🔬 Model Lab

## Step 1: Problem 1 - Restate the problem
We need to find the number of different ways to distribute 5 different awards to 3 students, with each student receiving at least one award.

## Step 2: Problem 1 - Calculate the total number of ways to distribute the awards without restrictions
First, we consider distributing the awards without any restrictions. For each award, there are 3 choices of students, so the total number of ways to distribute 5 awards is $3^5 = 243$.

## Step 3: Problem 1 - Apply the Principle of Inclusion-Exclusion to find the number of ways where at least one student receives no award
To ensure each student gets at least one award, we'll use the Principle of Inclusion-Exclusion. We calculate the number of ways to distribute the awards where at least one student gets no award. There are $3$ ways to choose a student to not receive an award, and for each such choice, there are $2^5$ ways to distribute the awards among the remaining 2 students. So, there are $3 \cdot 2^5 = 96$ ways. However, this counts cases where 2 students receive no awards twice, so we must adjust for that.

## Step 4: Problem 1 - Continue applying the Principle of Inclusion-Exclusion
There are $3$ ways to choose 2 students to not receive an award, and for each such choice, there's $1^5 = 1$ way for the remaining student to receive all awards. So, there are $3 \cdot 1 = 3$ such cases. The total number of ways where at least one student gets no award is $3 \cdot 2^5 - 3 \cdot 1 = 96 - 3 = 93$.

## Step 5: Problem 1 - Calculate the final number of ways to distribute the awards
The number of ways to distribute the awards such that each student gets at least one is $3^5 - 93 = 243 - 93 = 150$.

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

## Step 7: Problem 2 - Restate the problem
We are to determine the number of ways Alina can divide the numbers 1 through 9 into 3 groups of 3 such that the sum of the numbers in each group is the same.

## Step 8: Problem 2 - Calculate the total sum of the numbers and the required sum for each group
The sum of the numbers 1 through 9 is $1+2+3+...+9 = \frac{9(9+1)}{2} = 45$. For 3 groups to have the same sum, each group's sum must be $45 / 3 = 15$.

## Step 9: Problem 2 - Analyze the possible combinations for the groups
To achieve a sum of 15 with 3 numbers, we need to consider the combinations. The highest number is 9, and to reach 15, the other two numbers must sum to 6. Possible pairs with 9 are (9,5,1), (9,4,2), and (9,3,3), but since we can't repeat numbers, (9,3,3) is not valid. We continue this process for other high numbers.

## Step 10: Problem 2 - Enumerate valid combinations
Valid sets that sum to 15 include (9,5,1), (9,4,2), (8,6,1), (8,5,2), (8,4,3), (7,6,2), (7,5,3), (7,4,4) - which is not valid due to repetition, and (6,5,4). We need to find combinations of 3 such numbers that use each number from 1 to 9 exactly once.

## Step 11: Problem 2 - Check for valid groupings
After examining possible combinations and considering the restrictions, we find that there are limited ways to form 3 groups. One valid grouping is (9,2,4), (8,3,4) - which doesn't work because it repeats 4, and (7,6,2), (7,5,3), (8,1,6), (9,5,1), (7,4,4) - again, not valid. A valid example is (9,5,1), (8,6,1) - not valid, (8,4,3), (7,6,2), (7,5,3), and (9,4,2), (8,6,1), (7,5,3). Upon careful examination, one valid set is (9,4,2), (8,7,0) - not valid because 0 is not in the list, but (8,6,1), (7,5,3), and (9,4,2) is a valid grouping: {9,4,2}, {8,6,1}, {7,5,3}. Another is {9,5,1}, {8,4,3}, {7,6,2}.

## Step 12: Problem 2 - Count the distinct ways to form the groups
There are 2 distinct ways to form the groups as shown by the examples.

## Step 13: Problem 2 - State the answer
ANSWER 2: C

## Step 14: Problem 3 - Restate the problem
We need to find the sum of the digits when $10^{93} - 93$ is expressed as a whole number.

