and $Z = 2$:
  $\frac{3}{1} - \frac{4}{2} = 3 - 2 = 1$
This perfectly uses all four distinct digits exactly once.
The question asks for the sum of $W$ and $Y$:
$W + Y = 3 + 4 = 7$

*(Note: Checking $W \times X - Y \times Z = 1$ yields no solutions where all 4 digits are distinct, confirming the fractional interpretation).*

ANSWER 5: E

---

**Problem 6:**
An integer between 1000 and 9999, inclusive, is chosen at random. What is the probability that it is an odd integer whose digits are all distinct?

**Step-by-step solution:**
1. The total number of integers from 1000 to 9999 inclusive is $9999 - 1000 + 1 = 9000$.
2. We want to form a 4-digit odd number (let's call its digits A B C D) with all distinct digits.
3. Because it must be odd, the units digit (D) must be 1, 3, 5, 7, or 9. There are **5** choices for D.
4. The thousands digit (A) cannot be 0, and it cannot be the digit used for D. Thus, out of 10 possible digits, 2 are restricted, leaving **8** choices for A.
5. The hundreds digit (B) can be 0, but cannot be the digits used for A or D. This leaves **8** choices for B.
6. The tens digit (C) cannot be the digits used for A, B, or D. This leaves **7** choices for C.
7. The total number of such integers is $5 \times 8 \times 8 \times 7 = 2240$.
8. The probability is $\frac{2240}{9000} = \frac{224}{900} = \frac{56}{225}$.

ANSWER 6: B

---

**Problem 7:**
The manager of a company planned to give a $50 bonus to each employee from the company fund, but the fund was $5 short of what was needed. Instead the manager gave each employee a $45 bonus and kept the remaining $95 in the fund. How much money was in the company fund before any bonuses were paid?

**Step-by-step solution:**
Let $n$ be the number of employees.
1. If the manager gave $50 to each employee, the fund would be $5 short. So, the total fund is $5

## Step 1: Problem 1 - Restate the problem
We need to find how many whole numbers from 1 through 46 are divisible by either 3 or 5 or both.

## Step 2: Problem 1 - Calculate numbers divisible by 3
First, let's find the numbers divisible by 3. The largest number less than or equal to 46 that is divisible by 3 is 45, and the number of multiples of 3 from 1 to 46 is 45 / 3 = 15.

## Step 3: Problem 1 - Calculate numbers divisible by 5
Next, find the numbers divisible by 5. The largest number less than or equal to 46 that is divisible by 5 is 45, and the number of multiples of 5 from 1 to 46 is 45 / 5 = 9.

## Step 4: Problem 1 - Calculate numbers divisible by both 3 and 5
Numbers divisible by both 3 and 5 are divisible by 15. The largest number less than or equal to 46 that is divisible by 15 is 45, and the number of multiples of 15 from 1 to 46 is 45 / 15 = 3.

## Step 5: Problem 1 - Apply the principle of inclusion-exclusion
To find the total number of numbers divisible by either 3 or 5 or both, we use the principle of inclusion-exclusion: (numbers divisible by 3) + (numbers divisible by 5) - (numbers divisible by both) = 15 + 9 - 3 = 21.

## Step 6: Problem 1 - Provide the answer
ANSWER 1: B

## Step 7: Problem 2 - Restate the problem
Jamal has 6 green socks, 18 purple socks, and 12 orange socks. After adding more purple socks, the probability of picking a purple sock is 60%. We need to find how many purple socks Jamal added.

## Step 8: Problem 2 - Set up the initial conditions
Initially, Jamal has 6 + 18 + 12 = 36 socks.

## Step 9: Problem 2 - Set up the equation for the probability of picking a purple sock after adding x purple socks
After adding x purple socks, the total number of socks is 36 + x, and the number of purple socks is 18 + x. The probability of picking a purple sock is (18 + x) / (36 + x) = 0.6.

## Step 10: Problem 2 - Solve the equation for x
(18 + x) / (36 + x) = 0.6. Multiplying both sides by (36 + x) gives 18 + x = 0.6(36 + x). Expanding, we get 18 + x = 21.6 + 0.6x. Rearranging, x - 0.6x = 21.6 - 18, so 0.4x = 3.6. Dividing both sides by 0.4, x = 9.

## Step 11: Problem 2 - Provide the answer
ANSWER 2: B

## Step 12: Problem 3 - Restate the problem
We need to find which year is NOT a lucky year, where a lucky year has at least one date where the month times the day equals the last two digits of the year.

## Step 13: Problem 3 - Analyze the condition for a lucky year
For a year to be lucky, there must exist a month (1-12) and a day (1-31) such that month * day = last two digits of the year.

## Step 14: Problem 3 - Check each given year
Checking each year from 1990 to 1994 to see if any date satisfies the condition.

## Step 15: Problem 3 - Analyze 1990 to 1994
- For 1990, the last two digits are 90. It's difficult to get 90 by multiplying two numbers between 1 and 12 and 1 and 31, respectively, because the factors of 90 that could work are (1,90), (2,45), (3,30), (5,18), (6,15), (9,10). Only (9,10) is plausible, and it works as 9/10/90.
- For 1991, 91 can be achieved with 7/13/91, so it's lucky.
- For 1992, 92 can be achieved, for example, with 4/23/92, so it's lucky.
- For 1993, 93 can be achieved with 3/31/93, so it's lucky.
- For 1994, 94 is not easily achieved because its factors are (1,94), (2,47). No month-day combination gives 94.

## Step 16: Problem 3 - Provide the answer
Given the analysis, 1994 is not lucky because there's no month-day product equal to 94.

ANSWER 3: E

## Step 17: Problem 4 - Restate the problem
We need to find the smallest difference possible by placing the digits 2, 4, 5, 6, 9 in the boxes of a subtraction problem.

## Step 18: Problem 4 - Understand the structure of the subtraction problem
To minimize the difference, we should minimize the minuend and maximize the subtrahend.

## Step 19: Problem 4 - Arrange the digits to minimize the difference
To achieve the smallest difference, the hundreds place of the minuend should be as small as possible, and the hundreds place of the subtrahend should be as large as possible. So, the minuend could be 2** and the subtrahend 9**. The remaining digits are 4, 5, 6.

## Step 20: Problem 4 - Try different combinations
Let's try to make the minuend as small as possible and the subtrahend as large as possible: 245 - 986 is not valid because it results in a negative number. A valid attempt is 246 - 985 or similar combinations.

## Step 21: Problem 4 - Calculate the difference for a valid combination
For 964 - 852 or similar, the difference is not minimal. Let's directly check: 245 - 196 = 49 (not an option), but trying to minimize, we can get 246 - 195 = 51 (still not an option). A plausible arrangement is 254 - 196 = 58.

## Step 22: Problem 4 - Provide the answer
The smallest difference is achieved with an arrangement like 254 - 196.

ANSWER 4: A

## Step 23: Problem 5 - Restate the problem
Given W, X, Y, Z are different integers from 1 to 4, and WX - YZ = 1, we need to find W + Y.

## Step 24: Problem 5 - Understand the equation WX - YZ = 1
Since W, X, Y, Z are different integers from 1 to 4, we need to consider the possible values of WX and YZ.

## Step 25: Problem 5 - Analyze possible values for WX and YZ
The possible products are: 1*2=2, 1*3=3, 1*4=4, 2*1=2, 2*3=6, 2*4=8, 3*1=3, 3*2=6, 3*4=12, 4*1=4, 4*2=8, 4*3=12. We need a difference of 1 between two of these products.

## Step 26: Problem 5 - Check for a difference of 1
The products are 2, 3, 4, 6, 8, 12. The pairs that differ by 1 are (2,3), (3,4). So, either WX = 3 and YZ = 2 or WX = 4 and YZ = 3.

## Step 27: Problem 5 - Solve for W + Y
For WX = 3 and YZ = 2, W = 3, X = 1, Y = 2, Z = 1 (not valid because Z = X), or W = 1, X = 3, Y = 1, Z = 2 (not valid because W = Y). A valid case is W = 3, X = 1, Y = 1, Z = 2 (not valid), but if we consider W = 1, X = 3, Y = 2, Z = 1, it's valid and gives W + Y = 1 + 2 = 3.

## Step 28: Problem 5 - Provide the answer
For the valid case where W = 1, X = 3, Y = 2, Z = 1 doesn't work due to repetition, but

Problem 1:

We need count the whole numbers from 1 through 46 that are divisible by 3 or 5 or both.

Count multiples of 3:

\[
\left\lfloor \frac{46}{3} \right\rfloor = 15
\]

Count multiples of 5:

\[
\left\lfloor \frac{46}{5} \right\rfloor = 9
\]

But multiples of both 3 and 5, meaning multiples of 15, were counted twice, so subtract them once:

\[
\left\lfloor \frac{46}{15} \right\rfloor = 3
\]

Therefore the total is

\[
15+9-3=21
\]

So choices like 24 or 27 overcount because they do not remove the overlap.

ANSWER 1: B

Problem 2:

Initially Jamal has:

\[
6+18+12=36
\]

socks total, of which 18 are purple.

Suppose he adds \(x\) purple socks. Then the number of purple socks is

\[
18+x
\]

and the total number of socks is

\[
36+x
\]

The probability of choosing a purple sock is now 60%, or \(\frac35\). So

\[
\frac{18+x}{36+x}=\frac35
\]

Cross multiply:

\[
5(18+x)=3(36+x)
\]

\[
90+5x=108+3x
\]

\[
2x=18
\]

\[
x=9
\]

So Jamal added 9 purple socks.

ANSWER 2: B

Problem 3:

A lucky year has some valid date month/day/year where

\[
\text{month} \times \text{day} = \text{last two digits of the year}
\]

Check each year:

For 1990:

\[
9 \times 10 = 90
\]

So 9/10/90 works.

For 1991:

\[
7 \times 13 = 91
\]

So 7/13/91 works.

For 1992:

\[
4 \times 23 = 92
\]

So 4/23/92 works.

For 1993:

\[
3 \times 31 = 93
\]

So 3/31/93 works.

For 1994:

\[
94=2 \times 47
\]

But day 47 is not valid. The only factor pairs are \(1 \times 94\) and \(2 \times 47\), and neither gives a valid date.

Thus 1994 is not lucky.

ANSWER 3: E

Problem 4:

The subtraction problem uses the digits 2, 4, 5, 6, 9 exactly once in a 3-digit number minus a 2-digit number.

To make the difference as small as possible, the 3-digit number should be as small as possible, and the 2-digit number should be as large as possible.

The hundreds digit of the 3-digit number should be the smallest digit, 2.

Now we want the rest of the top number small and the bottom number large. Using the remaining digits 4, 5, 6, 9, the smallest possible two digits for the top number are 4 and 5, making 245, while the largest possible 2-digit number from the remaining digits is 96.

So try:

\[
245-96=149
\]

If we tried to get below 149, we would need either a smaller 3-digit number than 245 or a larger 2-digit number than 96, but neither is possible with these digits.

Therefore the smallest difference is 149.

ANSWER 4: C

Problem 5:

The equation is

\[
\frac{W}{X}-\frac{Y}{Z}=1
\]

where \(W, X, Y, Z\) are the numbers 1, 2, 3, 4 in some order.

We need find an arrangement that works.

Try to make the first fraction bigger than 1. One successful choice is

\[
\frac{3}{1}-\frac{4}{2}=1
\]

because

\[
3-2=1
\]

So

\[
W=3,\quad X=1,\quad Y=4,\quad Z=2
\]

Then

\[
W+Y=3+4=7
\]

The other choices are smaller sums, but the only working arrangement gives 7.

ANSWER 5: E

Problem 6:

There are four-digit integers from 1000 through 9999. The total number is

\[
9999-1000+1=9000
\]

We want odd integers whose digits are all distinct.

For a number to be odd, the units digit must be one of

\[
1,3,5,7,9
\]

So there are 5 choices for the units digit.

The thousands digit cannot be 0 and cannot equal the units digit. Since there are 9 possible nonzero thousands digits, and one is already used, there are

\[
8
\]

choices.

The hundreds digit can be any digit except the two already used:

\[
8
\]

choices.

The tens digit can be any digit except the three already used:

\[
7
\]

choices.

So the number of favorable integers is

\[
5 \cdot 8 \cdot 8 \cdot 7=2240
\]

The probability is

\[
\frac{2240}{9000}
\]

Reduce:

\[
\frac{2240}{9000}=\frac{224}{900}=\frac{56}{225}
\]

ANSWER 6: B

Problem 7:

Let \(n\) be the number of employees, and let \(F\) be the amount of money in the fund.

The manager planned to give each employee $50, but the fund was $5 short. So

\[
F=50n-5
\]

Instead, the manager gave each employee $45 and had $95 left. So

\[
F=45n+95
\]

Set these equal:

\[
50n-5=45n+95
\]

\[
5n=100
\]

\[
n=20
\]

Now find the fund:

\[
F=50(20)-5=1000-5=995
\]

So there were $995 in the fund.

ANSWER 7: E

Problem 8:

We are given

\[
ABA \times CD = CDCD
\]

The number \(ABA\) is

\[
101A+10B
\]

The number \(CDCD\) is a repeated two-digit number, so

\[
CDCD=101 \times CD
\]

Thus the equation becomes

\[
ABA \times CD = 101 \times CD
\]

Since \(CD\) is a two-digit number, it is not 0, so we can divide both sides by \(CD\):

\[
ABA=101
\]

Therefore

\[
A=1,\quad B=0
\]

So

\[
A+B=1+0=1
\]

ANSWER 8: A

Problem 9:

We need choose 3 numbers from

\[
\{89,95,99,132,166,173\}
\]

so that their sum is even.

Classify by parity.

Odd numbers:

\[
89,95,99,173
\]

There are 4 odds.

Even numbers:

\[
132,166
\]

There are 2 evens.

A sum of three numbers is even if it contains either:

- 0 odd numbers and 3 even numbers, or
- 2 odd numbers and 1 even number.

We cannot choose 3 even numbers because there are only 2 even numbers.

So we must choose 2 odds and 1 even.

Number of ways:

\[
\binom{4}{2}\binom{2}{1}=6 \cdot 2=12
\]

ANSWER 9: D

Problem 10:

The set has 5 positive integers with mean 5, so the total sum is

\[
5 \cdot 5=25
\]

The median is 5, so when the numbers are ordered, the middle number is 5.

Also, 8 is the only mode, meaning 8 appears more often than any other number. Since there are only five numbers and the median is 5, the two largest numbers must be 8 and 8:

\[
\_,\_,5,8,8
\]

Their sum so far is

\[
5+8+8=21
\]

So the first two numbers must sum to

\[
25-21=4
\]

They are positive integers less than or equal to 5. The possible pairs are:

\[
1,3 \quad \text{or} \quad 2,2
\]

But \(2,2\) would make 2 appear twice, tying 8 as a mode. Since 8 must be the only mode, the pair must be

\[
1,3
\]

So the set is

\[
1,3,5,8,8
\]

The difference between the largest and smallest integers is

\[
8-1=7
\]

ANSWER 10: D

Problem 11:

We need count three-digit whole numbers between 99 and 999 that contain exactly one 0.

A t

**Problem 1:**
How many whole numbers from 1 through 46 are divisible by either 3 or 5 or both?

**Step-by-step solution:**
To find the number of integers divisible by 3, 5, or both, we can use the Principle of Inclusion-Exclusion. 
1. First, we count the numbers from 1 to 46 that are divisible by 3:
   $\lfloor 46 \div 3 \rfloor = 15$
2. Next, we count the numbers from 1 to 46 that are divisible by 5:
   $\lfloor 46 \div 5 \rfloor = 9$
3. Then, we count the numbers that are divisible by BOTH 3 and 5 (which means they are divisible by their least common multiple, 15):
   $\lfloor 46 \div 15 \rfloor = 3$
4. Finally, we add the multiples of 3 and 5, and subtract the multiples of 15 so they aren't counted twice:
   $15 + 9 - 3 = 21$

ANSWER 1: B

---

**Problem 2:**
Jamal has a drawer containing 6 green socks, 18 purple socks, and 12 orange socks. After adding more purple socks, Jamal noticed that there is now a 60% chance that a sock randomly selected from the drawer is purple. How many purple socks did Jamal add?

**Step-by-step solution:**
1. Initially, Jamal has $6 + 18 + 12 = 36$ socks in total.
2. Let $x$ be the number of purple socks Jamal added. 
3. The new number of purple socks is $18 + x$.
4. The new total number of socks is $36 + x$.
5. We are told the new probability of selecting a purple sock is 60%, which simplifies to $\frac{60}{100} = \frac{3}{5}$.
6. We set up the equation:
   $\frac{18 + x}{36 + x} = \frac{3}{5}$
7. Cross-multiply to solve for $x$:
   $5(18 + x) = 3(36 + x)$
   $90 + 5x = 108 + 3x$
   $2x = 18$
   $x = 9$

Jamal added 9 purple socks.

ANSWER 2: B

---

**Problem 3:**
A lucky year is one in which at least one date, written as month/day/year, has the property that the month times the day equals the last two digits of the year. For example, 1956 is lucky because 7/8/56 has 7 × 8 = 56. Which of the following is NOT a lucky year?

**Step-by-step solution:**
We need to find if the last two digits of each given year can be factored into a valid month (1 through 12) and a valid day (1 through 31).
- **A. 1990:** $90 = 9 \times 10$. This corresponds to September 10th (9/10/90). Lucky.
- **B. 1991:** $91 = 7 \times 13$. This corresponds to July 13th (7/13/91). Lucky.
- **C. 1992:** $92 = 4 \times 23$. This corresponds to April 23rd (4/23/92). Lucky.
- **D. 1993:** $93 = 3 \times 31$. This corresponds to March 31st (3/31/93). Lucky.
- **E. 1994:** $94 = 2 \times 47$. Since a month cannot have 47 days, and 94 has no other integer pairs that fit the calendar constraints (1×94 is also invalid), 1994 is NOT a lucky year.

ANSWER 3: E

---

**Problem 4:**
When placing each of the digits 2, 4, 5, 6, 9 in exactly one of the boxes of this subtraction problem, what is the smallest difference that is possible?       −

**Step-by-step solution:**
We are given 5 digits to place in a subtraction problem. To yield a positive difference using 5 digits, the setup must be a 3-digit number minus a 2-digit number.
To get the smallest possible difference, we need the 3-digit number to be as small as possible and the 2-digit number to be as large as possible.
1. The smallest possible hundreds digit for the 3-digit number is 2. So the number is $2\_ \_$.
2. We are left with the digits 4, 5, 6, and 9. 
3. To minimize the 3-digit number, the remaining two digits should be the smallest available: 4 and 5, making the number 245.
4. To maximize the 2-digit number being subtracted, we use the largest remaining digits: 9 and 6, making the number 96.
5. Subtracting the two gives: $245 - 96 = 149$.
(Testing other close combinations like $254 - 96 = 158$ or $246 - 95 = 151$ confirms 149 is the absolute minimum difference.)

ANSWER 4: C

---

**Problem 5:**
Each of the letters W , X , Y , and Z represents a different integer in the set {1, 2, 3, 4}, but not necessarily in that order. If W X − Y Z = 1, then the sum of W and Y is

**Step-by-step solution:**
The equation given is meant to be read as fractions (a common formatting artifact drops the fraction bar): $\frac{W}{X} - \frac{Y}{Z} = 1$.
Let's test the values {1, 2, 3, 4} to satisfy this equation:
- If we set $W = 3$, $X = 1$, $Y = 4$, and $Z = 2$:
  $\frac{3}{1} - \frac{4}{2} = 3 - 2 = 1$
This perfectly uses all four distinct digits exactly once.
The question asks for the sum of $W$ and $Y$:
$W + Y = 3 + 4 = 7$

*(Note: Checking $W \times X - Y \times Z = 1$ yields no solutions where all 4 digits are distinct, confirming the fractional interpretation).*

ANSWER 5: E

---

**Problem 6:**
An integer between 1000 and 9999, inclusive, is chosen at random. What is the probability that it is an odd integer whose digits are all distinct?

**Step-by-step solution:**
1. The total number of integers from 1000 to 9999 inclusive is $9999 - 1000 + 1 = 9000$.
2. We want to form a 4-digit odd number (let's call its digits A B C D) with all distinct digits.
3. Because it must be odd, the units digit (D) must be 1, 3, 5, 7, or 9. There are **5** choices for D.
4. The thousands digit (A) cannot be 0, and it cannot be the digit used for D. Thus, out of 10 possible digits, 2 are restricted, leaving **8** choices for A.
5. The hundreds digit (B) can be 0, but cannot be the digits used for A or D. This leaves **8** choices for B.
6. The tens digit (C) cannot be the digits used for A, B, or D. This leaves **7** choices for C.
7. The total number of such integers is $5 \times 8 \times 8 \times 7 = 2240$.
8. The probability is $\frac{2240}{9000} = \frac{224}{900} = \frac{56}{225}$.

ANSWER 6: B

---

**Problem 7:**
The manager of a company planned to give a $50 bonus to each employee from the company fund, but the fund was $5 short of what was needed. Instead the manager gave each employee a $45 bonus and kept the remaining $95 in the fund. How much money was in the company fund before any bonuses were paid?

**Step-by-step solution:**
Let $n$ be the number of employees.
1. If the manager gave $50 to each employee, the fund would be $5 short. So, the total fund is $5

## Step 1: Problem 1 - Restate the problem
We need to find how many whole numbers from 1 through 46 are divisible by either 3 or 5 or both.

## Step 2: Problem 1 - Calculate numbers divisible by 3
First, let's find the numbers divisible by 3. The largest number less than or equal to 46 that is divisible by 3 is 45, and the number of multiples of 3 from 1 to 46 is 45 / 3 = 15.

## Step 3: Problem 1 - Calculate numbers divisible by 5
Next, find the numbers divisible by 5. The largest number less than or equal to 46 that is divisible by 5 is 45, and the number of multiples of 5 from 1 to 46 is 45 / 5 = 9.

## Step 4: Problem 1 - Calculate numbers divisible by both 3 and 5
Numbers divisible by both 3 and 5 are divisible by 15. The largest number less than or equal to 46 that is divisible by 15 is 45, and the number of multiples of 15 from 1 to 46 is 45 / 15 = 3.

## Step 5: Problem 1 - Apply the principle of inclusion-exclusion
To find the total number of numbers divisible by either 3 or 5 or both, we use the principle of inclusion-exclusion: (numbers divisible by 3) + (numbers divisible by 5) - (numbers divisible by both) = 15 + 9 - 3 = 21.

## Step 6: Problem 1 - Provide the answer
ANSWER 1: B

## Step 7: Problem 2 - Restate the problem
Jamal has 6 green socks, 18 purple socks, and 12 orange socks. After adding more purple socks, the probability of picking a purple sock is 60%. We need to find how many purple socks Jamal added.

## Step 8: Problem 2 - Set up the initial conditions
Initially, Jamal has 6 + 18 + 12 = 36 socks.

## Step 9: Problem 2 - Set up the equation for the probability of picking a purple sock after adding x purple socks
After adding x purple socks, the total number of socks is 36 + x, and the number of purple socks is 18 + x. The probability of picking a purple sock is (18 + x) / (36 + x) = 0.6.

## Step 10: Problem 2 - Solve the equation for x
(18 + x) / (36 + x) = 0.6. Multiplying both sides by (36 + x) gives 18 + x = 0.6(36 + x). Expanding, we get 18 + x = 21.6 + 0.6x. Rearranging, x - 0.6x = 21.6 - 18, so 0.4x = 3.6. Dividing both sides by 0.4, x = 9.

## Step 11: Problem 2 - Provide the answer
ANSWER 2: B

## Step 12: Problem 3 - Restate the problem
We need to find which year is NOT a lucky year, where a lucky year has at least one date where the month times the day equals the last two digits of the year.

## Step 13: Problem 3 - Analyze the condition for a lucky year
For a year to be lucky, there must exist a month (1-12) and a day (1-31) such that month * day = last two digits of the year.

## Step 14: Problem 3 - Check each given year
Checking each year from 1990 to 1994 to see if any date satisfies the condition.

## Step 15: Problem 3 - Analyze 1990 to 1994
- For 1990, the last two digits are 90. It's difficult to get 90 by multiplying two numbers between 1 and 12 and 1 and 31, respectively, because the factors of 90 that could work are (1,90), (2,45), (3,30), (5,18), (6,15), (9,10). Only (9,10) is plausible, and it works as 9/10/90.
- For 1991, 91 can be achieved with 7/13/91, so it's lucky.
- For 1992, 92 can be achieved, for example, with 4/23/92, so it's lucky.
- For 1993, 93 can be achieved with 3/31/93, so it's lucky.
- For 1994, 94 is not easily achieved because its factors are (1,94), (2,47). No month-day combination gives 94.

## Step 16: Problem 3 - Provide the answer
Given the analysis, 1994 is not lucky because there's no month-day product equal to 94.

ANSWER 3: E

## Step 17: Problem 4 - Restate the problem
We need to find the smallest difference possible by placing the digits 2, 4, 5, 6, 9 in the boxes of a subtraction problem.

## Step 18: Problem 4 - Understand the structure of the subtraction problem
To minimize the difference, we should minimize the minuend and maximize the subtrahend.

## Step 19: Problem 4 - Arrange the digits to minimize the difference
To achieve the smallest difference, the hundreds place of the minuend should be as small as possible, and the hundreds place of the subtrahend should be as large as possible. So, the minuend could be 2** and the subtrahend 9**. The remaining digits are 4, 5, 6.

## Step 20: Problem 4 - Try different combinations
Let's try to make the minuend as small as possible and the subtrahend as large as possible: 245 - 986 is not valid because it results in a negative number. A valid attempt is 246 - 985 or similar combinations.

## Step 21: Problem 4 - Calculate the difference for a valid combination
For 964 - 852 or similar, the difference is not minimal. Let's directly check: 245 - 196 = 49 (not an option), but trying to minimize, we can get 246 - 195 = 51 (still not an option). A plausible arrangement is 254 - 196 = 58.

## Step 22: Problem 4 - Provide the answer
The smallest difference is achieved with an arrangement like 254 - 196.

ANSWER 4: A

## Step 23: Problem 5 - Restate the problem
Given W, X, Y, Z are different integers from 1 to 4, and WX - YZ = 1, we need to find W + Y.

## Step 24: Problem 5 - Understand the equation WX - YZ = 1
Since W, X, Y, Z are different integers from 1 to 4, we need to consider the possible values of WX and YZ.

## Step 25: Problem 5 - Analyze possible values for WX and YZ
The possible products are: 1*2=2, 1*3=3, 1*4=4, 2*1=2, 2*3=6, 2*4=8, 3*1=3, 3*2=6, 3*4=12, 4*1=4, 4*2=8, 4*3=12. We need a difference of 1 between two of these products.

## Step 26: Problem 5 - Check for a difference of 1
The products are 2, 3, 4, 6, 8, 12. The pairs that differ by 1 are (2,3), (3,4). So, either WX = 3 and YZ = 2 or WX = 4 and YZ = 3.

## Step 27: Problem 5 - Solve for W + Y
For WX = 3 and YZ = 2, W = 3, X = 1, Y = 2, Z = 1 (not valid because Z = X), or W = 1, X = 3, Y = 1, Z = 2 (not valid because W = Y). A valid case is W = 3, X = 1, Y = 1, Z = 2 (not valid), but if we consider W = 1, X = 3, Y = 2, Z = 1, it's valid and gives W + Y = 1 + 2 = 3.

## Step 28: Problem 5 - Provide the answer
For the valid case where W = 1, X = 3, Y = 2, Z = 1 doesn't work due to repetition, but

Problem 1:

We need count the whole numbers from 1 through 46 that are divisible by 3 or 5 or both.

Count multiples of 3:

\[
\left\lfloor \frac{46}{3} \right\rfloor = 15
\]

Count multiples of 5:

\[
\left\lfloor \frac{46}{5} \right\rfloor = 9
\]

But multiples of both 3 and 5, meaning multiples of 15, were counted twice, so subtract them once:

\[
\left\lfloor \frac{46}{15} \right\rfloor = 3
\]

Therefore the total is

\[
15+9-3=21
\]

So choices like 24 or 27 overcount because they do not remove the overlap.

ANSWER 1: B

Problem 2:

Initially Jamal has:

\[
6+18+12=36
\]

socks total, of which 18 are purple.

Suppose he adds \(x\) purple socks. Then the number of purple socks is

\[
18+x
\]

and the total number of socks is

\[
36+x
\]

The probability of choosing a purple sock is now 60%, or \(\frac35\). So

\[
\frac{18+x}{36+x}=\frac35
\]

Cross multiply:

\[
5(18+x)=3(36+x)
\]

\[
90+5x=108+3x
\]

\[
2x=18
\]

\[
x=9
\]

So Jamal added 9 purple socks.

ANSWER 2: B

Problem 3:

A lucky year has some valid date month/day/year where

\[
\text{month} \times \text{day} = \text{last two digits of the year}
\]

Check each year:

For 1990:

\[
9 \times 10 = 90
\]

So 9/10/90 works.

For 1991:

\[
7 \times 13 = 91
\]

So 7/13/91 works.

For 1992:

\[
4 \times 23 = 92
\]

So 4/23/92 works.

For 1993:

\[
3 \times 31 = 93
\]

So 3/31/93 works.

For 1994:

\[
94=2 \times 47
\]

But day 47 is not valid. The only factor pairs are \(1 \times 94\) and \(2 \times 47\), and neither gives a valid date.

Thus 1994 is not lucky.

ANSWER 3: E

Problem 4:

The subtraction problem uses the digits 2, 4, 5, 6, 9 exactly once in a 3-digit number minus a 2-digit number.

To make the difference as small as possible, the 3-digit number should be as small as possible, and the 2-digit number should be as large as possible.

The hundreds digit of the 3-digit number should be the smallest digit, 2.

Now we want the rest of the top number small and the bottom number large. Using the remaining digits 4, 5, 6, 9, the smallest possible two digits for the top number are 4 and 5, making 245, while the largest possible 2-digit number from the remaining digits is 96.

So try:

\[
245-96=149
\]

If we tried to get below 149, we would need either a smaller 3-digit number than 245 or a larger 2-digit number than 96, but neither is possible with these digits.

Therefore the smallest difference is 149.

ANSWER 4: C

Problem 5:

The equation is

\[
\frac{W}{X}-\frac{Y}{Z}=1
\]

where \(W, X, Y, Z\) are the numbers 1, 2, 3, 4 in some order.

We need find an arrangement that works.

Try to make the first fraction bigger than 1. One successful choice is

\[
\frac{3}{1}-\frac{4}{2}=1
\]

because

\[
3-2=1
\]

So

\[
W=3,\quad X=1,\quad Y=4,\quad Z=2
\]

Then

\[
W+Y=3+4=7
\]

The other choices are smaller sums, but the only working arrangement gives 7.

ANSWER 5: E

Problem 6:

There are four-digit integers from 1000 through 9999. The total number is

\[
9999-1000+1=9000
\]

We want odd integers whose digits are all distinct.

For a number to be odd, the units digit must be one of

\[
1,3,5,7,9
\]

So there are 5 choices for the units digit.

The thousands digit cannot be 0 and cannot equal the units digit. Since there are 9 possible nonzero thousands digits, and one is already used, there are

\[
8
\]

choices.

The hundreds digit can be any digit except the two already used:

\[
8
\]

choices.

The tens digit can be any digit except the three already used:

\[
7
\]

choices.

So the number of favorable integers is

\[
5 \cdot 8 \cdot 8 \cdot 7=2240
\]

The probability is

\[
\frac{2240}{9000}
\]

Reduce:

\[
\frac{2240}{9000}=\frac{224}{900}=\frac{56}{225}
\]

ANSWER 6: B

Problem 7:

Let \(n\) be the number of employees, and let \(F\) be the amount of money in the fund.

The manager planned to give each employee $50, but the fund was $5 short. So

\[
F=50n-5
\]

Instead, the manager gave each employee $45 and had $95 left. So

\[
F=45n+95
\]

Set these equal:

\[
50n-5=45n+95
\]

\[
5n=100
\]

\[
n=20
\]

Now find the fund:

\[
F=50(20)-5=1000-5=995
\]

So there were $995 in the fund.

ANSWER 7: E

Problem 8:

We are given

\[
ABA \times CD = CDCD
\]

The number \(ABA\) is

\[
101A+10B
\]

The number \(CDCD\) is a repeated two-digit number, so

\[
CDCD=101 \times CD
\]

Thus the equation becomes

\[
ABA \times CD = 101 \times CD
\]

Since \(CD\) is a two-digit number, it is not 0, so we can divide both sides by \(CD\):

\[
ABA=101
\]

Therefore

\[
A=1,\quad B=0
\]

So

\[
A+B=1+0=1
\]

ANSWER 8: A

Problem 9:

We need choose 3 numbers from

\[
\{89,95,99,132,166,173\}
\]

so that their sum is even.

Classify by parity.

Odd numbers:

\[
89,95,99,173
\]

There are 4 odds.

Even numbers:

\[
132,166
\]

There are 2 evens.

A sum of three numbers is even if it contains either:

- 0 odd numbers and 3 even numbers, or
- 2 odd numbers and 1 even number.

We cannot choose 3 even numbers because there are only 2 even numbers.

So we must choose 2 odds and 1 even.

Number of ways:

\[
\binom{4}{2}\binom{2}{1}=6 \cdot 2=12
\]

ANSWER 9: D

Problem 10:

The set has 5 positive integers with mean 5, so the total sum is

\[
5 \cdot 5=25
\]

The median is 5, so when the numbers are ordered, the middle number is 5.

Also, 8 is the only mode, meaning 8 appears more often than any other number. Since there are only five numbers and the median is 5, the two largest numbers must be 8 and 8:

\[
\_,\_,5,8,8
\]

Their sum so far is

\[
5+8+8=21
\]

So the first two numbers must sum to

\[
25-21=4
\]

They are positive integers less than or equal to 5. The possible pairs are:

\[
1,3 \quad \text{or} \quad 2,2
\]

But \(2,2\) would make 2 appear twice, tying 8 as a mode. Since 8 must be the only mode, the pair must be

\[
1,3
\]

So the set is

\[
1,3,5,8,8
\]

The difference between the largest and smallest integers is

\[
8-1=7
\]

ANSWER 10: D

Problem 11:

We need count three-digit whole numbers between 99 and 999 that contain exactly one 0.

A t

**Problem 1:**
How many whole numbers from 1 through 46 are divisible by either 3 or 5 or both?

**Step-by-step solution:**
To find the number of integers divisible by 3, 5, or both, we can use the Principle of Inclusion-Exclusion. 
1. First, we count the numbers from 1 to 46 that are divisible by 3:
   $\lfloor 46 \div 3 \rfloor = 15$
2. Next, we count the numbers from 1 to 46 that are divisible by 5:
   $\lfloor 46 \div 5 \rfloor = 9$
3. Then, we count the numbers that are divisible by BOTH 3 and 5 (which means they are divisible by their least common multiple, 15):
   $\lfloor 46 \div 15 \rfloor = 3$
4. Finally, we add the multiples of 3 and 5, and subtract the multiples of 15 so they aren't counted twice:
   $15 + 9 - 3 = 21$

ANSWER 1: B

---

**Problem 2:**
Jamal has a drawer containing 6 green socks, 18 purple socks, and 12 orange socks. After adding more purple socks, Jamal noticed that there is now a 60% chance that a sock randomly selected from the drawer is purple. How many purple socks did Jamal add?

**Step-by-step solution:**
1. Initially, Jamal has $6 + 18 + 12 = 36$ socks in total.
2. Let $x$ be the number of purple socks Jamal added. 
3. The new number of purple socks is $18 + x$.
4. The new total number of socks is $36 + x$.
5. We are told the new probability of selecting a purple sock is 60%, which simplifies to $\frac{60}{100} = \frac{3}{5}$.
6. We set up the equation:
   $\frac{18 + x}{36 + x} = \frac{3}{5}$
7. Cross-multiply to solve for $x$:
   $5(18 + x) = 3(36 + x)$
   $90 + 5x = 108 + 3x$
   $2x = 18$
   $x = 9$

Jamal added 9 purple socks.

ANSWER 2: B

---

**Problem 3:**
A lucky year is one in which at least one date, written as month/day/year, has the property that the month times the day equals the last two digits of the year. For example, 1956 is lucky because 7/8/56 has 7 × 8 = 56. Which of the following is NOT a lucky year?

**Step-by-step solution:**
We need to find if the last two digits of each given year can be factored into a valid month (1 through 12) and a valid day (1 through 31).
- **A. 1990:** $90 = 9 \times 10$. This corresponds to September 10th (9/10/90). Lucky.
- **B. 1991:** $91 = 7 \times 13$. This corresponds to July 13th (7/13/91). Lucky.
- **C. 1992:** $92 = 4 \times 23$. This corresponds to April 23rd (4/23/92). Lucky.
- **D. 1993:** $93 = 3 \times 31$. This corresponds to March 31st (3/31/93). Lucky.
- **E. 1994:** $94 = 2 \times 47$. Since a month cannot have 47 days, and 94 has no other integer pairs that fit the calendar constraints (1×94 is also invalid), 1994 is NOT a lucky year.

ANSWER 3: E

---

**Problem 4:**
When placing each of the digits 2, 4, 5, 6, 9 in exactly one of the boxes of this subtraction problem, what is the smallest difference that is possible?       −

**Step-by-step solution:**
We are given 5 digits to place in a subtraction problem. To yield a positive difference using 5 digits, the setup must be a 3-digit number minus a 2-digit number.
To get the smallest possible difference, we need the 3-digit number to be as small as possible and the 2-digit number to be as large as possible.
1. The smallest possible hundreds digit for the 3-digit number is 2. So the number is $2\_ \_$.
2. We are left with the digits 4, 5, 6, and 9. 
3. To minimize the 3-digit number, the remaining two digits should be the smallest available: 4 and 5, making the number 245.
4. To maximize the 2-digit number being subtracted, we use the largest remaining digits: 9 and 6, making the number 96.
5. Subtracting the two gives: $245 - 96 = 149$.
(Testing other close combinations like $254 - 96 = 158$ or $246 - 95 = 151$ confirms 149 is the absolute minimum difference.)

ANSWER 4: C

---

**Problem 5:**
Each of the letters W , X , Y , and Z represents a different integer in the set {1, 2, 3, 4}, but not necessarily in that order. If W X − Y Z = 1, then the sum of W and Y is

**Step-by-step solution:**
The equation given is meant to be read as fractions (a common formatting artifact drops the fraction bar): $\frac{W}{X} - \frac{Y}{Z} = 1$.
Let's test the values {1, 2, 3, 4} to satisfy this equation:
- If we set $W = 3$, $X = 1$, $Y = 4$, and $Z = 2$:
  $\frac{3}{1} - \frac{4}{2} = 3 - 2 = 1$
This perfectly uses all four distinct digits exactly once.
The question asks for the sum of $W$ and $Y$:
$W + Y = 3 + 4 = 7$

*(Note: Checking $W \times X - Y \times Z = 1$ yields no solutions where all 4 digits are distinct, confirming the fractional interpretation).*

ANSWER 5: E

---

**Problem 6:**
An integer between 1000 and 9999, inclusive, is chosen at random. What is the probability that it is an odd integer whose digits are all distinct?

**Step-by-step solution:**
1. The total number of integers from 1000 to 9999 inclusive is $9999 - 1000 + 1 = 9000$.
2. We want to form a 4-digit odd number (let's call its digits A B C D) with all distinct digits.
3. Because it must be odd, the units digit (D) must be 1, 3, 5, 7, or 9. There are **5** choices for D.
4. The thousands digit (A) cannot be 0, and it cannot be the digit used for D. Thus, out of 10 possible digits, 2 are restricted, leaving **8** choices for A.
5. The hundreds digit (B) can be 0, but cannot be the digits used for A or D. This leaves **8** choices for B.
6. The tens digit (C) cannot be the digits used for A, B, or D. This leaves **7** choices for C.
7. The total number of such integers is $5 \times 8 \times 8 \times 7 = 2240$.
8. The probability is $\frac{2240}{9000} = \frac{224}{900} = \frac{56}{225}$.

ANSWER 6: B

---

**Problem 7:**
The manager of a company planned to give a $50 bonus to each employee from the company fund, but the fund was $5 short of what was needed. Instead the manager gave each employee a $45 bonus and kept the remaining $95 in the fund. How much money was in the company fund before any bonuses were paid?

**Step-by-step solution:**
Let $n$ be the number of employees.
1. If the manager gave $50 to each employee, the fund would be $5 short. So, the total fund is $5

## Step 1: Problem 1 - Restate the problem
We need to find how many whole numbers from 1 through 46 are divisible by either 3 or 5 or both.

## Step 2: Problem 1 - Calculate numbers divisible by 3
First, let's find the numbers divisible by 3. The largest number less than or equal to 46 that is divisible by 3 is 45, and the number of multiples of 3 from 1 to 46 is 45 / 3 = 15.

## Step 3: Problem 1 - Calculate numbers divisible by 5
Next, find the numbers divisible by 5. The largest number less than or equal to 46 that is divisible by 5 is 45, and the number of multiples of 5 from 1 to 46 is 45 / 5 = 9.

## Step 4: Problem 1 - Calculate numbers divisible by both 3 and 5
Numbers divisible by both 3 and 5 are divisible by 15. The largest number less than or equal to 46 that is divisible by 15 is 45, and the number of multiples of 15 from 1 to 46 is 45 / 15 = 3.

## Step 5: Problem 1 - Apply the principle of inclusion-exclusion
To find the total number of numbers divisible by either 3 or 5 or both, we use the principle of inclusion-exclusion: (numbers divisible by 3) + (numbers divisible by 5) - (numbers divisible by both) = 15 + 9 - 3 = 21.

## Step 6: Problem 1 - Provide the answer
ANSWER 1: B

## Step 7: Problem 2 - Restate the problem
Jamal has 6 green socks, 18 purple socks, and 12 orange socks. After adding more purple socks, the probability of picking a purple sock is 60%. We need to find how many purple socks Jamal added.

## Step 8: Problem 2 - Set up the initial conditions
Initially, Jamal has 6 + 18 + 12 = 36 socks.

## Step 9: Problem 2 - Set up the equation for the probability of picking a purple sock after adding x purple socks
After adding x purple socks, the total number of socks is 36 + x, and the number of purple socks is 18 + x. The probability of picking a purple sock is (18 + x) / (36 + x) = 0.6.

## Step 10: Problem 2 - Solve the equation for x
(18 + x) / (36 + x) = 0.6. Multiplying both sides by (36 + x) gives 18 + x = 0.6(36 + x). Expanding, we get 18 + x = 21.6 + 0.6x. Rearranging, x - 0.6x = 21.6 - 18, so 0.4x = 3.6. Dividing both sides by 0.4, x = 9.

## Step 11: Problem 2 - Provide the answer
ANSWER 2: B

## Step 12: Problem 3 - Restate the problem
We need to find which year is NOT a lucky year, where a lucky year has at least one date where the month times the day equals the last two digits of the year.

## Step 13: Problem 3 - Analyze the condition for a lucky year
For a year to be lucky, there must exist a month (1-12) and a day (1-31) such that month * day = last two digits of the year.

## Step 14: Problem 3 - Check each given year
Checking each year from 1990 to 1994 to see if any date satisfies the condition.

## Step 15: Problem 3 - Analyze 1990 to 1994
- For 1990, the last two digits are 90. It's difficult to get 90 by multiplying two numbers between 1 and 12 and 1 and 31, respectively, because the factors of 90 that could work are (1,90), (2,45), (3,30), (5,18), (6,15), (9,10). Only (9,10) is plausible, and it works as 9/10/90.
- For 1991, 91 can be achieved with 7/13/91, so it's lucky.
- For 1992, 92 can be achieved, for example, with 4/23/92, so it's lucky.
- For 1993, 93 can be achieved with 3/31/93, so it's lucky.
- For 1994, 94 is not easily achieved because its factors are (1,94), (2,47). No month-day combination gives 94.

## Step 16: Problem 3 - Provide the answer
Given the analysis, 1994 is not lucky because there's no month-day product equal to 94.

ANSWER 3: E

## Step 17: Problem 4 - Restate the problem
We need to find the smallest difference possible by placing the digits 2, 4, 5, 6, 9 in the boxes of a subtraction problem.

## Step 18: Problem 4 - Understand the structure of the subtraction problem
To minimize the difference, we should minimize the minuend and maximize the subtrahend.

## Step 19: Problem 4 - Arrange the digits to minimize the difference
To achieve the smallest difference, the hundreds place of the minuend should be as small as possible, and the hundreds place of the subtrahend should be as large as possible. So, the minuend could be 2** and the subtrahend 9**. The remaining digits are 4, 5, 6.

## Step 20: Problem 4 - Try different combinations
Let's try to make the minuend as small as possible and the subtrahend as large as possible: 245 - 986 is not valid because it results in a negative number. A valid attempt is 246 - 985 or similar combinations.

## Step 21: Problem 4 - Calculate the difference for a valid combination
For 964 - 852 or similar, the difference is not minimal. Let's directly check: 245 - 196 = 49 (not an option), but trying to minimize, we can get 246 - 195 = 51 (still not an option). A plausible arrangement is 254 - 196 = 58.

## Step 22: Problem 4 - Provide the answer
The smallest difference is achieved with an arrangement like 254 - 196.

ANSWER 4: A

## Step 23: Problem 5 - Restate the problem
Given W, X, Y, Z are different integers from 1 to 4, and WX - YZ = 1, we need to find W + Y.

## Step 24: Problem 5 - Understand the equation WX - YZ = 1
Since W, X, Y, Z are different integers from 1 to 4, we need to consider the possible values of WX and YZ.

## Step 25: Problem 5 - Analyze possible values for WX and YZ
The possible products are: 1*2=2, 1*3=3, 1*4=4, 2*1=2, 2*3=6, 2*4=8, 3*1=3, 3*2=6, 3*4=12, 4*1=4, 4*2=8, 4*3=12. We need a difference of 1 between two of these products.

## Step 26: Problem 5 - Check for a difference of 1
The products are 2, 3, 4, 6, 8, 12. The pairs that differ by 1 are (2,3), (3,4). So, either WX = 3 and YZ = 2 or WX = 4 and YZ = 3.

## Step 27: Problem 5 - Solve for W + Y
For WX = 3 and YZ = 2, W = 3, X = 1, Y = 2, Z = 1 (not valid because Z = X), or W = 1, X = 3, Y = 1, Z = 2 (not valid because W = Y). A valid case is W = 3, X = 1, Y = 1, Z = 2 (not valid), but if we consider W = 1, X = 3, Y = 2, Z = 1, it's valid and gives W + Y = 1 + 2 = 3.

## Step 28: Problem 5 - Provide the answer
For the valid case where W = 1, X = 3, Y = 2, Z = 1 doesn't work due to repetition, but

Problem 1:

We need count the whole numbers from 1 through 46 that are divisible by 3 or 5 or both.

Count multiples of 3:

\[
\left\lfloor \frac{46}{3} \right\rfloor = 15
\]

Count multiples of 5:

\[
\left\lfloor \frac{46}{5} \right\rfloor = 9
\]

But multiples of both 3 and 5, meaning multiples of 15, were counted twice, so subtract them once:

\[
\left\lfloor \frac{46}{15} \right\rfloor = 3
\]

Therefore the total is

\[
15+9-3=21
\]

So choices like 24 or 27 overcount because they do not remove the overlap.

ANSWER 1: B

Problem 2:

Initially Jamal has:

\[
6+18+12=36
\]

socks total, of which 18 are purple.

Suppose he adds \(x\) purple socks. Then the number of purple socks is

\[
18+x
\]

and the total number of socks is

\[
36+x
\]

The probability of choosing a purple sock is now 60%, or \(\frac35\). So

\[
\frac{18+x}{36+x}=\frac35
\]

Cross multiply:

\[
5(18+x)=3(36+x)
\]

\[
90+5x=108+3x
\]

\[
2x=18
\]

\[
x=9
\]

So Jamal added 9 purple socks.

ANSWER 2: B

Problem 3:

A lucky year has some valid date month/day/year where

\[
\text{month} \times \text{day} = \text{last two digits of the year}
\]

Check each year:

For 1990:

\[
9 \times 10 = 90
\]

So 9/10/90 works.

For 1991:

\[
7 \times 13 = 91
\]

So 7/13/91 works.

For 1992:

\[
4 \times 23 = 92
\]

So 4/23/92 works.

For 1993:

\[
3 \times 31 = 93
\]

So 3/31/93 works.

For 1994:

\[
94=2 \times 47
\]

But day 47 is not valid. The only factor pairs are \(1 \times 94\) and \(2 \times 47\), and neither gives a valid date.

Thus 1994 is not lucky.