## Step 15: Problem 3 - Analyze the number $10^{93}$
$10^{93}$ is a 1 followed by 93 zeros.

## Step 16: Problem 3 - Subtract 93 from $10^{93}$
Subtracting 93 from $10^{93}$ gives us a number that is 93 less than $10^{93}$. This results in a number with 93 digits, where the last two digits are 07, and the rest are 9s except for the third last digit which is 9 - 1 = 8 (due to borrowing), but since we're subtracting from a number ending in 00, it actually results in ...9907.

## Step 17: Problem 3 - Calculate the sum of the digits
The number $10^{93} - 93$ is ...9999907. It has 91 nines, an 8 (from the subtraction affecting the hundreds place due to borrowing), a 0, and a 7. So, the sum is $91*9 + 8 + 0 + 7$.

## Step 18: Problem 3 - Perform the sum of the digits calculation
$91*9 = 819$, $819 + 8 + 7 = 834$.

## Step 19: Problem 3 - State the answer
ANSWER 3: E

## Step 20: Problem 4 - Restate the problem
We need to find the total number of points scored by the other 7 team members in a basketball game given that Alexa scored 1/4 of the points, Brittany scored 2/7, and Chelsea scored 15 points.

## Step 21: Problem 4 - Set up an equation for the total points
Let $T$ be the total points. Then, $\frac{1}{4}T + \frac{2}{7}T + 15 + x = T$, where $x$ is the total points scored by the other 7 team members.

## Step 22: Problem 4 - Simplify the equation
$\frac{7T + 8T}{28} + 15 + x = T$, so $\frac{15T}{28} + 15 + x = T$.

## Step 23: Problem 4 - Solve for $x$
$T - \frac{15T}{28} = 15 + x$, which simplifies to $\frac{13T}{28} = 15 + x$. Thus, $T = \frac{28}{13}(15 + x)$.

## Step 24: Problem 4 - Use the information that none of the other 7 team members scored more than 2 points
$x \leq 14$ because $7 \times 2 = 14$.

## Step 25: Problem 4 - Test possible values of $T$ and $x$
Since $T$ must be a whole number, and given that $\frac{1}{4}T$ and $\frac{2}{7}T$ are also whole numbers, $T$ must be divisible by both 4 and 7, hence divisible by 28.

## Step 26: Problem 4 - Find the appropriate $T$
Let's start with the smallest $T$ that is divisible by 28 and satisfies the given condi

Problem 1:

We are distributing 5 **different** awards to 3 students, with each student receiving at least one award.

First count all possible distributions with no restriction:

- Each of 5 awards can go to any of 3 students.
- Total: \(3^5=243\)

Now subtract distributions where at least one student gets no award.

If a particular student gets no award, then all 5 awards go to the other 2 students:

\[
2^5=32
\]

There are 3 choices for which student gets no award, so subtract:

\[
3\cdot 32=96
\]

But distributions where two students get no awards were subtracted twice, so add them back. If only one student gets all awards, there are 3 such distributions:

\[
243-96+3=150
\]

So the number of ways is \(150\).

ANSWER 1: B

Problem 2:

The numbers are \(1,2,\dots,9\), whose total sum is

\[
1+2+\cdots+9=45
\]

They are divided into 3 groups with equal sums, so each group must sum to

\[
45\div 3=15
\]

Now list triples from \(1\) to \(9\) that sum to 15:

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

Now form complete partitions into 3 disjoint triples.

If the group containing 1 is \(\{1,5,9\}\), the remaining numbers are

\[
2,3,4,6,7,8
\]

These form

\[
\{2,6,7\},\{3,4,8\}
\]

So one partition is

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

If the group containing 1 is \(\{1,6,8\}\), the remaining numbers are

\[
2,3,4,5,7,9
\]

These form

\[
\{2,4,9\},\{3,5,7\}
\]

So another partition is

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

There are exactly 2 ways.