ANSWER 3: E

Problem 4:

The subtraction problem uses the digits 2, 4, 5, 6, 9 exactly once in a 3-digit number minus a 2-digit number.

To make the difference as small as possible, the 3-digit number should be as small as possible, and the 2-digit number should be as large as possible.

The hundreds digit of the 3-digit number should be the smallest digit, 2.

Now we want the rest of the top number small and the bottom number large. Using the remaining digits 4, 5, 6, 9, the smallest possible two digits for the top number are 4 and 5, making 245, while the largest possible 2-digit number from the remaining digits is 96.

So try:

\[
245-96=149
\]

If we tried to get below 149, we would need either a smaller 3-digit number than 245 or a larger 2-digit number than 96, but neither is possible with these digits.

Therefore the smallest difference is 149.

ANSWER 4: C

Problem 5:

The equation is

\[
\frac{W}{X}-\frac{Y}{Z}=1
\]

where \(W, X, Y, Z\) are the numbers 1, 2, 3, 4 in some order.

We need find an arrangement that works.

Try to make the first fraction bigger than 1. One successful choice is

\[
\frac{3}{1}-\frac{4}{2}=1
\]

because

\[
3-2=1
\]

So

\[
W=3,\quad X=1,\quad Y=4,\quad Z=2
\]

Then

\[
W+Y=3+4=7
\]

The other choices are smaller sums, but the only working arrangement gives 7.

ANSWER 5: E

Problem 6:

There are four-digit integers from 1000 through 9999. The total number is

\[
9999-1000+1=9000
\]

We want odd integers whose digits are all distinct.

For a number to be odd, the units digit must be one of

\[
1,3,5,7,9
\]

So there are 5 choices for the units digit.

The thousands digit cannot be 0 and cannot equal the units digit. Since there are 9 possible nonzero thousands digits, and one is already used, there are

\[
8
\]

choices.

The hundreds digit can be any digit except the two already used:

\[
8
\]

choices.

The tens digit can be any digit except the three already used:

\[
7
\]

choices.

So the number of favorable integers is

\[
5 \cdot 8 \cdot 8 \cdot 7=2240
\]

The probability is

\[
\frac{2240}{9000}
\]

Reduce:

\[
\frac{2240}{9000}=\frac{224}{900}=\frac{56}{225}
\]

ANSWER 6: B

Problem 7:

Let \(n\) be the number of employees, and let \(F\) be the amount of money in the fund.

The manager planned to give each employee $50, but the fund was $5 short. So

\[
F=50n-5
\]

Instead, the manager gave each employee $45 and had $95 left. So

\[
F=45n+95
\]

Set these equal:

\[
50n-5=45n+95
\]

\[
5n=100
\]

\[
n=20
\]

Now find the fund:

\[
F=50(20)-5=1000-5=995
\]

So there were $995 in the fund.

ANSWER 7: E

Problem 8:

We are given

\[
ABA \times CD = CDCD
\]

The number \(ABA\) is

\[
101A+10B
\]

The number \(CDCD\) is a repeated two-digit number, so

\[
CDCD=101 \times CD
\]

Thus the equation becomes

\[
ABA \times CD = 101 \times CD
\]

Since \(CD\) is a two-digit number, it is not 0, so we can divide both sides by \(CD\):

\[
ABA=101
\]

Therefore

\[
A=1,\quad B=0
\]

So

\[
A+B=1+0=1
\]

ANSWER 8: A

Problem 9:

We need choose 3 numbers from

\[
\{89,95,99,132,166,173\}
\]

so that their sum is even.

Classify by parity.

Odd numbers:

\[
89,95,99,173
\]

There are 4 odds.

Even numbers:

\[
132,166
\]

There are 2 evens.

A sum of three numbers is even if it contains either:

- 0 odd numbers and 3 even numbers, or
- 2 odd numbers and 1 even number.

We cannot choose 3 even numbers because there are only 2 even numbers.

So we must choose 2 odds and 1 even.

Number of ways:

\[
\binom{4}{2}\binom{2}{1}=6 \cdot 2=12
\]

ANSWER 9: D

Problem 10:

The set has 5 positive integers with mean 5, so the total sum is

\[
5 \cdot 5=25
\]

The median is 5, so when the numbers are ordered, the middle number is 5.

Also, 8 is the only mode, meaning 8 appears more often than any other number. Since there are only five numbers and the median is 5, the two largest numbers must be 8 and 8:

\[
\_,\_,5,8,8
\]

Their sum so far is

\[
5+8+8=21
\]

So the first two numbers must sum to

\[
25-21=4
\]

They are positive integers less than or equal to 5. The possible pairs are:

\[
1,3 \quad \text{or} \quad 2,2
\]

But \(2,2\) would make 2 appear twice, tying 8 as a mode. Since 8 must be the only mode, the pair must be

\[
1,3
\]

So the set is

\[
1,3,5,8,8
\]

The difference between the largest and smallest integers is

\[
8-1=7
\]

ANSWER 10: D

Problem 11:

We need count three-digit whole numbers between 99 and 999 that contain exactly one 0.

A t

**Problem 1:**
How many whole numbers from 1 through 46 are divisible by either 3 or 5 or both?

**Step-by-step solution:**
To find the number of integers divisible by 3, 5, or both, we can use the Principle of Inclusion-Exclusion. 
1. First, we count the numbers from 1 to 46 that are divisible by 3:
   $\lfloor 46 \div 3 \rfloor = 15$
2. Next, we count the numbers from 1 to 46 that are divisible by 5:
   $\lfloor 46 \div 5 \rfloor = 9$
3. Then, we count the numbers that are divisible by BOTH 3 and 5 (which means they are divisible by their least common multiple, 15):
   $\lfloor 46 \div 15 \rfloor = 3$
4. Finally, we add the multiples of 3 and 5, and subtract the multiples of 15 so they aren't counted twice:
   $15 + 9 - 3 = 21$

ANSWER 1: B

---

**Problem 2:**
Jamal has a drawer containing 6 green socks, 18 purple socks, and 12 orange socks. After adding more purple socks, Jamal noticed that there is now a 60% chance that a sock randomly selected from the drawer is purple. How many purple socks did Jamal add?

**Step-by-step solution:**
1. Initially, Jamal has $6 + 18 + 12 = 36$ socks in total.
2. Let $x$ be the number of purple socks Jamal added. 
3. The new number of purple socks is $18 + x$.
4. The new total number of socks is $36 + x$.
5. We are told the new probability of selecting a purple sock is 60%, which simplifies to $\frac{60}{100} = \frac{3}{5}$.
6. We set up the equation:
   $\frac{18 + x}{36 + x} = \frac{3}{5}$
7. Cross-multiply to solve for $x$:
   $5(18 + x) = 3(36 + x)$
   $90 + 5x = 108 + 3x$
   $2x = 18$
   $x = 9$

Jamal added 9 purple socks.

ANSWER 2: B

---

**Problem 3:**
A lucky year is one in which at least one date, written as month/day/year, has the property that the month times the day equals the last two digits of the year. For example, 1956 is lucky because 7/8/56 has 7 × 8 = 56. Which of the following is NOT a lucky year?

**Step-by-step solution:**
We need to find if the last two digits of each given year can be factored into a valid month (1 through 12) and a valid day (1 through 31).
- **A. 1990:** $90 = 9 \times 10$. This corresponds to September 10th (9/10/90). Lucky.
- **B. 1991:** $91 = 7 \times 13$. This corresponds to July 13th (7/13/91). Lucky.
- **C. 1992:** $92 = 4 \times 23$. This corresponds to April 23rd (4/23/92). Lucky.
- **D. 1993:** $93 = 3 \times 31$. This corresponds to March 31st (3/31/93). Lucky.
- **E. 1994:** $94 = 2 \times 47$. Since a month cannot have 47 days, and 94 has no other integer pairs that fit the calendar constraints (1×94 is also invalid), 1994 is NOT a lucky year.

ANSWER 3: E

---

**Problem 4:**
When placing each of the digits 2, 4, 5, 6, 9 in exactly one of the boxes of this subtraction problem, what is the smallest difference that is possible?       −

**Step-by-step solution:**
We are given 5 digits to place in a subtraction problem. To yield a positive difference using 5 digits, the setup must be a 3-digit number minus a 2-digit number.
To get the smallest possible difference, we need the 3-digit number to be as small as possible and the 2-digit number to be as large as possible.
1. The smallest possible hundreds digit for the 3-digit number is 2. So the number is $2\_ \_$.
2. We are left with the digits 4, 5, 6, and 9. 
3. To minimize the 3-digit number, the remaining two digits should be the smallest available: 4 and 5, making the number 245.
4. To maximize the 2-digit number being subtracted, we use the largest remaining digits: 9 and 6, making the number 96.
5. Subtracting the two gives: $245 - 96 = 149$.
(Testing other close combinations like $254 - 96 = 158$ or $246 - 95 = 151$ confirms 149 is the absolute minimum difference.)

ANSWER 4: C

---

**Problem 5:**
Each of the letters W , X , Y , and Z represents a different integer in the set {1, 2, 3, 4}, but not necessarily in that order. If W X − Y Z = 1, then the sum of W and Y is

**Step-by-step solution:**
The equation given is meant to be read as fractions (a common formatting artifact drops the fraction bar): $\frac{W}{X} - \frac{Y}{Z} = 1$.
Let's test the values {1, 2, 3, 4} to satisfy this equation:
- If we set $W = 3$, $X = 1$, $Y = 4$, and $Z = 2$:
  $\frac{3}{1} - \frac{4}{2} = 3 - 2 = 1$
This perfectly uses all four distinct digits exactly once.
The question asks for the sum of $W$ and $Y$:
$W + Y = 3 + 4 = 7$

*(Note: Checking $W \times X - Y \times Z = 1$ yields no solutions where all 4 digits are distinct, confirming the fractional interpretation).*

ANSWER 5: E

---

**Problem 6:**
An integer between 1000 and 9999, inclusive, is chosen at random. What is the probability that it is an odd integer whose digits are all distinct?

**Step-by-step solution:**
1. The total number of integers from 1000 to 9999 inclusive is $9999 - 1000 + 1 = 9000$.
2. We want to form a 4-digit odd number (let's call its digits A B C D) with all distinct digits.
3. Because it must be odd, the units digit (D) must be 1, 3, 5, 7, or 9. There are **5** choices for D.
4. The thousands digit (A) cannot be 0, and it cannot be the digit used for D. Thus, out of 10 possible digits, 2 are restricted, leaving **8** choices for A.
5. The hundreds digit (B) can be 0, but cannot be the digits used for A or D. This leaves **8** choices for B.
6. The tens digit (C) cannot be the digits used for A, B, or D. This leaves **7** choices for C.
7. The total number of such integers is $5 \times 8 \times 8 \times 7 = 2240$.
8. The probability is $\frac{2240}{9000} = \frac{224}{900} = \frac{56}{225}$.

ANSWER 6: B

---

**Problem 7:**
The manager of a company planned to give a $50 bonus to each employee from the company fund, but the fund was $5 short of what was needed. Instead the manager gave each employee a $45 bonus and kept the remaining $95 in the fund. How much money was in the company fund before any bonuses were paid?

**Step-by-step solution:**
Let $n$ be the number of employees.
1. If the manager gave $50 to each employee, the fund would be $5 short. So, the total fund is $5

## Step 1: Problem 1 - Restate the problem
We need to find how many whole numbers from 1 through 46 are divisible by either 3 or 5 or both.

## Step 2: Problem 1 - Calculate numbers divisible by 3
First, let's find the numbers divisible by 3. The largest number less than or equal to 46 that is divisible by 3 is 45, and the number of multiples of 3 from 1 to 46 is 45 / 3 = 15.

## Step 3: Problem 1 - Calculate numbers divisible by 5
Next, find the numbers divisible by 5. The largest number less than or equal to 46 that is divisible by 5 is 45, and the number of multiples of 5 from 1 to 46 is 45 / 5 = 9.

## Step 4: Problem 1 - Calculate numbers divisible by both 3 and 5
Numbers divisible by both 3 and 5 are divisible by 15. The largest number less than or equal to 46 that is divisible by 15 is 45, and the number of multiples of 15 from 1 to 46 is 45 / 15 = 3.

## Step 5: Problem 1 - Apply the principle of inclusion-exclusion
To find the total number of numbers divisible by either 3 or 5 or both, we use the principle of inclusion-exclusion: (numbers divisible by 3) + (numbers divisible by 5) - (numbers divisible by both) = 15 + 9 - 3 = 21.

## Step 6: Problem 1 - Provide the answer
ANSWER 1: B

## Step 7: Problem 2 - Restate the problem
Jamal has 6 green socks, 18 purple socks, and 12 orange socks. After adding more purple socks, the probability of picking a purple sock is 60%. We need to find how many purple socks Jamal added.

## Step 8: Problem 2 - Set up the initial conditions
Initially, Jamal has 6 + 18 + 12 = 36 socks.

## Step 9: Problem 2 - Set up the equation for the probability of picking a purple sock after adding x purple socks
After adding x purple socks, the total number of socks is 36 + x, and the number of purple socks is 18 + x. The probability of picking a purple sock is (18 + x) / (36 + x) = 0.6.

## Step 10: Problem 2 - Solve the equation for x
(18 + x) / (36 + x) = 0.6. Multiplying both sides by (36 + x) gives 18 + x = 0.6(36 + x). Expanding, we get 18 + x = 21.6 + 0.6x. Rearranging, x - 0.6x = 21.6 - 18, so 0.4x = 3.6. Dividing both sides by 0.4, x = 9.

## Step 11: Problem 2 - Provide the answer
ANSWER 2: B

## Step 12: Problem 3 - Restate the problem
We need to find which year is NOT a lucky year, where a lucky year has at least one date where the month times the day equals the last two digits of the year.

## Step 13: Problem 3 - Analyze the condition for a lucky year
For a year to be lucky, there must exist a month (1-12) and a day (1-31) such that month * day = last two digits of the year.

## Step 14: Problem 3 - Check each given year
Checking each year from 1990 to 1994 to see if any date satisfies the condition.

## Step 15: Problem 3 - Analyze 1990 to 1994
- For 1990, the last two digits are 90. It's difficult to get 90 by multiplying two numbers between 1 and 12 and 1 and 31, respectively, because the factors of 90 that could work are (1,90), (2,45), (3,30), (5,18), (6,15), (9,10). Only (9,10) is plausible, and it works as 9/10/90.
- For 1991, 91 can be achieved with 7/13/91, so it's lucky.
- For 1992, 92 can be achieved, for example, with 4/23/92, so it's lucky.
- For 1993, 93 can be achieved with 3/31/93, so it's lucky.
- For 1994, 94 is not easily achieved because its factors are (1,94), (2,47). No month-day combination gives 94.

## Step 16: Problem 3 - Provide the answer
Given the analysis, 1994 is not lucky because there's no month-day product equal to 94.

ANSWER 3: E

## Step 17: Problem 4 - Restate the problem
We need to find the smallest difference possible by placing the digits 2, 4, 5, 6, 9 in the boxes of a subtraction problem.

## Step 18: Problem 4 - Understand the structure of the subtraction problem
To minimize the difference, we should minimize the minuend and maximize the subtrahend.

## Step 19: Problem 4 - Arrange the digits to minimize the difference
To achieve the smallest difference, the hundreds place of the minuend should be as small as possible, and the hundreds place of the subtrahend should be as large as possible. So, the minuend could be 2** and the subtrahend 9**. The remaining digits are 4, 5, 6.

## Step 20: Problem 4 - Try different combinations
Let's try to make the minuend as small as possible and the subtrahend as large as possible: 245 - 986 is not valid because it results in a negative number. A valid attempt is 246 - 985 or similar combinations.

## Step 21: Problem 4 - Calculate the difference for a valid combination
For 964 - 852 or similar, the difference is not minimal. Let's directly check: 245 - 196 = 49 (not an option), but trying to minimize, we can get 246 - 195 = 51 (still not an option). A plausible arrangement is 254 - 196 = 58.

## Step 22: Problem 4 - Provide the answer
The smallest difference is achieved with an arrangement like 254 - 196.

ANSWER 4: A

## Step 23: Problem 5 - Restate the problem
Given W, X, Y, Z are different integers from 1 to 4, and WX - YZ = 1, we need to find W + Y.

## Step 24: Problem 5 - Understand the equation WX - YZ = 1
Since W, X, Y, Z are different integers from 1 to 4, we need to consider the possible values of WX and YZ.

## Step 25: Problem 5 - Analyze possible values for WX and YZ
The possible products are: 1*2=2, 1*3=3, 1*4=4, 2*1=2, 2*3=6, 2*4=8, 3*1=3, 3*2=6, 3*4=12, 4*1=4, 4*2=8, 4*3=12. We need a difference of 1 between two of these products.

## Step 26: Problem 5 - Check for a difference of 1
The products are 2, 3, 4, 6, 8, 12. The pairs that differ by 1 are (2,3), (3,4). So, either WX = 3 and YZ = 2 or WX = 4 and YZ = 3.

## Step 27: Problem 5 - Solve for W + Y
For WX = 3 and YZ = 2, W = 3, X = 1, Y = 2, Z = 1 (not valid because Z = X), or W = 1, X = 3, Y = 1, Z = 2 (not valid because W = Y). A valid case is W = 3, X = 1, Y = 1, Z = 2 (not valid), but if we consider W = 1, X = 3, Y = 2, Z = 1, it's valid and gives W + Y = 1 + 2 = 3.

## Step 28: Problem 5 - Provide the answer
For the valid case where W = 1, X = 3, Y = 2, Z = 1 doesn't work due to repetition, but

Problem 1:

We need count the whole numbers from 1 through 46 that are divisible by 3 or 5 or both.

Count multiples of 3:

\[
\left\lfloor \frac{46}{3} \right\rfloor = 15
\]

Count multiples of 5:

\[
\left\lfloor \frac{46}{5} \right\rfloor = 9
\]

But multiples of both 3 and 5, meaning multiples of 15, were counted twice, so subtract them once:

\[
\left\lfloor \frac{46}{15} \right\rfloor = 3
\]

Therefore the total is

\[
15+9-3=21
\]

So choices like 24 or 27 overcount because they do not remove the overlap.

ANSWER 1: B

Problem 2:

Initially Jamal has:

\[
6+18+12=36
\]

socks total, of which 18 are purple.

Suppose he adds \(x\) purple socks. Then the number of purple socks is

\[
18+x
\]

and the total number of socks is

\[
36+x
\]

The probability of choosing a purple sock is now 60%, or \(\frac35\). So

\[
\frac{18+x}{36+x}=\frac35
\]

Cross multiply:

\[
5(18+x)=3(36+x)
\]

\[
90+5x=108+3x
\]

\[
2x=18
\]

\[
x=9
\]

So Jamal added 9 purple socks.

ANSWER 2: B

Problem 3:

A lucky year has some valid date month/day/year where

\[
\text{month} \times \text{day} = \text{last two digits of the year}
\]

Check each year:

For 1990:

\[
9 \times 10 = 90
\]

So 9/10/90 works.

For 1991:

\[
7 \times 13 = 91
\]

So 7/13/91 works.

For 1992:

\[
4 \times 23 = 92
\]

So 4/23/92 works.

For 1993:

\[
3 \times 31 = 93
\]

So 3/31/93 works.

For 1994:

\[
94=2 \times 47
\]

But day 47 is not valid. The only factor pairs are \(1 \times 94\) and \(2 \times 47\), and neither gives a valid date.

Thus 1994 is not lucky.

ANSWER 3: E

Problem 4:

The subtraction problem uses the digits 2, 4, 5, 6, 9 exactly once in a 3-digit number minus a 2-digit number.

To make the difference as small as possible, the 3-digit number should be as small as possible, and the 2-digit number should be as large as possible.

The hundreds digit of the 3-digit number should be the smallest digit, 2.

Now we want the rest of the top number small and the bottom number large. Using the remaining digits 4, 5, 6, 9, the smallest possible two digits for the top number are 4 and 5, making 245, while the largest possible 2-digit number from the remaining digits is 96.

So try:

\[
245-96=149
\]

If we tried to get below 149, we would need either a smaller 3-digit number than 245 or a larger 2-digit number than 96, but neither is possible with these digits.

Therefore the smallest difference is 149.

ANSWER 4: C

Problem 5:

The equation is

\[
\frac{W}{X}-\frac{Y}{Z}=1
\]

where \(W, X, Y, Z\) are the numbers 1, 2, 3, 4 in some order.

We need find an arrangement that works.

Try to make the first fraction bigger than 1. One successful choice is

\[
\frac{3}{1}-\frac{4}{2}=1
\]

because

\[
3-2=1
\]

So

\[
W=3,\quad X=1,\quad Y=4,\quad Z=2
\]

Then

\[
W+Y=3+4=7
\]

The other choices are smaller sums, but the only working arrangement gives 7.

ANSWER 5: E

Problem 6:

There are four-digit integers from 1000 through 9999. The total number is

\[
9999-1000+1=9000
\]

We want odd integers whose digits are all distinct.

For a number to be odd, the units digit must be one of

\[
1,3,5,7,9
\]

So there are 5 choices for the units digit.

The thousands digit cannot be 0 and cannot equal the units digit. Since there are 9 possible nonzero thousands digits, and one is already used, there are

\[
8
\]

choices.

The hundreds digit can be any digit except the two already used:

\[
8
\]

choices.

The tens digit can be any digit except the three already used:

\[
7
\]

choices.

So the number of favorable integers is

\[
5 \cdot 8 \cdot 8 \cdot 7=2240
\]

The probability is

\[
\frac{2240}{9000}
\]

Reduce:

\[
\frac{2240}{9000}=\frac{224}{900}=\frac{56}{225}
\]

ANSWER 6: B

Problem 7:

Let \(n\) be the number of employees, and let \(F\) be the amount of money in the fund.

The manager planned to give each employee $50, but the fund was $5 short. So

\[
F=50n-5
\]

Instead, the manager gave each employee $45 and had $95 left. So

\[
F=45n+95
\]

Set these equal:

\[
50n-5=45n+95
\]

\[
5n=100
\]

\[
n=20
\]

Now find the fund:

\[
F=50(20)-5=1000-5=995
\]

So there were $995 in the fund.

ANSWER 7: E

Problem 8:

We are given

\[
ABA \times CD = CDCD
\]

The number \(ABA\) is

\[
101A+10B
\]

The number \(CDCD\) is a repeated two-digit number, so

\[
CDCD=101 \times CD
\]

Thus the equation becomes

\[
ABA \times CD = 101 \times CD
\]

Since \(CD\) is a two-digit number, it is not 0, so we can divide both sides by \(CD\):

\[
ABA=101
\]

Therefore

\[
A=1,\quad B=0
\]

So

\[
A+B=1+0=1
\]

ANSWER 8: A

Problem 9:

We need choose 3 numbers from

\[
\{89,95,99,132,166,173\}
\]

so that their sum is even.

Classify by parity.

Odd numbers:

\[
89,95,99,173
\]

There are 4 odds.

Even numbers:

\[
132,166
\]

There are 2 evens.

A sum of three numbers is even if it contains either:

- 0 odd numbers and 3 even numbers, or
- 2 odd numbers and 1 even number.

We cannot choose 3 even numbers because there are only 2 even numbers.

So we must choose 2 odds and 1 even.

Number of ways:

\[
\binom{4}{2}\binom{2}{1}=6 \cdot 2=12
\]

ANSWER 9: D

Problem 10:

The set has 5 positive integers with mean 5, so the total sum is

\[
5 \cdot 5=25
\]

The median is 5, so when the numbers are ordered, the middle number is 5.

Also, 8 is the only mode, meaning 8 appears more often than any other number. Since there are only five numbers and the median is 5, the two largest numbers must be 8 and 8:

\[
\_,\_,5,8,8
\]

Their sum so far is

\[
5+8+8=21
\]

So the first two numbers must sum to

\[
25-21=4
\]

They are positive integers less than or equal to 5. The possible pairs are:

\[
1,3 \quad \text{or} \quad 2,2
\]

But \(2,2\) would make 2 appear twice, tying 8 as a mode. Since 8 must be the only mode, the pair must be

\[
1,3
\]

So the set is

\[
1,3,5,8,8
\]

The difference between the largest and smallest integers is

\[
8-1=7
\]

ANSWER 10: D

Problem 11:

We need count three-digit whole numbers between 99 and 999 that contain exactly one 0.

A t

**Problem 1:**
How many whole numbers from 1 through 46 are divisible by either 3 or 5 or both?

**Step-by-step solution:**
To find the number of integers divisible by 3, 5, or both, we can use the Principle of Inclusion-Exclusion. 
1. First, we count the numbers from 1 to 46 that are divisible by 3:
   $\lfloor 46 \div 3 \rfloor = 15$
2. Next, we count the numbers from 1 to 46 that are divisible by 5:
   $\lfloor 46 \div 5 \rfloor = 9$
3. Then, we count the numbers that are divisible by BOTH 3 and 5 (which means they are divisible by their least common multiple, 15):
   $\lfloor 46 \div 15 \rfloor = 3$
4. Finally, we add the multiples of 3 and 5, and subtract the multiples of 15 so they aren't counted twice:
   $15 + 9 - 3 = 21$

ANSWER 1: B

---

**Problem 2:**
Jamal has a drawer containing 6 green socks, 18 purple socks, and 12 orange socks. After adding more purple socks, Jamal noticed that there is now a 60% chance that a sock randomly selected from the drawer is purple. How many purple socks did Jamal add?

**Step-by-step solution:**
1. Initially, Jamal has $6 + 18 + 12 = 36$ socks in total.
2. Let $x$ be the number of purple socks Jamal added. 
3. The new number of purple socks is $18 + x$.
4. The new total number of socks is $36 + x$.
5. We are told the new probability of selecting a purple sock is 60%, which simplifies to $\frac{60}{100} = \frac{3}{5}$.
6. We set up the equation:
   $\frac{18 + x}{36 + x} = \frac{3}{5}$
7. Cross-multiply to solve for $x$:
   $5(18 + x) = 3(36 + x)$
   $90 + 5x = 108 + 3x$
   $2x = 18$
   $x = 9$

Jamal added 9 purple socks.

ANSWER 2: B

---

**Problem 3:**
A lucky year is one in which at least one date, written as month/day/year, has the property that the month times the day equals the last two digits of the year. For example, 1956 is lucky because 7/8/56 has 7 × 8 = 56. Which of the following is NOT a lucky year?

**Step-by-step solution:**
We need to find if the last two digits of each given year can be factored into a valid month (1 through 12) and a valid day (1 through 31).
- **A. 1990:** $90 = 9 \times 10$. This corresponds to September 10th (9/10/90). Lucky.
- **B. 1991:** $91 = 7 \times 13$. This corresponds to July 13th (7/13/91). Lucky.
- **C. 1992:** $92 = 4 \times 23$. This corresponds to April 23rd (4/23/92). Lucky.
- **D. 1993:** $93 = 3 \times 31$. This corresponds to March 31st (3/31/93). Lucky.
- **E. 1994:** $94 = 2 \times 47$. Since a month cannot have 47 days, and 94 has no other integer pairs that fit the calendar constraints (1×94 is also invalid), 1994 is NOT a lucky year.

ANSWER 3: E

---

**Problem 4:**
When placing each of the digits 2, 4, 5, 6, 9 in exactly one of the boxes of this subtraction problem, what is the smallest difference that is possible?       −

**Step-by-step solution:**
We are given 5 digits to place in a subtraction problem. To yield a positive difference using 5 digits, the setup must be a 3-digit number minus a 2-digit number.
To get the smallest possible difference, we need the 3-digit number to be as small as possible and the 2-digit number to be as large as possible.
1. The smallest possible hundreds digit for the 3-digit number is 2. So the number is $2\_ \_$.
2. We are left with the digits 4, 5, 6, and 9. 
3. To minimize the 3-digit number, the remaining two digits should be the smallest available: 4 and 5, making the number 245.
4. To maximize the 2-digit number being subtracted, we use the largest remaining digits: 9 and 6, making the number 96.
5. Subtracting the two gives: $245 - 96 = 149$.
(Testing other close combinations like $254 - 96 = 158$ or $246 - 95 = 151$ confirms 149 is the absolute minimum difference.)

ANSWER 4: C

---

**Problem 5:**
Each of the letters W , X , Y , and Z represents a different integer in the set {1, 2, 3, 4}, but not necessarily in that order. If W X − Y Z = 1, then the sum of W and Y is

**Step-by-step solution:**
The equation given is meant to be read as fractions (a common formatting artifact drops the fraction bar): $\frac{W}{X} - \frac{Y}{Z} = 1$.
Let's test the values {1, 2, 3, 4} to satisfy this equation:
- If we set $W = 3$, $X = 1$, $Y = 4$, and $Z = 2$:
  $\frac{3}{1} - \frac{4}{2} = 3 - 2 = 1$
This perfectly uses all four distinct digits exactly once.
The question asks for the sum of $W$ and $Y$:
$W + Y = 3 + 4 = 7$

*(Note: Checking $W \times X - Y \times Z = 1$ yields no solutions where all 4 digits are distinct, confirming the fractional interpretation).*

ANSWER 5: E

---

**Problem 6:**
An integer between 1000 and 9999, inclusive, is chosen at random. What is the probability that it is an odd integer whose digits are all distinct?

**Step-by-step solution:**
1. The total number of integers from 1000 to 9999 inclusive is $9999 - 1000 + 1 = 9000$.
2. We want to form a 4-digit odd number (let's call its digits A B C D) with all distinct digits.
3. Because it must be odd, the units digit (D) must be 1, 3, 5, 7, or 9. There are **5** choices for D.
4. The thousands digit (A) cannot be 0, and it cannot be the digit used for D. Thus, out of 10 possible digits, 2 are restricted, leaving **8** choices for A.
5. The hundreds digit (B) can be 0, but cannot be the digits used for A or D. This leaves **8** choices for B.
6. The tens digit (C) cannot be the digits used for A, B, or D. This leaves **7** choices for C.
7. The total number of such integers is $5 \times 8 \times 8 \times 7 = 2240$.
8. The probability is $\frac{2240}{9000} = \frac{224}{900} = \frac{56}{225}$.

ANSWER 6: B

---

**Problem 7:**
The manager of a company planned to give a $50 bonus to each employee from the company fund, but the fund was $5 short of what was needed. Instead the manager gave each employee a $45 bonus and kept the remaining $95 in the fund. How much money was in the company fund before any bonuses were paid?

**Step-by-step solution:**
Let $n$ be the number of employees.
1. If the manager gave $50 to each employee, the fund would be $5 short. So, the total fund is $5

## Step 1: Problem 1 - Restate the problem
We need to find how many whole numbers from 1 through 46 are divisible by either 3 or 5 or both.

## Step 2: Problem 1 - Calculate numbers divisible by 3
First, let's find the numbers divisible by 3. The largest number less than or equal to 46 that is divisible by 3 is 45, and the number of multiples of 3 from 1 to 46 is 45 / 3 = 15.

## Step 3: Problem 1 - Calculate numbers divisible by 5
Next, find the numbers divisible by 5. The largest number less than or equal to 46 that is divisible by 5 is 45, and the number of multiples of 5 from 1 to 46 is 45 / 5 = 9.

## Step 4: Problem 1 - Calculate numbers divisible by both 3 and 5
Numbers divisible by both 3 and 5 are divisible by 15. The largest number less than or equal to 46 that is divisible by 15 is 45, and the number of multiples of 15 from 1 to 46 is 45 / 15 = 3.

## Step 5: Problem 1 - Apply the principle of inclusion-exclusion
To find the total number of numbers divisible by either 3 or 5 or both, we use the principle of inclusion-exclusion: (numbers divisible by 3) + (numbers divisible by 5) - (numbers divisible by both) = 15 + 9 - 3 = 21.

## Step 6: Problem 1 - Provide the answer
ANSWER 1: B

## Step 7: Problem 2 - Restate the problem
Jamal has 6 green socks, 18 purple socks, and 12 orange socks. After adding more purple socks, the probability of picking a purple sock is 60%. We need to find how many purple socks Jamal added.

## Step 8: Problem 2 - Set up the initial conditions
Initially, Jamal has 6 + 18 + 12 = 36 socks.

## Step 9: Problem 2 - Set up the equation for the probability of picking a purple sock after adding x purple socks
After adding x purple socks, the total number of socks is 36 + x, and the number of purple socks is 18 + x. The probability of picking a purple sock is (18 + x) / (36 + x) = 0.6.

## Step 10: Problem 2 - Solve the equation for x
(18 + x) / (36 + x) = 0.6. Multiplying both sides by (36 + x) gives 18 + x = 0.6(36 + x). Expanding, we get 18 + x = 21.6 + 0.6x. Rearranging, x - 0.6x = 21.6 - 18, so 0.4x = 3.6. Dividing both sides by 0.4, x = 9.

## Step 11: Problem 2 - Provide the answer
ANSWER 2: B

## Step 12: Problem 3 - Restate the problem
We need to find which year is NOT a lucky year, where a lucky year has at least one date where the month times the day equals the last two digits of the year.

## Step 13: Problem 3 - Analyze the condition for a lucky year
For a year to be lucky, there must exist a month (1-12) and a day (1-31) such that month * day = last two digits of the year.

## Step 14: Problem 3 - Check each given year
Checking each year from 1990 to 1994 to see if any date satisfies the condition.

## Step 15: Problem 3 - Analyze 1990 to 1994
- For 1990, the last two digits are 90. It's difficult to get 90 by multiplying two numbers between 1 and 12 and 1 and 31, respectively, because the factors of 90 that could work are (1,90), (2,45), (3,30), (5,18), (6,15), (9,10). Only (9,10) is plausible, and it works as 9/10/90.
- For 1991, 91 can be achieved with 7/13/91, so it's lucky.
- For 1992, 92 can be achieved, for example, with 4/23/92, so it's lucky.
- For 1993, 93 can be achieved with 3/31/93, so it's lucky.
- For 1994, 94 is not easily achieved because its factors are (1,94), (2,47). No month-day combination gives 94.

## Step 16: Problem 3 - Provide the answer
Given the analysis, 1994 is not lucky because there's no month-day product equal to 94.

ANSWER 3: E

## Step 17: Problem 4 - Restate the problem
We need to find the smallest difference possible by placing the digits 2, 4, 5, 6, 9 in the boxes of a subtraction problem.

## Step 18: Problem 4 - Understand the structure of the subtraction problem
To minimize the difference, we should minimize the minuend and maximize the subtrahend.

## Step 19: Problem 4 - Arrange the digits to minimize the difference
To achieve the smallest difference, the hundreds place of the minuend should be as small as possible, and the hundreds place of the subtrahend should be as large as possible. So, the minuend could be 2** and the subtrahend 9**. The remaining digits are 4, 5, 6.

## Step 20: Problem 4 - Try different combinations
Let's try to make the minuend as small as possible and the subtrahend as large as possible: 245 - 986 is not valid because it results in a negative number. A valid attempt is 246 - 985 or similar combinations.

## Step 21: Problem 4 - Calculate the difference for a valid combination
For 964 - 852 or similar, the difference is not minimal. Let's directly check: 245 - 196 = 49 (not an option), but trying to minimize, we can get 246 - 195 = 51 (still not an option). A plausible arrangement is 254 - 196 = 58.

## Step 22: Problem 4 - Provide the answer
The smallest difference is achieved with an arrangement like 254 - 196.

ANSWER 4: A

## Step 23: Problem 5 - Restate the problem
Given W, X, Y, Z are different integers from 1 to 4, and WX - YZ = 1, we need to find W + Y.

## Step 24: Problem 5 - Understand the equation WX - YZ = 1
Since W, X, Y, Z are different integers from 1 to 4, we need to consider the possible values of WX and YZ.

## Step 25: Problem 5 - Analyze possible values for WX and YZ
The possible products are: 1*2=2, 1*3=3, 1*4=4, 2*1=2, 2*3=6, 2*4=8, 3*1=3, 3*2=6, 3*4=12, 4*1=4, 4*2=8, 4*3=12. We need a difference of 1 between two of these products.

## Step 26: Problem 5 - Check for a difference of 1
The products are 2, 3, 4, 6, 8, 12. The pairs that differ by 1 are (2,3), (3,4). So, either WX = 3 and YZ = 2 or WX = 4 and YZ = 3.

## Step 27: Problem 5 - Solve for W + Y
For WX = 3 and YZ = 2, W = 3, X = 1, Y = 2, Z = 1 (not valid because Z = X), or W = 1, X = 3, Y = 1, Z = 2 (not valid because W = Y). A valid case is W = 3, X = 1, Y = 1, Z = 2 (not valid), but if we consider W = 1, X = 3, Y = 2, Z = 1, it's valid and gives W + Y = 1 + 2 = 3.

## Step 28: Problem 5 - Provide the answer
For the valid case where W = 1, X = 3, Y = 2, Z = 1 doesn't work due to repetition, but

Problem 1:

We need count the whole numbers from 1 through 46 that are divisible by 3 or 5 or both.

Count multiples of 3:

\[
\left\lfloor \frac{46}{3} \right\rfloor = 15
\]

Count multiples of 5:

\[
\left\lfloor \frac{46}{5} \right\rfloor = 9
\]

But multiples of both 3 and 5, meaning multiples of 15, were counted twice, so subtract them once:

\[
\left\lfloor \frac{46}{15} \right\rfloor = 3
\]

Therefore the total is

\[
15+9-3=21
\]

So choices like 24 or 27 overcount because they do not remove the overlap.

ANSWER 1: B

Problem 2:

Initially Jamal has:

\[
6+18+12=36
\]

socks total, of which 18 are purple.

Suppose he adds \(x\) purple socks. Then the number of purple socks is

\[
18+x
\]

and the total number of socks is

\[
36+x
\]

The probability of choosing a purple sock is now 60%, or \(\frac35\). So

\[
\frac{18+x}{36+x}=\frac35
\]

Cross multiply:

\[
5(18+x)=3(36+x)
\]

\[
90+5x=108+3x
\]

\[
2x=18
\]

\[
x=9
\]

So Jamal added 9 purple socks.

ANSWER 2: B

Problem 3:

A lucky year has some valid date month/day/year where

\[
\text{month} \times \text{day} = \text{last two digits of the year}
\]

Check each year:

For 1990:

\[
9 \times 10 = 90
\]

So 9/10/90 works.

For 1991:

\[
7 \times 13 = 91
\]

So 7/13/91 works.

For 1992:

\[
4 \times 23 = 92
\]

So 4/23/92 works.

For 1993:

\[
3 \times 31 = 93
\]

So 3/31/93 works.

For 1994:

\[
94=2 \times 47
\]

But day 47 is not valid. The only factor pairs are \(1 \times 94\) and \(2 \times 47\), and neither gives a valid date.

Thus 1994 is not lucky.

ANSWER 3: E

Problem 4:

The subtraction problem uses the digits 2, 4, 5, 6, 9 exactly once in a 3-digit number minus a 2-digit number.

To make the difference as small as possible, the 3-digit number should be as small as possible, and the 2-digit number should be as large as possible.

The hundreds digit of the 3-digit number should be the smallest digit, 2.

Now we want the rest of the top number small and the bottom number large. Using the remaining digits 4, 5, 6, 9, the smallest possible two digits for the top number are 4 and 5, making 245, while the largest possible 2-digit number from the remaining digits is 96.

So try:

\[
245-96=149
\]

If we tried to get below 149, we would need either a smaller 3-digit number than 245 or a larger 2-digit number than 96, but neither is possible with these digits.

Therefore the smallest difference is 149.

ANSWER 4: C

Problem 5:

The equation is

\[
\frac{W}{X}-\frac{Y}{Z}=1
\]

where \(W, X, Y, Z\) are the numbers 1, 2, 3, 4 in some order.

We need find an arrangement that works.

Try to make the first fraction bigger than 1. One successful choice is

\[
\frac{3}{1}-\frac{4}{2}=1
\]

because

\[
3-2=1
\]

So

\[
W=3,\quad X=1,\quad Y=4,\quad Z=2
\]

Then

\[
W+Y=3+4=7
\]

The other choices are smaller sums, but the only working arrangement gives 7.

ANSWER 5: E

Problem 6:

There are four-digit integers from 1000 through 9999. The total number is

\[
9999-1000+1=9000
\]

We want odd integers whose digits are all distinct.

For a number to be odd, the units digit must be one of

\[
1,3,5,7,9
\]

So there are 5 choices for the units digit.

The thousands digit cannot be 0 and cannot equal the units digit. Since there are 9 possible nonzero thousands digits, and one is already used, there are

\[
8
\]

choices.

The hundreds digit can be any digit except the two already used:

\[
8
\]

choices.

The tens digit can be any digit except the three already used:

\[
7
\]

choices.

So the number of favorable integers is

\[
5 \cdot 8 \cdot 8 \cdot 7=2240
\]

The probability is

\[
\frac{2240}{9000}
\]

Reduce:

\[
\frac{2240}{9000}=\frac{224}{900}=\frac{56}{225}
\]

ANSWER 6: B

Problem 7:

Let \(n\) be the number of employees, and let \(F\) be the amount of money in the fund.

The manager planned to give each employee $50, but the fund was $5 short. So

\[
F=50n-5
\]

Instead, the manager gave each employee $45 and had $95 left. So

\[
F=45n+95
\]

Set these equal:

\[
50n-5=45n+95
\]

\[
5n=100
\]

\[
n=20
\]

Now find the fund:

\[
F=50(20)-5=1000-5=995
\]

So there were $995 in the fund.

ANSWER 7: E

Problem 8:

We are given

\[
ABA \times CD = CDCD
\]

The number \(ABA\) is

\[
101A+10B
\]

The number \(CDCD\) is a repeated two-digit number, so

\[
CDCD=101 \times CD
\]

Thus the equation becomes

\[
ABA \times CD = 101 \times CD
\]

Since \(CD\) is a two-digit number, it is not 0, so we can divide both sides by \(CD\):

\[
ABA=101
\]

Therefore

\[
A=1,\quad B=0
\]

So

\[
A+B=1+0=1
\]

ANSWER 8: A

Problem 9:

We need choose 3 numbers from

\[
\{89,95,99,132,166,173\}
\]

so that their sum is even.

Classify by parity.

Odd numbers:

\[
89,95,99,173
\]

There are 4 odds.

Even numbers:

\[
132,166
\]

There are 2 evens.

A sum of three numbers is even if it contains either:

- 0 odd numbers and 3 even numbers, or
- 2 odd numbers and 1 even number.

We cannot choose 3 even numbers because there are only 2 even numbers.

So we must choose 2 odds and 1 even.

Number of ways:

\[
\binom{4}{2}\binom{2}{1}=6 \cdot 2=12
\]

ANSWER 9: D

Problem 10:

The set has 5 positive integers with mean 5, so the total sum is

\[
5 \cdot 5=25
\]

The median is 5, so when the numbers are ordered, the middle number is 5.

Also, 8 is the only mode, meaning 8 appears more often than any other number. Since there are only five numbers and the median is 5, the two largest numbers must be 8 and 8:

\[
\_,\_,5,8,8
\]

Their sum so far is

\[
5+8+8=21
\]

So the first two numbers must sum to

\[
25-21=4
\]

They are positive integers less than or equal to 5. The possible pairs are:

\[
1,3 \quad \text{or} \quad 2,2
\]

But \(2,2\) would make 2 appear twice, tying 8 as a mode. Since 8 must be the only mode, the pair must be

\[
1,3
\]

So the set is

\[
1,3,5,8,8
\]

The difference between the largest and smallest integers is

\[
8-1=7
\]

ANSWER 10: D

Problem 11:

We need count three-digit whole numbers between 99 and 999 that contain exactly one 0.

A t

**Problem 1:**
How many whole numbers from 1 through 46 are divisible by either 3 or 5 or both?

**Step-by-step solution:**
To find the number of integers divisible by 3, 5, or both, we can use the Principle of Inclusion-Exclusion. 
1. First, we count the numbers from 1 to 46 that are divisible by 3:
   $\lfloor 46 \div 3 \rfloor = 15$
2. Next, we count the numbers from 1 to 46 that are divisible by 5:
   $\lfloor 46 \div 5 \rfloor = 9$
3. Then, we count the numbers that are divisible by BOTH 3 and 5 (which means they are divisible by their least common multiple, 15):
   $\lfloor 46 \div 15 \rfloor = 3$
4. Finally, we add the multiples of 3 and 5, and subtract the multiples of 15 so they aren't counted twice:
   $15 + 9 - 3 = 21$

ANSWER 1: B

---

**Problem 2:**
Jamal has a drawer containing 6 green socks, 18 purple socks, and 12 orange socks. After adding more purple socks, Jamal noticed that there is now a 60% chance that a sock randomly selected from the drawer is purple. How many purple socks did Jamal add?

**Step-by-step solution:**
1. Initially, Jamal has $6 + 18 + 12 = 36$ socks in total.
2. Let $x$ be the number of purple socks Jamal added. 
3. The new number of purple socks is $18 + x$.
4. The new total number of socks is $36 + x$.
5. We are told the new probability of selecting a purple sock is 60%, which simplifies to $\frac{60}{100} = \frac{3}{5}$.
6. We set up the equation:
   $\frac{18 + x}{36 + x} = \frac{3}{5}$
7. Cross-multiply to solve for $x$:
   $5(18 + x) = 3(36 + x)$
   $90 + 5x = 108 + 3x$
   $2x = 18$
   $x = 9$

Jamal added 9 purple socks.

ANSWER 2: B

---

**Problem 3:**
A lucky year is one in which at least one date, written as month/day/year, has the property that the month times the day equals the last two digits of the year. For example, 1956 is lucky because 7/8/56 has 7 × 8 = 56. Which of the following is NOT a lucky year?

**Step-by-step solution:**
We need to find if the last two digits of each given year can be factored into a valid month (1 through 12) and a valid day (1 through 31).
- **A. 1990:** $90 = 9 \times 10$. This corresponds to September 10th (9/10/90). Lucky.
- **B. 1991:** $91 = 7 \times 13$. This corresponds to July 13th (7/13/91). Lucky.
- **C. 1992:** $92 = 4 \times 23$. This corresponds to April 23rd (4/23/92). Lucky.
- **D. 1993:** $93 = 3 \times 31$. This corresponds to March 31st (3/31/93). Lucky.
- **E. 1994:** $94 = 2 \times 47$. Since a month cannot have 47 days, and 94 has no other integer pairs that fit the calendar constraints (1×94 is also invalid), 1994 is NOT a lucky year.

ANSWER 3: E

---

**Problem 4:**
When placing each of the digits 2, 4, 5, 6, 9 in exactly one of the boxes of this subtraction problem, what is the smallest difference that is possible?       −

**Step-by-step solution:**
We are given 5 digits to place in a subtraction problem. To yield a positive difference using 5 digits, the setup must be a 3-digit number minus a 2-digit number.
To get the smallest possible difference, we need the 3-digit number to be as small as possible and the 2-digit number to be as large as possible.
1. The smallest possible hundreds digit for the 3-digit number is 2. So the number is $2\_ \_$.
2. We are left with the digits 4, 5, 6, and 9. 
3. To minimize the 3-digit number, the remaining two digits should be the smallest available: 4 and 5, making the number 245.
4. To maximize the 2-digit number being subtracted, we use the largest remaining digits: 9 and 6, making the number 96.
5. Subtracting the two gives: $245 - 96 = 149$.
(Testing other close combinations like $254 - 96 = 158$ or $246 - 95 = 151$ confirms 149 is the absolute minimum difference.)

ANSWER 4: C

---

**Problem 5:**
Each of the letters W , X , Y , and Z represents a different integer in the set {1, 2, 3, 4}, but not necessarily in that order. If W X − Y Z = 1, then the sum of W and Y is

**Step-by-step solution:**
The equation given is meant to be read as fractions (a common formatting artifact drops the fraction bar): $\frac{W}{X} - \frac{Y}{Z} = 1$.
Let's test the values {1, 2, 3, 4} to satisfy this equation:
- If we set $W = 3$, $X = 1$, $Y = 4$, and $Z = 2$:
  $\frac{3}{1} - \frac{4}{2} = 3 - 2 = 1$
This perfectly uses all four distinct digits exactly once.
The question asks for the sum of $W$ and $Y$:
$W + Y = 3 + 4 = 7$

*(Note: Checking $W \times X - Y \times Z = 1$ yields no solutions where all 4 digits are distinct, confirming the fractional interpretation).*

ANSWER 5: E

---

**Problem 6:**
An integer between 1000 and 9999, inclusive, is chosen at random. What is the probability that it is an odd integer whose digits are all distinct?

**Step-by-step solution:**
1. The total number of integers from 1000 to 9999 inclusive is $9999 - 1000 + 1 = 9000$.
2. We want to form a 4-digit odd number (let's call its digits A B C D) with all distinct digits.
3. Because it must be odd, the units digit (D) must be 1, 3, 5, 7, or 9. There are **5** choices for D.
4. The thousands digit (A) cannot be 0, and it cannot be the digit used for D. Thus, out of 10 possible digits, 2 are restricted, leaving **8** choices for A.
5. The hundreds digit (B) can be 0, but cannot be the digits used for A or D. This leaves **8** choices for B.
6. The tens digit (C) cannot be the digits used for A, B, or D. This leaves **7** choices for C.
7. The total number of such integers is $5 \times 8 \times 8 \times 7 = 2240$.
8. The probability is $\frac{2240}{9000} = \frac{224}{900} = \frac{56}{225}$.

ANSWER 6: B

---

**Problem 7:**
The manager of a company planned to give a $50 bonus to each employee from the company fund, but the fund was $5 short of what was needed. Instead the manager gave each employee a $45 bonus and kept the remaining $95 in the fund. How much money was in the company fund before any bonuses were paid?

**Step-by-step solution:**
Let $n$ be the number of employees.
1. If the manager gave $50 to each employee, the fund would be $5 short. So, the total fund is $5

## Step 1: Problem 1 - Restate the problem
We need to find how many whole numbers from 1 through 46 are divisible by either 3 or 5 or both.

## Step 2: Problem 1 - Calculate numbers divisible by 3
First, let's find the numbers divisible by 3. The largest number less than or equal to 46 that is divisible by 3 is 45, and the number of multiples of 3 from 1 to 46 is 45 / 3 = 15.

## Step 3: Problem 1 - Calculate numbers divisible by 5
Next, find the numbers divisible by 5. The largest number less than or equal to 46 that is divisible by 5 is 45, and the number of multiples of 5 from 1 to 46 is 45 / 5 = 9.

## Step 4: Problem 1 - Calculate numbers divisible by both 3 and 5
Numbers divisible by both 3 and 5 are divisible by 15. The largest number less than or equal to 46 that is divisible by 15 is 45, and the number of multiples of 15 from 1 to 46 is 45 / 15 = 3.

## Step 5: Problem 1 - Apply the principle of inclusion-exclusion
To find the total number of numbers divisible by either 3 or 5 or both, we use the principle of inclusion-exclusion: (numbers divisible by 3) + (numbers divisible by 5) - (numbers divisible by both) = 15 + 9 - 3 = 21.

## Step 6: Problem 1 - Provide the answer
ANSWER 1: B

## Step 7: Problem 2 - Restate the problem
Jamal has 6 green socks, 18 purple socks, and 12 orange socks. After adding more purple socks, the probability of picking a purple sock is 60%. We need to find how many purple socks Jamal added.

## Step 8: Problem 2 - Set up the initial conditions
Initially, Jamal has 6 + 18 + 12 = 36 socks.

## Step 9: Problem 2 - Set up the equation for the probability of picking a purple sock after adding x purple socks
After adding x purple socks, the total number of socks is 36 + x, and the number of purple socks is 18 + x. The probability of picking a purple sock is (18 + x) / (36 + x) = 0.6.

## Step 10: Problem 2 - Solve the equation for x
(18 + x) / (36 + x) = 0.6. Multiplying both sides by (36 + x) gives 18 + x = 0.6(36 + x). Expanding, we get 18 + x = 21.6 + 0.6x. Rearranging, x - 0.6x = 21.6 - 18, so 0.4x = 3.6. Dividing both sides by 0.4, x = 9.

## Step 11: Problem 2 - Provide the answer
ANSWER 2: B

## Step 12: Problem 3 - Restate the problem
We need to find which year is NOT a lucky year, where a lucky year has at least one date where the month times the day equals the last two digits of the year.

## Step 13: Problem 3 - Analyze the condition for a lucky year
For a year to be lucky, there must exist a month (1-12) and a day (1-31) such that month * day = last two digits of the year.

## Step 14: Problem 3 - Check each given year
Checking each year from 1990 to 1994 to see if any date satisfies the condition.

## Step 15: Problem 3 - Analyze 1990 to 1994
- For 1990, the last two digits are 90. It's difficult to get 90 by multiplying two numbers between 1 and 12 and 1 and 31, respectively, because the factors of 90 that could work are (1,90), (2,45), (3,30), (5,18), (6,15), (9,10). Only (9,10) is plausible, and it works as 9/10/90.
- For 1991, 91 can be achieved with 7/13/91, so it's lucky.
- For 1992, 92 can be achieved, for example, with 4/23/92, so it's lucky.
- For 1993, 93 can be achieved with 3/31/93, so it's lucky.
- For 1994, 94 is not easily achieved because its factors are (1,94), (2,47). No month-day combination gives 94.

## Step 16: Problem 3 - Provide the answer
Given the analysis, 1994 is not lucky because there's no month-day product equal to 94.

ANSWER 3: E

## Step 17: Problem 4 - Restate the problem
We need to find the smallest difference possible by placing the digits 2, 4, 5, 6, 9 in the boxes of a subtraction problem.

## Step 18: Problem 4 - Understand the structure of the subtraction problem
To minimize the difference, we should minimize the minuend and maximize the subtrahend.

## Step 19: Problem 4 - Arrange the digits to minimize the difference
To achieve the smallest difference, the hundreds place of the minuend should be as small as possible, and the hundreds place of the subtrahend should be as large as possible. So, the minuend could be 2** and the subtrahend 9**. The remaining digits are 4, 5, 6.

## Step 20: Problem 4 - Try different combinations
Let's try to make the minuend as small as possible and the subtrahend as large as possible: 245 - 986 is not valid because it results in a negative number. A valid attempt is 246 - 985 or similar combinations.

## Step 21: Problem 4 - Calculate the difference for a valid combination
For 964 - 852 or similar, the difference is not minimal. Let's directly check: 245 - 196 = 49 (not an option), but trying to minimize, we can get 246 - 195 = 51 (still not an option). A plausible arrangement is 254 - 196 = 58.

## Step 22: Problem 4 - Provide the answer
The smallest difference is achieved with an arrangement like 254 - 196.

ANSWER 4: A

## Step 23: Problem 5 - Restate the problem
Given W, X, Y, Z are different integers from 1 to 4, and WX - YZ = 1, we need to find W + Y.

## Step 24: Problem 5 - Understand the equation WX - YZ = 1
Since W, X, Y, Z are different integers from 1 to 4, we need to consider the possible values of WX and YZ.

## Step 25: Problem 5 - Analyze possible values for WX and YZ
The possible products are: 1*2=2, 1*3=3, 1*4=4, 2*1=2, 2*3=6, 2*4=8, 3*1=3, 3*2=6, 3*4=12, 4*1=4, 4*2=8, 4*3=12. We need a difference of 1 between two of these products.

## Step 26: Problem 5 - Check for a difference of 1
The products are 2, 3, 4, 6, 8, 12. The pairs that differ by 1 are (2,3), (3,4). So, either WX = 3 and YZ = 2 or WX = 4 and YZ = 3.

## Step 27: Problem 5 - Solve for W + Y
For WX = 3 and YZ = 2, W = 3, X = 1, Y = 2, Z = 1 (not valid because Z = X), or W = 1, X = 3, Y = 1, Z = 2 (not valid because W = Y). A valid case is W = 3, X = 1, Y = 1, Z = 2 (not valid), but if we consider W = 1, X = 3, Y = 2, Z = 1, it's valid and gives W + Y = 1 + 2 = 3.

## Step 28: Problem 5 - Provide the answer
For the valid case where W = 1, X = 3, Y = 2, Z = 1 doesn't work due to repetition, but

Problem 1:

We need count the whole numbers from 1 through 46 that are divisible by 3 or 5 or both.

Count multiples of 3:

\[
\left\lfloor \frac{46}{3} \right\rfloor = 15
\]

Count multiples of 5:

\[
\left\lfloor \frac{46}{5} \right\rfloor = 9
\]

But multiples of both 3 and 5, meaning multiples of 15, were counted twice, so subtract them once:

\[
\left\lfloor \frac{46}{15} \right\rfloor = 3
\]

Therefore the total is

\[
15+9-3=21
\]

So choices like 24 or 27 overcount because they do not remove the overlap.

ANSWER 1: B

Problem 2:

Initially Jamal has:

\[
6+18+12=36
\]

socks total, of which 18 are purple.

Suppose he adds \(x\) purple socks. Then the number of purple socks is

\[
18+x
\]

and the total number of socks is

\[
36+x
\]

The probability of choosing a purple sock is now 60%, or \(\frac35\). So

\[
\frac{18+x}{36+x}=\frac35
\]

Cross multiply:

\[
5(18+x)=3(36+x)
\]

\[
90+5x=108+3x
\]

\[
2x=18
\]

\[
x=9
\]

So Jamal added 9 purple socks.

ANSWER 2: B

Problem 3:

A lucky year has some valid date month/day/year where

\[
\text{month} \times \text{day} = \text{last two digits of the year}
\]

Check each year:

For 1990:

\[
9 \times 10 = 90
\]

So 9/10/90 works.

For 1991:

\[
7 \times 13 = 91
\]

So 7/13/91 works.

For 1992:

\[
4 \times 23 = 92
\]

So 4/23/92 works.

For 1993:

\[
3 \times 31 = 93
\]

So 3/31/93 works.

For 1994:

\[
94=2 \times 47
\]

But day 47 is not valid. The only factor pairs are \(1 \times 94\) and \(2 \times 47\), and neither gives a valid date.

Thus 1994 is not lucky.

ANSWER 3: E

Problem 4:

The subtraction problem uses the digits 2, 4, 5, 6, 9 exactly once in a 3-digit number minus a 2-digit number.

To make the difference as small as possible, the 3-digit number should be as small as possible, and the 2-digit number should be as large as possible.

The hundreds digit of the 3-digit number should be the smallest digit, 2.

Now we want the rest of the top number small and the bottom number large. Using the remaining digits 4, 5, 6, 9, the smallest possible two digits for the top number are 4 and 5, making 245, while the largest possible 2-digit number from the remaining digits is 96.

So try:

\[
245-96=149
\]

If we tried to get below 149, we would need either a smaller 3-digit number than 245 or a larger 2-digit number than 96, but neither is possible with these digits.

Therefore the smallest difference is 149.

ANSWER 4: C

Problem 5:

The equation is

\[
\frac{W}{X}-\frac{Y}{Z}=1
\]

where \(W, X, Y, Z\) are the numbers 1, 2, 3, 4 in some order.

We need find an arrangement that works.

Try to make the first fraction bigger than 1. One successful choice is

\[
\frac{3}{1}-\frac{4}{2}=1
\]

because

\[
3-2=1
\]

So

\[
W=3,\quad X=1,\quad Y=4,\quad Z=2
\]

Then

\[
W+Y=3+4=7
\]

The other choices are smaller sums, but the only working arrangement gives 7.

ANSWER 5: E

Problem 6:

There are four-digit integers from 1000 through 9999. The total number is

\[
9999-1000+1=9000
\]

We want odd integers whose digits are all distinct.

For a number to be odd, the units digit must be one of

\[
1,3,5,7,9
\]

So there are 5 choices for the units digit.

The thousands digit cannot be 0 and cannot equal the units digit. Since there are 9 possible nonzero thousands digits, and one is already used, there are

\[
8
\]

choices.

The hundreds digit can be any digit except the two already used:

\[
8
\]

choices.

The tens digit can be any digit except the three already used:

\[
7
\]

choices.

So the number of favorable integers is

\[
5 \cdot 8 \cdot 8 \cdot 7=2240
\]

The probability is

\[
\frac{2240}{9000}
\]

Reduce:

\[
\frac{2240}{9000}=\frac{224}{900}=\frac{56}{225}
\]

ANSWER 6: B

Problem 7:

Let \(n\) be the number of employees, and let \(F\) be the amount of money in the fund.

The manager planned to give each employee $50, but the fund was $5 short. So

\[
F=50n-5
\]

Instead, the manager gave each employee $45 and had $95 left. So

\[
F=45n+95
\]

Set these equal:

\[
50n-5=45n+95
\]

\[
5n=100
\]

\[
n=20
\]

Now find the fund:

\[
F=50(20)-5=1000-5=995
\]

So there were $995 in the fund.

ANSWER 7: E

Problem 8:

We are given

\[
ABA \times CD = CDCD
\]

The number \(ABA\) is

\[
101A+10B
\]

The number \(CDCD\) is a repeated two-digit number, so

\[
CDCD=101 \times CD
\]

Thus the equation becomes

\[
ABA \times CD = 101 \times CD
\]

Since \(CD\) is a two-digit number, it is not 0, so we can divide both sides by \(CD\):

\[
ABA=101
\]

Therefore

\[
A=1,\quad B=0
\]

So

\[
A+B=1+0=1
\]

ANSWER 8: A

Problem 9:

We need choose 3 numbers from

\[
\{89,95,99,132,166,173\}
\]

so that their sum is even.

Classify by parity.

Odd numbers:

\[
89,95,99,173
\]

There are 4 odds.

Even numbers:

\[
132,166
\]

There are 2 evens.

A sum of three numbers is even if it contains either:

- 0 odd numbers and 3 even numbers, or
- 2 odd numbers and 1 even number.

We cannot choose 3 even numbers because there are only 2 even numbers.

So we must choose 2 odds and 1 even.

Number of ways:

\[
\binom{4}{2}\binom{2}{1}=6 \cdot 2=12
\]

ANSWER 9: D

Problem 10:

The set has 5 positive integers with mean 5, so the total sum is

\[
5 \cdot 5=25
\]

The median is 5, so when the numbers are ordered, the middle number is 5.

Also, 8 is the only mode, meaning 8 appears more often than any other number. Since there are only five numbers and the median is 5, the two largest numbers must be 8 and 8:

\[
\_,\_,5,8,8
\]

Their sum so far is

\[
5+8+8=21
\]

So the first two numbers must sum to

\[
25-21=4
\]

They are positive integers less than or equal to 5. The possible pairs are:

\[
1,3 \quad \text{or} \quad 2,2
\]

But \(2,2\) would make 2 appear twice, tying 8 as a mode. Since 8 must be the only mode, the pair must be

\[
1,3
\]

So the set is

\[
1,3,5,8,8
\]

The difference between the largest and smallest integers is

\[
8-1=7
\]

ANSWER 10: D

Problem 11:

We need count three-digit whole numbers between 99 and 999 that contain exactly one 0.

A t

**Problem 1:**
How many whole numbers from 1 through 46 are divisible by either 3 or 5 or both?

**Step-by-step solution:**
To find the number of integers divisible by 3, 5, or both, we can use the Principle of Inclusion-Exclusion. 
1. First, we count the numbers from 1 to 46 that are divisible by 3:
   $\lfloor 46 \div 3 \rfloor = 15$
2. Next, we count the numbers from 1 to 46 that are divisible by 5:
   $\lfloor 46 \div 5 \rfloor = 9$
3. Then, we count the numbers that are divisible by BOTH 3 and 5 (which means they are divisible by their least common multiple, 15):
   $\lfloor 46 \div 15 \rfloor = 3$
4. Finally, we add the multiples of 3 and 5, and subtract the multiples of 15 so they aren't counted twice:
   $15 + 9 - 3 = 21$

ANSWER 1: B

---

**Problem 2:**
Jamal has a drawer containing 6 green socks, 18 purple socks, and 12 orange socks. After adding more purple socks, Jamal noticed that there is now a 60% chance that a sock randomly selected from the drawer is purple. How many purple socks did Jamal add?

**Step-by-step solution:**
1. Initially, Jamal has $6 + 18 + 12 = 36$ socks in total.
2. Let $x$ be the number of purple socks Jamal added. 
3. The new number of purple socks is $18 + x$.
4. The new total number of socks is $36 + x$.
5. We are told the new probability of selecting a purple sock is 60%, which simplifies to $\frac{60}{100} = \frac{3}{5}$.
6. We set up the equation:
   $\frac{18 + x}{36 + x} = \frac{3}{5}$
7. Cross-multiply to solve for $x$:
   $5(18 + x) = 3(36 + x)$
   $90 + 5x = 108 + 3x$
   $2x = 18$
   $x = 9$

Jamal added 9 purple socks.

ANSWER 2: B

---

**Problem 3:**
A lucky year is one in which at least one date, written as month/day/year, has the property that the month times the day equals the last two digits of the year. For example, 1956 is lucky because 7/8/56 has 7 × 8 = 56. Which of the following is NOT a lucky year?

**Step-by-step solution:**
We need to find if the last two digits of each given year can be factored into a valid month (1 through 12) and a valid day (1 through 31).
- **A. 1990:** $90 = 9 \times 10$. This corresponds to September 10th (9/10/90). Lucky.
- **B. 1991:** $91 = 7 \times 13$. This corresponds to July 13th (7/13/91). Lucky.
- **C. 1992:** $92 = 4 \times 23$. This corresponds to April 23rd (4/23/92). Lucky.
- **D. 1993:** $93 = 3 \times 31$. This corresponds to March 31st (3/31/93). Lucky.
- **E. 1994:** $94 = 2 \times 47$. Since a month cannot have 47 days, and 94 has no other integer pairs that fit the calendar constraints (1×94 is also invalid), 1994 is NOT a lucky year.

ANSWER 3: E

---

**Problem 4:**
When placing each of the digits 2, 4, 5, 6, 9 in exactly one of the boxes of this subtraction problem, what is the smallest difference that is possible?       −

**Step-by-step solution:**
We are given 5 digits to place in a subtraction problem. To yield a positive difference using 5 digits, the setup must be a 3-digit number minus a 2-digit number.
To get the smallest possible difference, we need the 3-digit number to be as small as possible and the 2-digit number to be as large as possible.
1. The smallest possible hundreds digit for the 3-digit number is 2. So the number is $2\_ \_$.
2. We are left with the digits 4, 5, 6, and 9. 
3. To minimize the 3-digit number, the remaining two digits should be the smallest available: 4 and 5, making the number 245.
4. To maximize the 2-digit number being subtracted, we use the largest remaining digits: 9 and 6, making the number 96.
5. Subtracting the two gives: $245 - 96 = 149$.
(Testing other close combinations like $254 - 96 = 158$ or $246 - 95 = 151$ confirms 149 is the absolute minimum difference.)

ANSWER 4: C

---

**Problem 5:**
Each of the letters W , X , Y , and Z represents a different integer in the set {1, 2, 3, 4}, but not necessarily in that order. If W X − Y Z = 1, then the sum of W and Y is

**Step-by-step solution:**
The equation given is meant to be read as fractions (a common formatting artifact drops the fraction bar): $\frac{W}{X} - \frac{Y}{Z} = 1$.
Let's test the values {1, 2, 3, 4} to satisfy this equation:
- If we set $W = 3$, $X = 1$, $Y = 4$, and $Z = 2$:
  $\frac{3}{1} - \frac{4}{2} = 3 - 2 = 1$
This perfectly uses all four distinct digits exactly once.
The question asks for the sum of $W$ and $Y$:
$W + Y = 3 + 4 = 7$

*(Note: Checking $W \times X - Y \times Z = 1$ yields no solutions where all 4 digits are distinct, confirming the fractional interpretation).*

ANSWER 5: E

---

**Problem 6:**
An integer between 1000 and 9999, inclusive, is chosen at random. What is the probability that it is an odd integer whose digits are all distinct?

**Step-by-step solution:**
1. The total number of integers from 1000 to 9999 inclusive is $9999 - 1000 + 1 = 9000$.
2. We want to form a 4-digit odd number (let's call its digits A B C D) with all distinct digits.
3. Because it must be odd, the units digit (D) must be 1, 3, 5, 7, or 9. There are **5** choices for D.
4. The thousands digit (A) cannot be 0, and it cannot be the digit used for D. Thus, out of 10 possible digits, 2 are restricted, leaving **8** choices for A.
5. The hundreds digit (B) can be 0, but cannot be the digits used for A or D. This leaves **8** choices for B.
6. The tens digit (C) cannot be the digits used for A, B, or D. This leaves **7** choices for C.
7. The total number of such integers is $5 \times 8 \times 8 \times 7 = 2240$.
8. The probability is $\frac{2240}{9000} = \frac{224}{900} = \frac{56}{225}$.

ANSWER 6: B

---

**Problem 7:**
The manager of a company planned to give a $50 bonus to each employee from the company fund, but the fund was $5 short of what was needed. Instead the manager gave each employee a $45 bonus and kept the remaining $95 in the fund. How much money was in the company fund before any bonuses were paid?

**Step-by-step solution:**
Let $n$ be the number of employees.
1. If the manager gave $50 to each employee, the fund would be $5 short. So, the total fund is $5

## Step 1: Problem 1 - Restate the problem
We need to find how many whole numbers from 1 through 46 are divisible by either 3 or 5 or both.

## Step 2: Problem 1 - Calculate numbers divisible by 3
First, let's find the numbers divisible by 3. The largest number less than or equal to 46 that is divisible by 3 is 45, and the number of multiples of 3 from 1 to 46 is 45 / 3 = 15.

## Step 3: Problem 1 - Calculate numbers divisible by 5
Next, find the numbers divisible by 5. The largest number less than or equal to 46 that is divisible by 5 is 45, and the number of multiples of 5 from 1 to 46 is 45 / 5 = 9.

## Step 4: Problem 1 - Calculate numbers divisible by both 3 and 5
Numbers divisible by both 3 and 5 are divisible by 15. The largest number less than or equal to 46 that is divisible by 15 is 45, and the number of multiples of 15 from 1 to 46 is 45 / 15 = 3.

## Step 5: Problem 1 - Apply the principle of inclusion-exclusion
To find the total number of numbers divisible by either 3 or 5 or both, we use the principle of inclusion-exclusion: (numbers divisible by 3) + (numbers divisible by 5) - (numbers divisible by both) = 15 + 9 - 3 = 21.

## Step 6: Problem 1 - Provide the answer
ANSWER 1: B

## Step 7: Problem 2 - Restate the problem
Jamal has 6 green socks, 18 purple socks, and 12 orange socks. After adding more purple socks, the probability of picking a purple sock is 60%. We need to find how many purple socks Jamal added.

## Step 8: Problem 2 - Set up the initial conditions
Initially, Jamal has 6 + 18 + 12 = 36 socks.

## Step 9: Problem 2 - Set up the equation for the probability of picking a purple sock after adding x purple socks
After adding x purple socks, the total number of socks is 36 + x, and the number of purple socks is 18 + x. The probability of picking a purple sock is (18 + x) / (36 + x) = 0.6.

## Step 10: Problem 2 - Solve the equation for x
(18 + x) / (36 + x) = 0.6. Multiplying both sides by (36 + x) gives 18 + x = 0.6(36 + x). Expanding, we get 18 + x = 21.6 + 0.6x. Rearranging, x - 0.6x = 21.6 - 18, so 0.4x = 3.6. Dividing both sides by 0.4, x = 9.

## Step 11: Problem 2 - Provide the answer
ANSWER 2: B

## Step 12: Problem 3 - Restate the problem
We need to find which year is NOT a lucky year, where a lucky year has at least one date where the month times the day equals the last two digits of the year.

## Step 13: Problem 3 - Analyze the condition for a lucky year
For a year to be lucky, there must exist a month (1-12) and a day (1-31) such that month * day = last two digits of the year.

## Step 14: Problem 3 - Check each given year
Checking each year from 1990 to 1994 to see if any date satisfies the condition.

## Step 15: Problem 3 - Analyze 1990 to 1994
- For 1990, the last two digits are 90. It's difficult to get 90 by multiplying two numbers between 1 and 12 and 1 and 31, respectively, because the factors of 90 that could work are (1,90), (2,45), (3,30), (5,18), (6,15), (9,10). Only (9,10) is plausible, and it works as 9/10/90.
- For 1991, 91 can be achieved with 7/13/91, so it's lucky.
- For 1992, 92 can be achieved, for example, with 4/23/92, so it's lucky.
- For 1993, 93 can be achieved with 3/31/93, so it's lucky.
- For 1994, 94 is not easily achieved because its factors are (1,94), (2,47). No month-day combination gives 94.

## Step 16: Problem 3 - Provide the answer
Given the analysis, 1994 is not lucky because there's no month-day product equal to 94.

ANSWER 3: E

## Step 17: Problem 4 - Restate the problem
We need to find the smallest difference possible by placing the digits 2, 4, 5, 6, 9 in the boxes of a subtraction problem.

## Step 18: Problem 4 - Understand the structure of the subtraction problem
To minimize the difference, we should minimize the minuend and maximize the subtrahend.

## Step 19: Problem 4 - Arrange the digits to minimize the difference
To achieve the smallest difference, the hundreds place of the minuend should be as small as possible, and the hundreds place of the subtrahend should be as large as possible. So, the minuend could be 2** and the subtrahend 9**. The remaining digits are 4, 5, 6.

## Step 20: Problem 4 - Try different combinations
Let's try to make the minuend as small as possible and the subtrahend as large as possible: 245 - 986 is not valid because it results in a negative number. A valid attempt is 246 - 985 or similar combinations.

## Step 21: Problem 4 - Calculate the difference for a valid combination
For 964 - 852 or similar, the difference is not minimal. Let's directly check: 245 - 196 = 49 (not an option), but trying to minimize, we can get 246 - 195 = 51 (still not an option). A plausible arrangement is 254 - 196 = 58.

## Step 22: Problem 4 - Provide the answer
The smallest difference is achieved with an arrangement like 254 - 196.

ANSWER 4: A

## Step 23: Problem 5 - Restate the problem
Given W, X, Y, Z are different integers from 1 to 4, and WX - YZ = 1, we need to find W + Y.

## Step 24: Problem 5 - Understand the equation WX - YZ = 1
Since W, X, Y, Z are different integers from 1 to 4, we need to consider the possible values of WX and YZ.

## Step 25: Problem 5 - Analyze possible values for WX and YZ
The possible products are: 1*2=2, 1*3=3, 1*4=4, 2*1=2, 2*3=6, 2*4=8, 3*1=3, 3*2=6, 3*4=12, 4*1=4, 4*2=8, 4*3=12. We need a difference of 1 between two of these products.

## Step 26: Problem 5 - Check for a difference of 1
The products are 2, 3, 4, 6, 8, 12. The pairs that differ by 1 are (2,3), (3,4). So, either WX = 3 and YZ = 2 or WX = 4 and YZ = 3.

## Step 27: Problem 5 - Solve for W + Y
For WX = 3 and YZ = 2, W = 3, X = 1, Y = 2, Z = 1 (not valid because Z = X), or W = 1, X = 3, Y = 1, Z = 2 (not valid because W = Y). A valid case is W = 3, X = 1, Y = 1, Z = 2 (not valid), but if we consider W = 1, X = 3, Y = 2, Z = 1, it's valid and gives W + Y = 1 + 2 = 3.

## Step 28: Problem 5 - Provide the answer
For the valid case where W = 1, X = 3, Y = 2, Z = 1 doesn't work due to repetition, but

Problem 1:

We need count the whole numbers from 1 through 46 that are divisible by 3 or 5 or both.

Count multiples of 3:

\[
\left\lfloor \frac{46}{3} \right\rfloor = 15
\]

Count multiples of 5:

\[
\left\lfloor \frac{46}{5} \right\rfloor = 9
\]

But multiples of both 3 and 5, meaning multiples of 15, were counted twice, so subtract them once:

\[
\left\lfloor \frac{46}{15} \right\rfloor = 3
\]

Therefore the total is

\[
15+9-3=21
\]

So choices like 24 or 27 overcount because they do not remove the overlap.

ANSWER 1: B

Problem 2:

Initially Jamal has:

\[
6+18+12=36
\]

socks total, of which 18 are purple.

Suppose he adds \(x\) purple socks. Then the number of purple socks is

\[
18+x
\]

and the total number of socks is

\[
36+x
\]

The probability of choosing a purple sock is now 60%, or \(\frac35\). So

\[
\frac{18+x}{36+x}=\frac35
\]

Cross multiply:

\[
5(18+x)=3(36+x)
\]

\[
90+5x=108+3x
\]

\[
2x=18
\]

\[
x=9
\]

So Jamal added 9 purple socks.

ANSWER 2: B

Problem 3:

A lucky year has some valid date month/day/year where

\[
\text{month} \times \text{day} = \text{last two digits of the year}
\]

Check each year:

For 1990:

\[
9 \times 10 = 90
\]

So 9/10/90 works.

For 1991:

\[
7 \times 13 = 91
\]

So 7/13/91 works.

For 1992:

\[
4 \times 23 = 92
\]

So 4/23/92 works.

For 1993:

\[
3 \times 31 = 93
\]

So 3/31/93 works.

For 1994:

\[
94=2 \times 47
\]

But day 47 is not valid. The only factor pairs are \(1 \times 94\) and \(2 \times 47\), and neither gives a valid date.

Thus 1994 is not lucky.

ANSWER 3: E

Problem 4:

The subtraction problem uses the digits 2, 4, 5, 6, 9 exactly once in a 3-digit number minus a 2-digit number.

To make the difference as small as possible, the 3-digit number should be as small as possible, and the 2-digit number should be as large as possible.

The hundreds digit of the 3-digit number should be the smallest digit, 2.

Now we want the rest of the top number small and the bottom number large. Using the remaining digits 4, 5, 6, 9, the smallest possible two digits for the top number are 4 and 5, making 245, while the largest possible 2-digit number from the remaining digits is 96.

So try:

\[
245-96=149
\]

If we tried to get below 149, we would need either a smaller 3-digit number than 245 or a larger 2-digit number than 96, but neither is possible with these digits.

Therefore the smallest difference is 149.

ANSWER 4: C

Problem 5:

The equation is

\[
\frac{W}{X}-\frac{Y}{Z}=1
\]

where \(W, X, Y, Z\) are the numbers 1, 2, 3, 4 in some order.

We need find an arrangement that works.

Try to make the first fraction bigger than 1. One successful choice is

\[
\frac{3}{1}-\frac{4}{2}=1
\]

because

\[
3-2=1
\]

So

\[
W=3,\quad X=1,\quad Y=4,\quad Z=2
\]

Then

\[
W+Y=3+4=7
\]

The other choices are smaller sums, but the only working arrangement gives 7.

ANSWER 5: E

Problem 6:

There are four-digit integers from 1000 through 9999. The total number is

\[
9999-1000+1=9000
\]

We want odd integers whose digits are all distinct.

For a number to be odd, the units digit must be one of

\[
1,3,5,7,9
\]

So there are 5 choices for the units digit.

The thousands digit cannot be 0 and cannot equal the units digit. Since there are 9 possible nonzero thousands digits, and one is already used, there are

\[
8
\]

choices.

The hundreds digit can be any digit except the two already used:

\[
8
\]

choices.

The tens digit can be any digit except the three already used:

\[
7
\]

choices.

So the number of favorable integers is

\[
5 \cdot 8 \cdot 8 \cdot 7=2240
\]

The probability is

\[
\frac{2240}{9000}
\]

Reduce:

\[
\frac{2240}{9000}=\frac{224}{900}=\frac{56}{225}
\]

ANSWER 6: B

Problem 7:

Let \(n\) be the number of employees, and let \(F\) be the amount of money in the fund.

The manager planned to give each employee $50, but the fund was $5 short. So

\[
F=50n-5
\]

Instead, the manager gave each employee $45 and had $95 left. So

\[
F=45n+95
\]

Set these equal:

\[
50n-5=45n+95
\]

\[
5n=100
\]

\[
n=20
\]

Now find the fund:

\[
F=50(20)-5=1000-5=995
\]

So there were $995 in the fund.

ANSWER 7: E

Problem 8:

We are given

\[
ABA \times CD = CDCD
\]

The number \(ABA\) is

\[
101A+10B
\]

The number \(CDCD\) is a repeated two-digit number, so

\[
CDCD=101 \times CD
\]

Thus the equation becomes

\[
ABA \times CD = 101 \times CD
\]

Since \(CD\) is a two-digit number, it is not 0, so we can divide both sides by \(CD\):

\[
ABA=101
\]

Therefore

\[
A=1,\quad B=0
\]

So

\[
A+B=1+0=1
\]

ANSWER 8: A

Problem 9:

We need choose 3 numbers from

\[
\{89,95,99,132,166,173\}
\]

so that their sum is even.

Classify by parity.

Odd numbers:

\[
89,95,99,173
\]

There are 4 odds.

Even numbers:

\[
132,166
\]

There are 2 evens.

A sum of three numbers is even if it contains either:

- 0 odd numbers and 3 even numbers, or
- 2 odd numbers and 1 even number.

We cannot choose 3 even numbers because there are only 2 even numbers.

So we must choose 2 odds and 1 even.

Number of ways:

\[
\binom{4}{2}\binom{2}{1}=6 \cdot 2=12
\]

ANSWER 9: D

Problem 10:

The set has 5 positive integers with mean 5, so the total sum is

\[
5 \cdot 5=25
\]

The median is 5, so when the numbers are ordered, the middle number is 5.

Also, 8 is the only mode, meaning 8 appears more often than any other number. Since there are only five numbers and the median is 5, the two largest numbers must be 8 and 8:

\[
\_,\_,5,8,8
\]

Their sum so far is

\[
5+8+8=21
\]

So the first two numbers must sum to

\[
25-21=4
\]

They are positive integers less than or equal to 5. The possible pairs are:

\[
1,3 \quad \text{or} \quad 2,2
\]

But \(2,2\) would make 2 appear twice, tying 8 as a mode. Since 8 must be the only mode, the pair must be

\[
1,3
\]

So the set is

\[
1,3,5,8,8
\]

The difference between the largest and smallest integers is

\[
8-1=7
\]

ANSWER 10: D

Problem 11:

We need count three-digit whole numbers between 99 and 999 that contain exactly one 0.

A t

**Problem 1:**
How many whole numbers from 1 through 46 are divisible by either 3 or 5 or both?

**Step-by-step solution:**
To find the number of integers divisible by 3, 5, or both, we can use the Principle of Inclusion-Exclusion. 
1. First, we count the numbers from 1 to 46 that are divisible by 3:
   $\lfloor 46 \div 3 \rfloor = 15$
2. Next, we count the numbers from 1 to 46 that are divisible by 5:
   $\lfloor 46 \div 5 \rfloor = 9$
3. Then, we count the numbers that are divisible by BOTH 3 and 5 (which means they are divisible by their least common multiple, 15):
   $\lfloor 46 \div 15 \rfloor = 3$
4. Finally, we add the multiples of 3 and 5, and subtract the multiples of 15 so they aren't counted twice:
   $15 + 9 - 3 = 21$

ANSWER 1: B

---

**Problem 2:**
Jamal has a drawer containing 6 green socks, 18 purple socks, and 12 orange socks. After adding more purple socks, Jamal noticed that there is now a 60% chance that a sock randomly selected from the drawer is purple. How many purple socks did Jamal add?

**Step-by-step solution:**
1. Initially, Jamal has $6 + 18 + 12 = 36$ socks in total.
2. Let $x$ be the number of purple socks Jamal added. 
3. The new number of purple socks is $18 + x$.
4. The new total number of socks is $36 + x$.
5. We are told the new probability of selecting a purple sock is 60%, which simplifies to $\frac{60}{100} = \frac{3}{5}$.
6. We set up the equation:
   $\frac{18 + x}{36 + x} = \frac{3}{5}$
7. Cross-multiply to solve for $x$:
   $5(18 + x) = 3(36 + x)$
   $90 + 5x = 108 + 3x$
   $2x = 18$
   $x = 9$

Jamal added 9 purple socks.

ANSWER 2: B

---

**Problem 3:**
A lucky year is one in which at least one date, written as month/day/year, has the property that the month times the day equals the last two digits of the year. For example, 1956 is lucky because 7/8/56 has 7 × 8 = 56. Which of the following is NOT a lucky year?

**Step-by-step solution:**
We need to find if the last two digits of each given year can be factored into a valid month (1 through 12) and a valid day (1 through 31).
- **A. 1990:** $90 = 9 \times 10$. This corresponds to September 10th (9/10/90). Lucky.
- **B. 1991:** $91 = 7 \times 13$. This corresponds to July 13th (7/13/91). Lucky.
- **C. 1992:** $92 = 4 \times 23$. This corresponds to April 23rd (4/23/92). Lucky.
- **D. 1993:** $93 = 3 \times 31$. This corresponds to March 31st (3/31/93). Lucky.
- **E. 1994:** $94 = 2 \times 47$. Since a month cannot have 47 days, and 94 has no other integer pairs that fit the calendar constraints (1×94 is also invalid), 1994 is NOT a lucky year.

ANSWER 3: E

---

**Problem 4:**
When placing each of the digits 2, 4, 5, 6, 9 in exactly one of the boxes of this subtraction problem, what is the smallest difference that is possible?       −

**Step-by-step solution:**
We are given 5 digits to place in a subtraction problem. To yield a positive difference using 5 digits, the setup must be a 3-digit number minus a 2-digit number.
To get the smallest possible difference, we need the 3-digit number to be as small as possible and the 2-digit number to be as large as possible.
1. The smallest possible hundreds digit for the 3-digit number is 2. So the number is $2\_ \_$.
2. We are left with the digits 4, 5, 6, and 9. 
3. To minimize the 3-digit number, the remaining two digits should be the smallest available: 4 and 5, making the number 245.
4. To maximize the 2-digit number being subtracted, we use the largest remaining digits: 9 and 6, making the number 96.
5. Subtracting the two gives: $245 - 96 = 149$.
(Testing other close combinations like $254 - 96 = 158$ or $246 - 95 = 151$ confirms 149 is the absolute minimum difference.)

ANSWER 4: C

---

**Problem 5:**
Each of the letters W , X , Y , and Z represents a different integer in the set {1, 2, 3, 4}, but not necessarily in that order. If W X − Y Z = 1, then the sum of W and Y is

**Step-by-step solution:**
The equation given is meant to be read as fractions (a common formatting artifact drops the fraction bar): $\frac{W}{X} - \frac{Y}{Z} = 1$.
Let's test the values {1, 2, 3, 4} to satisfy this equation:
- If we set $W = 3$, $X = 1$, $Y = 4$, and $Z = 2$:
  $\frac{3}{1} - \frac{4}{2} = 3 - 2 = 1$
This perfectly uses all four distinct digits exactly once.
The question asks for the sum of $W$ and $Y$:
$W + Y = 3 + 4 = 7$

*(Note: Checking $W \times X - Y \times Z = 1$ yields no solutions where all 4 digits are distinct, confirming the fractional interpretation).*

ANSWER 5: E

---

**Problem 6:**
An integer between 1000 and 9999, inclusive, is chosen at random. What is the probability that it is an odd integer whose digits are all distinct?

**Step-by-step solution:**
1. The total number of integers from 1000 to 9999 inclusive is $9999 - 1000 + 1 = 9000$.
2. We want to form a 4-digit odd number (let's call its digits A B C D) with all distinct digits.
3. Because it must be odd, the units digit (D) must be 1, 3, 5, 7, or 9. There are **5** choices for D.
4. The thousands digit (A) cannot be 0, and it cannot be the digit used for D. Thus, out of 10 possible digits, 2 are restricted, leaving **8** choices for A.
5. The hundreds digit (B) can be 0, but cannot be the digits used for A or D. This leaves **8** choices for B.
6. The tens digit (C) cannot be the digits used for A, B, or D. This leaves **7** choices for C.
7. The total number of such integers is $5 \times 8 \times 8 \times 7 = 2240$.
8. The probability is $\frac{2240}{9000} = \frac{224}{900} = \frac{56}{225}$.

ANSWER 6: B

---

**Problem 7:**
The manager of a company planned to give a $50 bonus to each employee from the company fund, but the fund was $5 short of what was needed. Instead the manager gave each employee a $45 bonus and kept the remaining $95 in the fund. How much money was in the company fund before any bonuses were paid?

**Step-by-step solution:**
Let $n$ be the number of employees.
1. If the manager gave $50 to each employee, the fund would be $5 short. So, the total fund is $5

## Step 1: Problem 1 - Restate the problem
We need to find how many whole numbers from 1 through 46 are divisible by either 3 or 5 or both.

## Step 2: Problem 1 - Calculate numbers divisible by 3
First, let's find the numbers divisible by 3. The largest number less than or equal to 46 that is divisible by 3 is 45, and the number of multiples of 3 from 1 to 46 is 45 / 3 = 15.

## Step 3: Problem 1 - Calculate numbers divisible by 5
Next, find the numbers divisible by 5. The largest number less than or equal to 46 that is divisible by 5 is 45, and the number of multiples of 5 from 1 to 46 is 45 / 5 = 9.

## Step 4: Problem 1 - Calculate numbers divisible by both 3 and 5
Numbers divisible by both 3 and 5 are divisible by 15. The largest number less than or equal to 46 that is divisible by 15 is 45, and the number of multiples of 15 from 1 to 46 is 45 / 15 = 3.

## Step 5: Problem 1 - Apply the principle of inclusion-exclusion
To find the total number of numbers divisible by either 3 or 5 or both, we use the principle of inclusion-exclusion: (numbers divisible by 3) + (numbers divisible by 5) - (numbers divisible by both) = 15 + 9 - 3 = 21.

## Step 6: Problem 1 - Provide the answer
ANSWER 1: B

## Step 7: Problem 2 - Restate the problem
Jamal has 6 green socks, 18 purple socks, and 12 orange socks. After adding more purple socks, the probability of picking a purple sock is 60%. We need to find how many purple socks Jamal added.

## Step 8: Problem 2 - Set up the initial conditions
Initially, Jamal has 6 + 18 + 12 = 36 socks.

## Step 9: Problem 2 - Set up the equation for the probability of picking a purple sock after adding x purple socks
After adding x purple socks, the total number of socks is 36 + x, and the number of purple socks is 18 + x. The probability of picking a purple sock is (18 + x) / (36 + x) = 0.6.

## Step 10: Problem 2 - Solve the equation for x
(18 + x) / (36 + x) = 0.6. Multiplying both sides by (36 + x) gives 18 + x = 0.6(36 + x). Expanding, we get 18 + x = 21.6 + 0.6x. Rearranging, x - 0.6x = 21.6 - 18, so 0.4x = 3.6. Dividing both sides by 0.4, x = 9.

## Step 11: Problem 2 - Provide the answer
ANSWER 2: B

## Step 12: Problem 3 - Restate the problem
We need to find which year is NOT a lucky year, where a lucky year has at least one date where the month times the day equals the last two digits of the year.

## Step 13: Problem 3 - Analyze the condition for a lucky year
For a year to be lucky, there must exist a month (1-12) and a day (1-31) such that month * day = last two digits of the year.

## Step 14: Problem 3 - Check each given year
Checking each year from 1990 to 1994 to see if any date satisfies the condition.

## Step 15: Problem 3 - Analyze 1990 to 1994
- For 1990, the last two digits are 90. It's difficult to get 90 by multiplying two numbers between 1 and 12 and 1 and 31, respectively, because the factors of 90 that could work are (1,90), (2,45), (3,30), (5,18), (6,15), (9,10). Only (9,10) is plausible, and it works as 9/10/90.
- For 1991, 91 can be achieved with 7/13/91, so it's lucky.
- For 1992, 92 can be achieved, for example, with 4/23/92, so it's lucky.
- For 1993, 93 can be achieved with 3/31/93, so it's lucky.
- For 1994, 94 is not easily achieved because its factors are (1,94), (2,47). No month-day combination gives 94.

## Step 16: Problem 3 - Provide the answer
Given the analysis, 1994 is not lucky because there's no month-day product equal to 94.

ANSWER 3: E

## Step 17: Problem 4 - Restate the problem
We need to find the smallest difference possible by placing the digits 2, 4, 5, 6, 9 in the boxes of a subtraction problem.

## Step 18: Problem 4 - Understand the structure of the subtraction problem
To minimize the difference, we should minimize the minuend and maximize the subtrahend.

## Step 19: Problem 4 - Arrange the digits to minimize the difference
To achieve the smallest difference, the hundreds place of the minuend should be as small as possible, and the hundreds place of the subtrahend should be as large as possible. So, the minuend could be 2** and the subtrahend 9**. The remaining digits are 4, 5, 6.

## Step 20: Problem 4 - Try different combinations
Let's try to make the minuend as small as possible and the subtrahend as large as possible: 245 - 986 is not valid because it results in a negative number. A valid attempt is 246 - 985 or similar combinations.

## Step 21: Problem 4 - Calculate the difference for a valid combination
For 964 - 852 or similar, the difference is not minimal. Let's directly check: 245 - 196 = 49 (not an option), but trying to minimize, we can get 246 - 195 = 51 (still not an option). A plausible arrangement is 254 - 196 = 58.

## Step 22: Problem 4 - Provide the answer
The smallest difference is achieved with an arrangement like 254 - 196.

ANSWER 4: A

## Step 23: Problem 5 - Restate the problem
Given W, X, Y, Z are different integers from 1 to 4, and WX - YZ = 1, we need to find W + Y.

## Step 24: Problem 5 - Understand the equation WX - YZ = 1
Since W, X, Y, Z are different integers from 1 to 4, we need to consider the possible values of WX and YZ.

## Step 25: Problem 5 - Analyze possible values for WX and YZ
The possible products are: 1*2=2, 1*3=3, 1*4=4, 2*1=2, 2*3=6, 2*4=8, 3*1=3, 3*2=6, 3*4=12, 4*1=4, 4*2=8, 4*3=12. We need a difference of 1 between two of these products.

## Step 26: Problem 5 - Check for a difference of 1
The products are 2, 3, 4, 6, 8, 12. The pairs that differ by 1 are (2,3), (3,4). So, either WX = 3 and YZ = 2 or WX = 4 and YZ = 3.

## Step 27: Problem 5 - Solve for W + Y
For WX = 3 and YZ = 2, W = 3, X = 1, Y = 2, Z = 1 (not valid because Z = X), or W = 1, X = 3, Y = 1, Z = 2 (not valid because W = Y). A valid case is W = 3, X = 1, Y = 1, Z = 2 (not valid), but if we consider W = 1, X = 3, Y = 2, Z = 1, it's valid and gives W + Y = 1 + 2 = 3.

## Step 28: Problem 5 - Provide the answer
For the valid case where W = 1, X = 3, Y = 2, Z = 1 doesn't work due to repetition, but

Problem 1:

We need count the whole numbers from 1 through 46 that are divisible by 3 or 5 or both.

Count multiples of 3:

\[
\left\lfloor \frac{46}{3} \right\rfloor = 15
\]

Count multiples of 5:

\[
\left\lfloor \frac{46}{5} \right\rfloor = 9
\]

But multiples of both 3 and 5, meaning multiples of 15, were counted twice, so subtract them once:

\[
\left\lfloor \frac{46}{15} \right\rfloor = 3
\]

Therefore the total is

\[
15+9-3=21
\]

So choices like 24 or 27 overcount because they do not remove the overlap.

ANSWER 1: B

Problem 2:

Initially Jamal has:

\[
6+18+12=36
\]

socks total, of which 18 are purple.

Suppose he adds \(x\) purple socks. Then the number of purple socks is

\[
18+x
\]

and the total number of socks is

\[
36+x
\]

The probability of choosing a purple sock is now 60%, or \(\frac35\). So

\[
\frac{18+x}{36+x}=\frac35
\]

Cross multiply:

\[
5(18+x)=3(36+x)
\]

\[
90+5x=108+3x
\]

\[
2x=18
\]

\[
x=9
\]

So Jamal added 9 purple socks.

ANSWER 2: B

Problem 3:

A lucky year has some valid date month/day/year where

\[
\text{month} \times \text{day} = \text{last two digits of the year}
\]

Check each year:

For 1990:

\[
9 \times 10 = 90
\]

So 9/10/90 works.

For 1991:

\[
7 \times 13 = 91
\]

So 7/13/91 works.

For 1992:

\[
4 \times 23 = 92
\]

So 4/23/92 works.

For 1993:

\[
3 \times 31 = 93
\]

So 3/31/93 works.

For 1994:

\[
94=2 \times 47
\]

But day 47 is not valid. The only factor pairs are \(1 \times 94\) and \(2 \times 47\), and neither gives a valid date.

Thus 1994 is not lucky.

ANSWER 3: E

Problem 4:

The subtraction problem uses the digits 2, 4, 5, 6, 9 exactly once in a 3-digit number minus a 2-digit number.

To make the difference as small as possible, the 3-digit number should be as small as possible, and the 2-digit number should be as large as possible.

The hundreds digit of the 3-digit number should be the smallest digit, 2.

Now we want the rest of the top number small and the bottom number large. Using the remaining digits 4, 5, 6, 9, the smallest possible two digits for the top number are 4 and 5, making 245, while the largest possible 2-digit number from the remaining digits is 96.

So try:

\[
245-96=149
\]

If we tried to get below 149, we would need either a smaller 3-digit number than 245 or a larger 2-digit number than 96, but neither is possible with these digits.

Therefore the smallest difference is 149.

ANSWER 4: C

Problem 5:

The equation is

\[
\frac{W}{X}-\frac{Y}{Z}=1
\]

where \(W, X, Y, Z\) are the numbers 1, 2, 3, 4 in some order.

We need find an arrangement that works.

Try to make the first fraction bigger than 1. One successful choice is

\[
\frac{3}{1}-\frac{4}{2}=1
\]

because

\[
3-2=1
\]

So

\[
W=3,\quad X=1,\quad Y=4,\quad Z=2
\]

Then

\[
W+Y=3+4=7
\]

The other choices are smaller sums, but the only working arrangement gives 7.

ANSWER 5: E

Problem 6:

There are four-digit integers from 1000 through 9999. The total number is

\[
9999-1000+1=9000
\]

We want odd integers whose digits are all distinct.

For a number to be odd, the units digit must be one of

\[
1,3,5,7,9
\]

So there are 5 choices for the units digit.

The thousands digit cannot be 0 and cannot equal the units digit. Since there are 9 possible nonzero thousands digits, and one is already used, there are

\[
8
\]

choices.

The hundreds digit can be any digit except the two already used:

\[
8
\]

choices.

The tens digit can be any digit except the three already used:

\[
7
\]

choices.

So the number of favorable integers is

\[
5 \cdot 8 \cdot 8 \cdot 7=2240
\]

The probability is

\[
\frac{2240}{9000}
\]

Reduce:

\[
\frac{2240}{9000}=\frac{224}{900}=\frac{56}{225}
\]

ANSWER 6: B

Problem 7:

Let \(n\) be the number of employees, and let \(F\) be the amount of money in the fund.

The manager planned to give each employee $50, but the fund was $5 short. So

\[
F=50n-5
\]

Instead, the manager gave each employee $45 and had $95 left. So

\[
F=45n+95
\]

Set these equal:

\[
50n-5=45n+95
\]

\[
5n=100
\]

\[
n=20
\]

Now find the fund:

\[
F=50(20)-5=1000-5=995
\]

So there were $995 in the fund.

ANSWER 7: E

Problem 8:

We are given

\[
ABA \times CD = CDCD
\]

The number \(ABA\) is

\[
101A+10B
\]

The number \(CDCD\) is a repeated two-digit number, so

\[
CDCD=101 \times CD
\]

Thus the equation becomes

\[
ABA \times CD = 101 \times CD
\]

Since \(CD\) is a two-digit number, it is not 0, so we can divide both sides by \(CD\):

\[
ABA=101
\]

Therefore

\[
A=1,\quad B=0
\]

So

\[
A+B=1+0=1
\]

ANSWER 8: A

Problem 9:

We need choose 3 numbers from

\[
\{89,95,99,132,166,173\}
\]

so that their sum is even.

Classify by parity.

Odd numbers:

\[
89,95,99,173
\]

There are 4 odds.

Even numbers:

\[
132,166
\]

There are 2 evens.

A sum of three numbers is even if it contains either:

- 0 odd numbers and 3 even numbers, or
- 2 odd numbers and 1 even number.

We cannot choose 3 even numbers because there are only 2 even numbers.

So we must choose 2 odds and 1 even.

Number of ways:

\[
\binom{4}{2}\binom{2}{1}=6 \cdot 2=12
\]

ANSWER 9: D

Problem 10:

The set has 5 positive integers with mean 5, so the total sum is

\[
5 \cdot 5=25
\]

The median is 5, so when the numbers are ordered, the middle number is 5.

Also, 8 is the only mode, meaning 8 appears more often than any other number. Since there are only five numbers and the median is 5, the two largest numbers must be 8 and 8:

\[
\_,\_,5,8,8
\]

Their sum so far is

\[
5+8+8=21
\]

So the first two numbers must sum to

\[
25-21=4
\]

They are positive integers less than or equal to 5. The possible pairs are:

\[
1,3 \quad \text{or} \quad 2,2
\]

But \(2,2\) would make 2 appear twice, tying 8 as a mode. Since 8 must be the only mode, the pair must be

\[
1,3
\]

So the set is

\[
1,3,5,8,8
\]

The difference between the largest and smallest integers is

\[
8-1=7
\]

ANSWER 10: D

Problem 11:

We need count three-digit whole numbers between 99 and 999 that contain exactly one 0.

A t

**Problem 1:**
How many whole numbers from 1 through 46 are divisible by either 3 or 5 or both?

**Step-by-step solution:**
To find the number of integers divisible by 3, 5, or both, we can use the Principle of Inclusion-Exclusion. 
1. First, we count the numbers from 1 to 46 that are divisible by 3:
   $\lfloor 46 \div 3 \rfloor = 15$
2. Next, we count the numbers from 1 to 46 that are divisible by 5:
   $\lfloor 46 \div 5 \rfloor = 9$
3. Then, we count the numbers that are divisible by BOTH 3 and 5 (which means they are divisible by their least common multiple, 15):
   $\lfloor 46 \div 15 \rfloor = 3$
4. Finally, we add the multiples of 3 and 5, and subtract the multiples of 15 so they aren't counted twice:
   $15 + 9 - 3 = 21$

ANSWER 1: B

---

**Problem 2:**
Jamal has a drawer containing 6 green socks, 18 purple socks, and 12 orange socks. After adding more purple socks, Jamal noticed that there is now a 60% chance that a sock randomly selected from the drawer is purple. How many purple socks did Jamal add?

**Step-by-step solution:**
1. Initially, Jamal has $6 + 18 + 12 = 36$ socks in total.
2. Let $x$ be the number of purple socks Jamal added. 
3. The new number of purple socks is $18 + x$.
4. The new total number of socks is $36 + x$.
5. We are told the new probability of selecting a purple sock is 60%, which simplifies to $\frac{60}{100} = \frac{3}{5}$.
6. We set up the equation:
   $\frac{18 + x}{36 + x} = \frac{3}{5}$
7. Cross-multiply to solve for $x$:
   $5(18 + x) = 3(36 + x)$
   $90 + 5x = 108 + 3x$
   $2x = 18$
   $x = 9$

Jamal added 9 purple socks.

ANSWER 2: B

---

**Problem 3:**
A lucky year is one in which at least one date, written as month/day/year, has the property that the month times the day equals the last two digits of the year. For example, 1956 is lucky because 7/8/56 has 7 × 8 = 56. Which of the following is NOT a lucky year?

**Step-by-step solution:**
We need to find if the last two digits of each given year can be factored into a valid month (1 through 12) and a valid day (1 through 31).
- **A. 1990:** $90 = 9 \times 10$. This corresponds to September 10th (9/10/90). Lucky.
- **B. 1991:** $91 = 7 \times 13$. This corresponds to July 13th (7/13/91). Lucky.
- **C. 1992:** $92 = 4 \times 23$. This corresponds to April 23rd (4/23/92). Lucky.
- **D. 1993:** $93 = 3 \times 31$. This corresponds to March 31st (3/31/93). Lucky.
- **E. 1994:** $94 = 2 \times 47$. Since a month cannot have 47 days, and 94 has no other integer pairs that fit the calendar constraints (1×94 is also invalid), 1994 is NOT a lucky year.

ANSWER 3: E

---

**Problem 4:**
When placing each of the digits 2, 4, 5, 6, 9 in exactly one of the boxes of this subtraction problem, what is the smallest difference that is possible?       −

**Step-by-step solution:**
We are given 5 digits to place in a subtraction problem. To yield a positive difference using 5 digits, the setup must be a 3-digit number minus a 2-digit number.
To get the smallest possible difference, we need the 3-digit number to be as small as possible and the 2-digit number to be as large as possible.
1. The smallest possible hundreds digit for the 3-digit number is 2. So the number is $2\_ \_$.
2. We are left with the digits 4, 5, 6, and 9. 
3. To minimize the 3-digit number, the remaining two digits should be the smallest available: 4 and 5, making the number 245.
4. To maximize the 2-digit number being subtracted, we use the largest remaining digits: 9 and 6, making the number 96.
5. Subtracting the two gives: $245 - 96 = 149$.
(Testing other close combinations like $254 - 96 = 158$ or $246 - 95 = 151$ confirms 149 is the absolute minimum difference.)

ANSWER 4: C

---

**Problem 5:**
Each of the letters W , X , Y , and Z represents a different integer in the set {1, 2, 3, 4}, but not necessarily in that order. If W X − Y Z = 1, then the sum of W and Y is

**Step-by-step solution:**
The equation given is meant to be read as fractions (a common formatting artifact drops the fraction bar): $\frac{W}{X} - \frac{Y}{Z} = 1$.
Let's test the values {1, 2, 3, 4} to satisfy this equation:
- If we set $W = 3$, $X = 1$, $Y = 4$, and $Z = 2$:
  $\frac{3}{1} - \frac{4}{2} = 3 - 2 = 1$
This perfectly uses all four distinct digits exactly once.
The question asks for the sum of $W$ and $Y$:
$W + Y = 3 + 4 = 7$

*(Note: Checking $W \times X - Y \times Z = 1$ yields no solutions where all 4 digits are distinct, confirming the fractional interpretation).*

ANSWER 5: E

---

**Problem 6:**
An integer between 1000 and 9999, inclusive, is chosen at random. What is the probability that it is an odd integer whose digits are all distinct?

**Step-by-step solution:**
1. The total number of integers from 1000 to 9999 inclusive is $9999 - 1000 + 1 = 9000$.
2. We want to form a 4-digit odd number (let's call its digits A B C D) with all distinct digits.
3. Because it must be odd, the units digit (D) must be 1, 3, 5, 7, or 9. There are **5** choices for D.
4. The thousands digit (A) cannot be 0, and it cannot be the digit used for D. Thus, out of 10 possible digits, 2 are restricted, leaving **8** choices for A.
5. The hundreds digit (B) can be 0, but cannot be the digits used for A or D. This leaves **8** choices for B.
6. The tens digit (C) cannot be the digits used for A, B, or D. This leaves **7** choices for C.
7. The total number of such integers is $5 \times 8 \times 8 \times 7 = 2240$.
8. The probability is $\frac{2240}{9000} = \frac{224}{900} = \frac{56}{225}$.

ANSWER 6: B

---

**Problem 7:**
The manager of a company planned to give a $50 bonus to each employee from the company fund, but the fund was $5 short of what was needed. Instead the manager gave each employee a $45 bonus and kept the remaining $95 in the fund. How much money was in the company fund before any bonuses were paid?

**Step-by-step solution:**
Let $n$ be the number of employees.
1. If the manager gave $50 to each employee, the fund would be $5 short. So, the total fund is $5

## Step 1: Problem 1 - Restate the problem
We need to find how many whole numbers from 1 through 46 are divisible by either 3 or 5 or both.

## Step 2: Problem 1 - Calculate numbers divisible by 3
First, let's find the numbers divisible by 3. The largest number less than or equal to 46 that is divisible by 3 is 45, and the number of multiples of 3 from 1 to 46 is 45 / 3 = 15.

## Step 3: Problem 1 - Calculate numbers divisible by 5
Next, find the numbers divisible by 5. The largest number less than or equal to 46 that is divisible by 5 is 45, and the number of multiples of 5 from 1 to 46 is 45 / 5 = 9.

## Step 4: Problem 1 - Calculate numbers divisible by both 3 and 5
Numbers divisible by both 3 and 5 are divisible by 15. The largest number less than or equal to 46 that is divisible by 15 is 45, and the number of multiples of 15 from 1 to 46 is 45 / 15 = 3.

## Step 5: Problem 1 - Apply the principle of inclusion-exclusion
To find the total number of numbers divisible by either 3 or 5 or both, we use the principle of inclusion-exclusion: (numbers divisible by 3) + (numbers divisible by 5) - (numbers divisible by both) = 15 + 9 - 3 = 21.

## Step 6: Problem 1 - Provide the answer
ANSWER 1: B

## Step 7: Problem 2 - Restate the problem
Jamal has 6 green socks, 18 purple socks, and 12 orange socks. After adding more purple socks, the probability of picking a purple sock is 60%. We need to find how many purple socks Jamal added.

## Step 8: Problem 2 - Set up the initial conditions
Initially, Jamal has 6 + 18 + 12 = 36 socks.

## Step 9: Problem 2 - Set up the equation for the probability of picking a purple sock after adding x purple socks
After adding x purple socks, the total number of socks is 36 + x, and the number of purple socks is 18 + x. The probability of picking a purple sock is (18 + x) / (36 + x) = 0.6.

## Step 10: Problem 2 - Solve the equation for x
(18 + x) / (36 + x) = 0.6. Multiplying both sides by (36 + x) gives 18 + x = 0.6(36 + x). Expanding, we get 18 + x = 21.6 + 0.6x. Rearranging, x - 0.6x = 21.6 - 18, so 0.4x = 3.6. Dividing both sides by 0.4, x = 9.

## Step 11: Problem 2 - Provide the answer
ANSWER 2: B

## Step 12: Problem 3 - Restate the problem
We need to find which year is NOT a lucky year, where a lucky year has at least one date where the month times the day equals the last two digits of the year.

## Step 13: Problem 3 - Analyze the condition for a lucky year
For a year to be lucky, there must exist a month (1-12) and a day (1-31) such that month * day = last two digits of the year.

## Step 14: Problem 3 - Check each given year
Checking each year from 1990 to 1994 to see if any date satisfies the condition.

## Step 15: Problem 3 - Analyze 1990 to 1994
- For 1990, the last two digits are 90. It's difficult to get 90 by multiplying two numbers between 1 and 12 and 1 and 31, respectively, because the factors of 90 that could work are (1,90), (2,45), (3,30), (5,18), (6,15), (9,10). Only (9,10) is plausible, and it works as 9/10/90.
- For 1991, 91 can be achieved with 7/13/91, so it's lucky.
- For 1992, 92 can be achieved, for example, with 4/23/92, so it's lucky.
- For 1993, 93 can be achieved with 3/31/93, so it's lucky.
- For 1994, 94 is not easily achieved because its factors are (1,94), (2,47). No month-day combination gives 94.

## Step 16: Problem 3 - Provide the answer
Given the analysis, 1994 is not lucky because there's no month-day product equal to 94.

ANSWER 3: E

## Step 17: Problem 4 - Restate the problem
We need to find the smallest difference possible by placing the digits 2, 4, 5, 6, 9 in the boxes of a subtraction problem.

## Step 18: Problem 4 - Understand the structure of the subtraction problem
To minimize the difference, we should minimize the minuend and maximize the subtrahend.

## Step 19: Problem 4 - Arrange the digits to minimize the difference
To achieve the smallest difference, the hundreds place of the minuend should be as small as possible, and the hundreds place of the subtrahend should be as large as possible. So, the minuend could be 2** and the subtrahend 9**. The remaining digits are 4, 5, 6.

## Step 20: Problem 4 - Try different combinations
Let's try to make the minuend as small as possible and the subtrahend as large as possible: 245 - 986 is not valid because it results in a negative number. A valid attempt is 246 - 985 or similar combinations.

## Step 21: Problem 4 - Calculate the difference for a valid combination
For 964 - 852 or similar, the difference is not minimal. Let's directly check: 245 - 196 = 49 (not an option), but trying to minimize, we can get 246 - 195 = 51 (still not an option). A plausible arrangement is 254 - 196 = 58.

## Step 22: Problem 4 - Provide the answer
The smallest difference is achieved with an arrangement like 254 - 196.

ANSWER 4: A

## Step 23: Problem 5 - Restate the problem
Given W, X, Y, Z are different integers from 1 to 4, and WX - YZ = 1, we need to find W + Y.

## Step 24: Problem 5 - Understand the equation WX - YZ = 1
Since W, X, Y, Z are different integers from 1 to 4, we need to consider the possible values of WX and YZ.

## Step 25: Problem 5 - Analyze possible values for WX and YZ
The possible products are: 1*2=2, 1*3=3, 1*4=4, 2*1=2, 2*3=6, 2*4=8, 3*1=3, 3*2=6, 3*4=12, 4*1=4, 4*2=8, 4*3=12. We need a difference of 1 between two of these products.

## Step 26: Problem 5 - Check for a difference of 1
The products are 2, 3, 4, 6, 8, 12. The pairs that differ by 1 are (2,3), (3,4). So, either WX = 3 and YZ = 2 or WX = 4 and YZ = 3.

## Step 27: Problem 5 - Solve for W + Y
For WX = 3 and YZ = 2, W = 3, X = 1, Y = 2, Z = 1 (not valid because Z = X), or W = 1, X = 3, Y = 1, Z = 2 (not valid because W = Y). A valid case is W = 3, X = 1, Y = 1, Z = 2 (not valid), but if we consider W = 1, X = 3, Y = 2, Z = 1, it's valid and gives W + Y = 1 + 2 = 3.

## Step 28: Problem 5 - Provide the answer
For the valid case where W = 1, X = 3, Y = 2, Z = 1 doesn't work due to repetition, but

Problem 1:

We need count the whole numbers from 1 through 46 that are divisible by 3 or 5 or both.

Count multiples of 3:

\[
\left\lfloor \frac{46}{3} \right\rfloor = 15
\]

Count multiples of 5:

\[
\left\lfloor \frac{46}{5} \right\rfloor = 9
\]

But multiples of both 3 and 5, meaning multiples of 15, were counted twice, so subtract them once:

\[
\left\lfloor \frac{46}{15} \right\rfloor = 3
\]

Therefore the total is

\[
15+9-3=21
\]

So choices like 24 or 27 overcount because they do not remove the overlap.

ANSWER 1: B

Problem 2:

Initially Jamal has:

\[
6+18+12=36
\]

socks total, of which 18 are purple.

Suppose he adds \(x\) purple socks. Then the number of purple socks is

\[
18+x
\]

and the total number of socks is

\[
36+x
\]

The probability of choosing a purple sock is now 60%, or \(\frac35\). So

\[
\frac{18+x}{36+x}=\frac35
\]

Cross multiply:

\[
5(18+x)=3(36+x)
\]

\[
90+5x=108+3x
\]

\[
2x=18
\]

\[
x=9
\]

So Jamal added 9 purple socks.

ANSWER 2: B

Problem 3:

A lucky year has some valid date month/day/year where

\[
\text{month} \times \text{day} = \text{last two digits of the year}
\]

Check each year:

For 1990:

\[
9 \times 10 = 90
\]

So 9/10/90 works.

For 1991:

\[
7 \times 13 = 91
\]

So 7/13/91 works.

For 1992:

\[
4 \times 23 = 92
\]

So 4/23/92 works.

For 1993:

\[
3 \times 31 = 93
\]

So 3/31/93 works.

For 1994:

\[
94=2 \times 47
\]

But day 47 is not valid. The only factor pairs are \(1 \times 94\) and \(2 \times 47\), and neither gives a valid date.

Thus 1994 is not lucky.

ANSWER 3: E

Problem 4:

The subtraction problem uses the digits 2, 4, 5, 6, 9 exactly once in a 3-digit number minus a 2-digit number.

To make the difference as small as possible, the 3-digit number should be as small as possible, and the 2-digit number should be as large as possible.

The hundreds digit of the 3-digit number should be the smallest digit, 2.

Now we want the rest of the top number small and the bottom number large. Using the remaining digits 4, 5, 6, 9, the smallest possible two digits for the top number are 4 and 5, making 245, while the largest possible 2-digit number from the remaining digits is 96.

So try:

\[
245-96=149
\]

If we tried to get below 149, we would need either a smaller 3-digit number than 245 or a larger 2-digit number than 96, but neither is possible with these digits.

Therefore the smallest difference is 149.

ANSWER 4: C

Problem 5:

The equation is

\[
\frac{W}{X}-\frac{Y}{Z}=1
\]

where \(W, X, Y, Z\) are the numbers 1, 2, 3, 4 in some order.

We need find an arrangement that works.

Try to make the first fraction bigger than 1. One successful choice is

\[
\frac{3}{1}-\frac{4}{2}=1
\]

because

\[
3-2=1
\]

So

\[
W=3,\quad X=1,\quad Y=4,\quad Z=2
\]

Then

\[
W+Y=3+4=7
\]

The other choices are smaller sums, but the only working arrangement gives 7.

ANSWER 5: E

Problem 6:

There are four-digit integers from 1000 through 9999. The total number is

\[
9999-1000+1=9000
\]

We want odd integers whose digits are all distinct.

For a number to be odd, the units digit must be one of

\[
1,3,5,7,9
\]

So there are 5 choices for the units digit.

The thousands digit cannot be 0 and cannot equal the units digit. Since there are 9 possible nonzero thousands digits, and one is already used, there are

\[
8
\]

choices.

The hundreds digit can be any digit except the two already used:

\[
8
\]

choices.

The tens digit can be any digit except the three already used:

\[
7
\]

choices.

So the number of favorable integers is

\[
5 \cdot 8 \cdot 8 \cdot 7=2240
\]

The probability is

\[
\frac{2240}{9000}
\]

Reduce:

\[
\frac{2240}{9000}=\frac{224}{900}=\frac{56}{225}
\]

ANSWER 6: B

Problem 7:

Let \(n\) be the number of employees, and let \(F\) be the amount of money in the fund.

The manager planned to give each employee $50, but the fund was $5 short. So

\[
F=50n-5
\]

Instead, the manager gave each employee $45 and had $95 left. So

\[
F=45n+95
\]

Set these equal:

\[
50n-5=45n+95
\]

\[
5n=100
\]

\[
n=20
\]

Now find the fund:

\[
F=50(20)-5=1000-5=995
\]

So there were $995 in the fund.

ANSWER 7: E

Problem 8:

We are given

\[
ABA \times CD = CDCD
\]

The number \(ABA\) is

\[
101A+10B
\]

The number \(CDCD\) is a repeated two-digit number, so

\[
CDCD=101 \times CD
\]

Thus the equation becomes

\[
ABA \times CD = 101 \times CD
\]

Since \(CD\) is a two-digit number, it is not 0, so we can divide both sides by \(CD\):

\[
ABA=101
\]

Therefore

\[
A=1,\quad B=0
\]

So

\[
A+B=1+0=1
\]

ANSWER 8: A

Problem 9:

We need choose 3 numbers from

\[
\{89,95,99,132,166,173\}
\]

so that their sum is even.

Classify by parity.

Odd numbers:

\[
89,95,99,173
\]

There are 4 odds.

Even numbers:

\[
132,166
\]

There are 2 evens.

A sum of three numbers is even if it contains either:

- 0 odd numbers and 3 even numbers, or
- 2 odd numbers and 1 even number.

We cannot choose 3 even numbers because there are only 2 even numbers.

So we must choose 2 odds and 1 even.

Number of ways:

\[
\binom{4}{2}\binom{2}{1}=6 \cdot 2=12
\]

ANSWER 9: D

Problem 10:

The set has 5 positive integers with mean 5, so the total sum is

\[
5 \cdot 5=25
\]

The median is 5, so when the numbers are ordered, the middle number is 5.

Also, 8 is the only mode, meaning 8 appears more often than any other number. Since there are only five numbers and the median is 5, the two largest numbers must be 8 and 8:

\[
\_,\_,5,8,8
\]

Their sum so far is

\[
5+8+8=21
\]

So the first two numbers must sum to

\[
25-21=4
\]

They are positive integers less than or equal to 5. The possible pairs are:

\[
1,3 \quad \text{or} \quad 2,2
\]

But \(2,2\) would make 2 appear twice, tying 8 as a mode. Since 8 must be the only mode, the pair must be

\[
1,3
\]

So the set is

\[
1,3,5,8,8
\]

The difference between the largest and smallest integers is

\[
8-1=7
\]

ANSWER 10: D

Problem 11:

We need count three-digit whole numbers between 99 and 999 that contain exactly one 0.

A t

**Problem 1:**
How many whole numbers from 1 through 46 are divisible by either 3 or 5 or both?

**Step-by-step solution:**
To find the number of integers divisible by 3, 5, or both, we can use the Principle of Inclusion-Exclusion. 
1. First, we count the numbers from 1 to 46 that are divisible by 3:
   $\lfloor 46 \div 3 \rfloor = 15$
2. Next, we count the numbers from 1 to 46 that are divisible by 5:
   $\lfloor 46 \div 5 \rfloor = 9$
3. Then, we count the numbers that are divisible by BOTH 3 and 5 (which means they are divisible by their least common multiple, 15):
   $\lfloor 46 \div 15 \rfloor = 3$
4. Finally, we add the multiples of 3 and 5, and subtract the multiples of 15 so they aren't counted twice:
   $15 + 9 - 3 = 21$

ANSWER 1: B

---

**Problem 2:**
Jamal has a drawer containing 6 green socks, 18 purple socks, and 12 orange socks. After adding more purple socks, Jamal noticed that there is now a 60% chance that a sock randomly selected from the drawer is purple. How many purple socks did Jamal add?

**Step-by-step solution:**
1. Initially, Jamal has $6 + 18 + 12 = 36$ socks in total.
2. Let $x$ be the number of purple socks Jamal added. 
3. The new number of purple socks is $18 + x$.
4. The new total number of socks is $36 + x$.
5. We are told the new probability of selecting a purple sock is 60%, which simplifies to $\frac{60}{100} = \frac{3}{5}$.
6. We set up the equation:
   $\frac{18 + x}{36 + x} = \frac{3}{5}$
7. Cross-multiply to solve for $x$:
   $5(18 + x) = 3(36 + x)$
   $90 + 5x = 108 + 3x$
   $2x = 18$
   $x = 9$

Jamal added 9 purple socks.

ANSWER 2: B

---

**Problem 3:**
A lucky year is one in which at least one date, written as month/day/year, has the property that the month times the day equals the last two digits of the year. For example, 1956 is lucky because 7/8/56 has 7 × 8 = 56. Which of the following is NOT a lucky year?

**Step-by-step solution:**
We need to find if the last two digits of each given year can be factored into a valid month (1 through 12) and a valid day (1 through 31).
- **A. 1990:** $90 = 9 \times 10$. This corresponds to September 10th (9/10/90). Lucky.
- **B. 1991:** $91 = 7 \times 13$. This corresponds to July 13th (7/13/91). Lucky.
- **C. 1992:** $92 = 4 \times 23$. This corresponds to April 23rd (4/23/92). Lucky.
- **D. 1993:** $93 = 3 \times 31$. This corresponds to March 31st (3/31/93). Lucky.
- **E. 1994:** $94 = 2 \times 47$. Since a month cannot have 47 days, and 94 has no other integer pairs that fit the calendar constraints (1×94 is also invalid), 1994 is NOT a lucky year.

ANSWER 3: E

---

**Problem 4:**
When placing each of the digits 2, 4, 5, 6, 9 in exactly one of the boxes of this subtraction problem, what is the smallest difference that is possible?       −

**Step-by-step solution:**
We are given 5 digits to place in a subtraction problem. To yield a positive difference using 5 digits, the setup must be a 3-digit number minus a 2-digit number.
To get the smallest possible difference, we need the 3-digit number to be as small as possible and the 2-digit number to be as large as possible.
1. The smallest possible hundreds digit for the 3-digit number is 2. So the number is $2\_ \_$.
2. We are left with the digits 4, 5, 6, and 9. 
3. To minimize the 3-digit number, the remaining two digits should be the smallest available: 4 and 5, making the number 245.
4. To maximize the 2-digit number being subtracted, we use the largest remaining digits: 9 and 6, making the number 96.
5. Subtracting the two gives: $245 - 96 = 149$.
(Testing other close combinations like $254 - 96 = 158$ or $246 - 95 = 151$ confirms 149 is the absolute minimum difference.)

ANSWER 4: C

---

**Problem 5:**
Each of the letters W , X , Y , and Z represents a different integer in the set {1, 2, 3, 4}, but not necessarily in that order. If W X − Y Z = 1, then the sum of W and Y is

**Step-by-step solution:**
The equation given is meant to be read as fractions (a common formatting artifact drops the fraction bar): $\frac{W}{X} - \frac{Y}{Z} = 1$.
Let's test the values {1, 2, 3, 4} to satisfy this equation:
- If we set $W = 3$, $X = 1$, $Y = 4$, and $Z = 2$:
  $\frac{3}{1} - \frac{4}{2} = 3 - 2 = 1$
This perfectly uses all four distinct digits exactly once.
The question asks for the sum of $W$ and $Y$:
$W + Y = 3 + 4 = 7$

*(Note: Checking $W \times X - Y \times Z = 1$ yields no solutions where all 4 digits are distinct, confirming the fractional interpretation).*

ANSWER 5: E

---

**Problem 6:**
An integer between 1000 and 9999, inclusive, is chosen at random. What is the probability that it is an odd integer whose digits are all distinct?

**Step-by-step solution:**
1. The total number of integers from 1000 to 9999 inclusive is $9999 - 1000 + 1 = 9000$.
2. We want to form a 4-digit odd number (let's call its digits A B C D) with all distinct digits.
3. Because it must be odd, the units digit (D) must be 1, 3, 5, 7, or 9. There are **5** choices for D.
4. The thousands digit (A) cannot be 0, and it cannot be the digit used for D. Thus, out of 10 possible digits, 2 are restricted, leaving **8** choices for A.
5. The hundreds digit (B) can be 0, but cannot be the digits used for A or D. This leaves **8** choices for B.
6. The tens digit (C) cannot be the digits used for A, B, or D. This leaves **7** choices for C.
7. The total number of such integers is $5 \times 8 \times 8 \times 7 = 2240$.
8. The probability is $\frac{2240}{9000} = \frac{224}{900} = \frac{56}{225}$.

ANSWER 6: B

---

**Problem 7:**
The manager of a company planned to give a $50 bonus to each employee from the company fund, but the fund was $5 short of what was needed. Instead the manager gave each employee a $45 bonus and kept the remaining $95 in the fund. How much money was in the company fund before any bonuses were paid?

**Step-by-step solution:**
Let $n$ be the number of employees.
1. If the manager gave $50 to each employee, the fund would be $5 short. So, the total fund is $5

## Step 1: Problem 1 - Restate the problem
We need to find how many whole numbers from 1 through 46 are divisible by either 3 or 5 or both.

## Step 2: Problem 1 - Calculate numbers divisible by 3
First, let's find the numbers divisible by 3. The largest number less than or equal to 46 that is divisible by 3 is 45, and the number of multiples of 3 from 1 to 46 is 45 / 3 = 15.

## Step 3: Problem 1 - Calculate numbers divisible by 5
Next, find the numbers divisible by 5. The largest number less than or equal to 46 that is divisible by 5 is 45, and the number of multiples of 5 from 1 to 46 is 45 / 5 = 9.

## Step 4: Problem 1 - Calculate numbers divisible by both 3 and 5
Numbers divisible by both 3 and 5 are divisible by 15. The largest number less than or equal to 46 that is divisible by 15 is 45, and the number of multiples of 15 from 1 to 46 is 45 / 15 = 3.

## Step 5: Problem 1 - Apply the principle of inclusion-exclusion
To find the total number of numbers divisible by either 3 or 5 or both, we use the principle of inclusion-exclusion: (numbers divisible by 3) + (numbers divisible by 5) - (numbers divisible by both) = 15 + 9 - 3 = 21.

## Step 6: Problem 1 - Provide the answer
ANSWER 1: B

## Step 7: Problem 2 - Restate the problem
Jamal has 6 green socks, 18 purple socks, and 12 orange socks. After adding more purple socks, the probability of picking a purple sock is 60%. We need to find how many purple socks Jamal added.

## Step 8: Problem 2 - Set up the initial conditions
Initially, Jamal has 6 + 18 + 12 = 36 socks.

## Step 9: Problem 2 - Set up the equation for the probability of picking a purple sock after adding x purple socks
After adding x purple socks, the total number of socks is 36 + x, and the number of purple socks is 18 + x. The probability of picking a purple sock is (18 + x) / (36 + x) = 0.6.

## Step 10: Problem 2 - Solve the equation for x
(18 + x) / (36 + x) = 0.6. Multiplying both sides by (36 + x) gives 18 + x = 0.6(36 + x). Expanding, we get 18 + x = 21.6 + 0.6x. Rearranging, x - 0.6x = 21.6 - 18, so 0.4x = 3.6. Dividing both sides by 0.4, x = 9.

## Step 11: Problem 2 - Provide the answer
ANSWER 2: B

## Step 12: Problem 3 - Restate the problem
We need to find which year is NOT a lucky year, where a lucky year has at least one date where the month times the day equals the last two digits of the year.

## Step 13: Problem 3 - Analyze the condition for a lucky year
For a year to be lucky, there must exist a month (1-12) and a day (1-31) such that month * day = last two digits of the year.

## Step 14: Problem 3 - Check each given year
Checking each year from 1990 to 1994 to see if any date satisfies the condition.

## Step 15: Problem 3 - Analyze 1990 to 1994
- For 1990, the last two digits are 90. It's difficult to get 90 by multiplying two numbers between 1 and 12 and 1 and 31, respectively, because the factors of 90 that could work are (1,90), (2,45), (3,30), (5,18), (6,15), (9,10). Only (9,10) is plausible, and it works as 9/10/90.
- For 1991, 91 can be achieved with 7/13/91, so it's lucky.
- For 1992, 92 can be achieved, for example, with 4/23/92, so it's lucky.
- For 1993, 93 can be achieved with 3/31/93, so it's lucky.
- For 1994, 94 is not easily achieved because its factors are (1,94), (2,47). No month-day combination gives 94.

## Step 16: Problem 3 - Provide the answer
Given the analysis, 1994 is not lucky because there's no month-day product equal to 94.

ANSWER 3: E

## Step 17: Problem 4 - Restate the problem
We need to find the smallest difference possible by placing the digits 2, 4, 5, 6, 9 in the boxes of a subtraction problem.

## Step 18: Problem 4 - Understand the structure of the subtraction problem
To minimize the difference, we should minimize the minuend and maximize the subtrahend.

## Step 19: Problem 4 - Arrange the digits to minimize the difference
To achieve the smallest difference, the hundreds place of the minuend should be as small as possible, and the hundreds place of the subtrahend should be as large as possible. So, the minuend could be 2** and the subtrahend 9**. The remaining digits are 4, 5, 6.

## Step 20: Problem 4 - Try different combinations
Let's try to make the minuend as small as possible and the subtrahend as large as possible: 245 - 986 is not valid because it results in a negative number. A valid attempt is 246 - 985 or similar combinations.

## Step 21: Problem 4 - Calculate the difference for a valid combination
For 964 - 852 or similar, the difference is not minimal. Let's directly check: 245 - 196 = 49 (not an option), but trying to minimize, we can get 246 - 195 = 51 (still not an option). A plausible arrangement is 254 - 196 = 58.

## Step 22: Problem 4 - Provide the answer
The smallest difference is achieved with an arrangement like 254 - 196.

ANSWER 4: A

## Step 23: Problem 5 - Restate the problem
Given W, X, Y, Z are different integers from 1 to 4, and WX - YZ = 1, we need to find W + Y.

## Step 24: Problem 5 - Understand the equation WX - YZ = 1
Since W, X, Y, Z are different integers from 1 to 4, we need to consider the possible values of WX and YZ.

## Step 25: Problem 5 - Analyze possible values for WX and YZ
The possible products are: 1*2=2, 1*3=3, 1*4=4, 2*1=2, 2*3=6, 2*4=8, 3*1=3, 3*2=6, 3*4=12, 4*1=4, 4*2=8, 4*3=12. We need a difference of 1 between two of these products.

## Step 26: Problem 5 - Check for a difference of 1
The products are 2, 3, 4, 6, 8, 12. The pairs that differ by 1 are (2,3), (3,4). So, either WX = 3 and YZ = 2 or WX = 4 and YZ = 3.

## Step 27: Problem 5 - Solve for W + Y
For WX = 3 and YZ = 2, W = 3, X = 1, Y = 2, Z = 1 (not valid because Z = X), or W = 1, X = 3, Y = 1, Z = 2 (not valid because W = Y). A valid case is W = 3, X = 1, Y = 1, Z = 2 (not valid), but if we consider W = 1, X = 3, Y = 2, Z = 1, it's valid and gives W + Y = 1 + 2 = 3.

## Step 28: Problem 5 - Provide the answer
For the valid case where W = 1, X = 3, Y = 2, Z = 1 doesn't work due to repetition, but

Problem 1:

We need count the whole numbers from 1 through 46 that are divisible by 3 or 5 or both.

Count multiples of 3:

\[
\left\lfloor \frac{46}{3} \right\rfloor = 15
\]

Count multiples of 5:

\[
\left\lfloor \frac{46}{5} \right\rfloor = 9
\]

But multiples of both 3 and 5, meaning multiples of 15, were counted twice, so subtract them once:

\[
\left\lfloor \frac{46}{15} \right\rfloor = 3
\]

Therefore the total is

\[
15+9-3=21
\]

So choices like 24 or 27 overcount because they do not remove the overlap.

ANSWER 1: B

Problem 2:

Initially Jamal has:

\[
6+18+12=36
\]

socks total, of which 18 are purple.

Suppose he adds \(x\) purple socks. Then the number of purple socks is

\[
18+x
\]

and the total number of socks is

\[
36+x
\]

The probability of choosing a purple sock is now 60%, or \(\frac35\). So

\[
\frac{18+x}{36+x}=\frac35
\]

Cross multiply:

\[
5(18+x)=3(36+x)
\]

\[
90+5x=108+3x
\]

\[
2x=18
\]

\[
x=9
\]

So Jamal added 9 purple socks.

ANSWER 2: B

Problem 3:

A lucky year has some valid date month/day/year where

\[
\text{month} \times \text{day} = \text{last two digits of the year}
\]

Check each year:

For 1990:

\[
9 \times 10 = 90
\]

So 9/10/90 works.

For 1991:

\[
7 \times 13 = 91
\]

So 7/13/91 works.

For 1992:

\[
4 \times 23 = 92
\]

So 4/23/92 works.

For 1993:

\[
3 \times 31 = 93
\]

So 3/31/93 works.

For 1994:

\[
94=2 \times 47
\]

But day 47 is not valid. The only factor pairs are \(1 \times 94\) and \(2 \times 47\), and neither gives a valid date.

Thus 1994 is not lucky.

ANSWER 3: E

Problem 4:

The subtraction problem uses the digits 2, 4, 5, 6, 9 exactly once in a 3-digit number minus a 2-digit number.

To make the difference as small as possible, the 3-digit number should be as small as possible, and the 2-digit number should be as large as possible.

The hundreds digit of the 3-digit number should be the smallest digit, 2.

Now we want the rest of the top number small and the bottom number large. Using the remaining digits 4, 5, 6, 9, the smallest possible two digits for the top number are 4 and 5, making 245, while the largest possible 2-digit number from the remaining digits is 96.

So try:

\[
245-96=149
\]

If we tried to get below 149, we would need either a smaller 3-digit number than 245 or a larger 2-digit number than 96, but neither is possible with these digits.

Therefore the smallest difference is 149.

ANSWER 4: C

Problem 5:

The equation is

\[
\frac{W}{X}-\frac{Y}{Z}=1
\]

where \(W, X, Y, Z\) are the numbers 1, 2, 3, 4 in some order.

We need find an arrangement that works.

Try to make the first fraction bigger than 1. One successful choice is

\[
\frac{3}{1}-\frac{4}{2}=1
\]

because

\[
3-2=1
\]

So

\[
W=3,\quad X=1,\quad Y=4,\quad Z=2
\]

Then

\[
W+Y=3+4=7
\]

The other choices are smaller sums, but the only working arrangement gives 7.

ANSWER 5: E

Problem 6:

There are four-digit integers from 1000 through 9999. The total number is

\[
9999-1000+1=9000
\]

We want odd integers whose digits are all distinct.

For a number to be odd, the units digit must be one of

\[
1,3,5,7,9
\]

So there are 5 choices for the units digit.

The thousands digit cannot be 0 and cannot equal the units digit. Since there are 9 possible nonzero thousands digits, and one is already used, there are

\[
8
\]

choices.

The hundreds digit can be any digit except the two already used:

\[
8
\]

choices.

The tens digit can be any digit except the three already used:

\[
7
\]

choices.

So the number of favorable integers is

\[
5 \cdot 8 \cdot 8 \cdot 7=2240
\]

The probability is

\[
\frac{2240}{9000}
\]

Reduce:

\[
\frac{2240}{9000}=\frac{224}{900}=\frac{56}{225}
\]

ANSWER 6: B

Problem 7:

Let \(n\) be the number of employees, and let \(F\) be the amount of money in the fund.

The manager planned to give each employee $50, but the fund was $5 short. So

\[
F=50n-5
\]

Instead, the manager gave each employee $45 and had $95 left. So

\[
F=45n+95
\]

Set these equal:

\[
50n-5=45n+95
\]

\[
5n=100
\]

\[
n=20
\]

Now find the fund:

\[
F=50(20)-5=1000-5=995
\]

So there were $995 in the fund.

ANSWER 7: E

Problem 8:

We are given

\[
ABA \times CD = CDCD
\]

The number \(ABA\) is

\[
101A+10B
\]

The number \(CDCD\) is a repeated two-digit number, so

\[
CDCD=101 \times CD
\]

Thus the equation becomes

\[
ABA \times CD = 101 \times CD
\]

Since \(CD\) is a two-digit number, it is not 0, so we can divide both sides by \(CD\):

\[
ABA=101
\]

Therefore

\[
A=1,\quad B=0
\]

So

\[
A+B=1+0=1
\]

ANSWER 8: A

Problem 9:

We need choose 3 numbers from

\[
\{89,95,99,132,166,173\}
\]

so that their sum is even.

Classify by parity.

Odd numbers:

\[
89,95,99,173
\]

There are 4 odds.

Even numbers:

\[
132,166
\]

There are 2 evens.

A sum of three numbers is even if it contains either:

- 0 odd numbers and 3 even numbers, or
- 2 odd numbers and 1 even number.

We cannot choose 3 even numbers because there are only 2 even numbers.

So we must choose 2 odds and 1 even.

Number of ways:

\[
\binom{4}{2}\binom{2}{1}=6 \cdot 2=12
\]

ANSWER 9: D

Problem 10:

The set has 5 positive integers with mean 5, so the total sum is

\[
5 \cdot 5=25
\]

The median is 5, so when the numbers are ordered, the middle number is 5.

Also, 8 is the only mode, meaning 8 appears more often than any other number. Since there are only five numbers and the median is 5, the two largest numbers must be 8 and 8:

\[
\_,\_,5,8,8
\]

Their sum so far is

\[
5+8+8=21
\]

So the first two numbers must sum to

\[
25-21=4
\]

They are positive integers less than or equal to 5. The possible pairs are:

\[
1,3 \quad \text{or} \quad 2,2
\]

But \(2,2\) would make 2 appear twice, tying 8 as a mode. Since 8 must be the only mode, the pair must be

\[
1,3
\]

So the set is

\[
1,3,5,8,8
\]

The difference between the largest and smallest integers is

\[
8-1=7
\]

ANSWER 10: D

Problem 11:

We need count three-digit whole numbers between 99 and 999 that contain exactly one 0.

A t

**Problem 1:**
How many whole numbers from 1 through 46 are divisible by either 3 or 5 or both?

**Step-by-step solution:**
To find the number of integers divisible by 3, 5, or both, we can use the Principle of Inclusion-Exclusion. 
1. First, we count the numbers from 1 to 46 that are divisible by 3:
   $\lfloor 46 \div 3 \rfloor = 15$
2. Next, we count the numbers from 1 to 46 that are divisible by 5:
   $\lfloor 46 \div 5 \rfloor = 9$
3. Then, we count the numbers that are divisible by BOTH 3 and 5 (which means they are divisible by their least common multiple, 15):
   $\lfloor 46 \div 15 \rfloor = 3$
4. Finally, we add the multiples of 3 and 5, and subtract the multiples of 15 so they aren't counted twice:
   $15 + 9 - 3 = 21$

ANSWER 1: B

---

**Problem 2:**
Jamal has a drawer containing 6 green socks, 18 purple socks, and 12 orange socks. After adding more purple socks, Jamal noticed that there is now a 60% chance that a sock randomly selected from the drawer is purple. How many purple socks did Jamal add?

**Step-by-step solution:**
1. Initially, Jamal has $6 + 18 + 12 = 36$ socks in total.
2. Let $x$ be the number of purple socks Jamal added. 
3. The new number of purple socks is $18 + x$.
4. The new total number of socks is $36 + x$.
5. We are told the new probability of selecting a purple sock is 60%, which simplifies to $\frac{60}{100} = \frac{3}{5}$.
6. We set up the equation:
   $\frac{18 + x}{36 + x} = \frac{3}{5}$
7. Cross-multiply to solve for $x$:
   $5(18 + x) = 3(36 + x)$
   $90 + 5x = 108 + 3x$
   $2x = 18$
   $x = 9$

Jamal added 9 purple socks.

ANSWER 2: B

---

**Problem 3:**
A lucky year is one in which at least one date, written as month/day/year, has the property that the month times the day equals the last two digits of the year. For example, 1956 is lucky because 7/8/56 has 7 × 8 = 56. Which of the following is NOT a lucky year?

**Step-by-step solution:**
We need to find if the last two digits of each given year can be factored into a valid month (1 through 12) and a valid day (1 through 31).
- **A. 1990:** $90 = 9 \times 10$. This corresponds to September 10th (9/10/90). Lucky.
- **B. 1991:** $91 = 7 \times 13$. This corresponds to July 13th (7/13/91). Lucky.
- **C. 1992:** $92 = 4 \times 23$. This corresponds to April 23rd (4/23/92). Lucky.
- **D. 1993:** $93 = 3 \times 31$. This corresponds to March 31st (3/31/93). Lucky.
- **E. 1994:** $94 = 2 \times 47$. Since a month cannot have 47 days, and 94 has no other integer pairs that fit the calendar constraints (1×94 is also invalid), 1994 is NOT a lucky year.

ANSWER 3: E

---

**Problem 4:**
When placing each of the digits 2, 4, 5, 6, 9 in exactly one of the boxes of this subtraction problem, what is the smallest difference that is possible?       −

**Step-by-step solution:**
We are given 5 digits to place in a subtraction problem. To yield a positive difference using 5 digits, the setup must be a 3-digit number minus a 2-digit number.
To get the smallest possible difference, we need the 3-digit number to be as small as possible and the 2-digit number to be as large as possible.
1. The smallest possible hundreds digit for the 3-digit number is 2. So the number is $2\_ \_$.
2. We are left with the digits 4, 5, 6, and 9. 
3. To minimize the 3-digit number, the remaining two digits should be the smallest available: 4 and 5, making the number 245.
4. To maximize the 2-digit number being subtracted, we use the largest remaining digits: 9 and 6, making the number 96.
5. Subtracting the two gives: $245 - 96 = 149$.
(Testing other close combinations like $254 - 96 = 158$ or $246 - 95 = 151$ confirms 149 is the absolute minimum difference.)

ANSWER 4: C

---

**Problem 5:**
Each of the letters W , X , Y , and Z represents a different integer in the set {1, 2, 3, 4}, but not necessarily in that order. If W X − Y Z = 1, then the sum of W and Y is

**Step-by-step solution:**
The equation given is meant to be read as fractions (a common formatting artifact drops the fraction bar): $\frac{W}{X} - \frac{Y}{Z} = 1$.
Let's test the values {1, 2, 3, 4} to satisfy this equation:
- If we set $W = 3$, $X = 1$, $Y = 4$, and $Z = 2$:
  $\frac{3}{1} - \frac{4}{2} = 3 - 2 = 1$
This perfectly uses all four distinct digits exactly once.
The question asks for the sum of $W$ and $Y$:
$W + Y = 3 + 4 = 7$

*(Note: Checking $W \times X - Y \times Z = 1$ yields no solutions where all 4 digits are distinct, confirming the fractional interpretation).*

ANSWER 5: E

---

**Problem 6:**
An integer between 1000 and 9999, inclusive, is chosen at random. What is the probability that it is an odd integer whose digits are all distinct?

**Step-by-step solution:**
1. The total number of integers from 1000 to 9999 inclusive is $9999 - 1000 + 1 = 9000$.
2. We want to form a 4-digit odd number (let's call its digits A B C D) with all distinct digits.
3. Because it must be odd, the units digit (D) must be 1, 3, 5, 7, or 9. There are **5** choices for D.
4. The thousands digit (A) cannot be 0, and it cannot be the digit used for D. Thus, out of 10 possible digits, 2 are restricted, leaving **8** choices for A.
5. The hundreds digit (B) can be 0, but cannot be the digits used for A or D. This leaves **8** choices for B.
6. The tens digit (C) cannot be the digits used for A, B, or D. This leaves **7** choices for C.
7. The total number of such integers is $5 \times 8 \times 8 \times 7 = 2240$.
8. The probability is $\frac{2240}{9000} = \frac{224}{900} = \frac{56}{225}$.

ANSWER 6: B

---

**Problem 7:**
The manager of a company planned to give a $50 bonus to each employee from the company fund, but the fund was $5 short of what was needed. Instead the manager gave each employee a $45 bonus and kept the remaining $95 in the fund. How much money was in the company fund before any bonuses were paid?

**Step-by-step solution:**
Let $n$ be the number of employees.
1. If the manager gave $50 to each employee, the fund would be $5 short. So, the total fund is $5