ANSWER 2: C

Problem 3:

We need the sum of the digits of

\[
10^{93}-93
\]

Think of subtracting 93 from a 1 followed by 93 zeros.

A useful way:

\[
10^{93}-93=(10^{93}-100)+7
\]

Now \(10^{93}-100\) is a number with 91 nines followed by two zeros:

\[
999\cdots 99900
\]

with 91 nines. Adding 7 gives

\[
999\cdots 99907
\]

So the digit sum is

\[
91\cdot 9+0+7=819+7=826
\]

ANSWER 3: D

Problem 4:

Let the total number of points be \(T\).

Alexa scored

\[
\frac14T
\]

Brittany scored

\[
\frac27T
\]

Chelsea scored 15 points.

The fraction of the total scored by Alexa and Brittany together is

\[
\frac14+\frac27=\frac7{28}+\frac8{28}=\frac{15}{28}
\]

So the remaining fraction is

\[
1-\frac{15}{28}=\frac{13}{28}
\]

Chelsea plus the other 7 players scored

\[
\frac{13}{28}T
\]

Since \(\frac14T\) and \(\frac27T\) are whole numbers, \(T\) must be a multiple of 28. Let

\[
T=28k
\]

Then Chelsea plus the other players scored

\[
\frac{13}{28}\cdot 28k=13k
\]

So the other 7 players scored

\[
13k-15
\]

Each of the other 7 players scored at most 2 points, so together they scored at most

\[
7\cdot 2=14
\]

Thus

\[
0\le 13k-15\le 14
\]

Testing positive integers \(k\):

- \(k=1\): \(13-15=-2\), impossible.
- \(k=2\): \(26-15=11\), possible.
- \(k=3\): \(39-15=24\), too large.

So the other 7 players scored 11 points.

ANSWER 4: B

Problem 5:

The expression is

\[
\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}
\]

Each fraction is just the sign of the number:

\[
\frac{x}{|x|}=
\begin{cases}
1, & x>0\\
-1, & x<0
\end{cases}
\]

Since \(a+b+c=0\) and none of \(a,b,c\) is zero, they cannot all be positive or all be negative.

So there are two cases.

Case 1: Two are positive and one is negative.

Then

\[
\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}=1+1-1=1
\]

The product \(abc\) is negative, so

\[
\frac{abc}{|abc|}=-1
\]

Total:

\[
1+(-1)=0
\]

Case 2: One is positive and two are negative.

Then

\[
1-1-1=-1
\]

The product \(abc\) is positive, so

\[
\frac{abc}{|abc|}=1
\]

Total:

\[
-1+1=0
\]

So the only possible value is \(0\).

ANSWER 5: A

Problem 6:

The numbers

\[
a_1,a_2,\dots,a_{15}
\]

are equally spaced integers, so they form an arithmetic sequence with integer common difference \(d\).

Thus

\[
a_{15}=a_1+14d
\]

We know

\[
1\le a_1\le 10
\]

and

\[
241\le a_{15}\le 250
\]

So

\[
a_{15}-a_1
\]

is between

\[
241-10=231
\]

and

\[
250-1=249
\]

But

\[
a_{15}-a_1=14d
\]

So \(14d\) must be a multiple of 14 between 231 and 249. The only such multiple is

\[
238=14\cdot 17
\]

Thus

\[
d=17
\]

Now

\[
a_2=a_1+17
\]

and we are told

\[
13\le a_2\le 20
\]

So

\[
13\le a_1+17\le 20
\]

which gives

\[
-4\le a_1\le 3
\]

Combining with \(1\le a_1\le 10\), we get

\[
a_1=3
\]

Now

\[
a_{14}=a_1+13d=3+13\cdot 17=3+221=224
\]

The digit sum of 224 is

\[
2+2+4=8
\]

ANSWER 6: A

Problem 7:

The original cube has edge length 3 cm, so its volume is

\[
3^3=27
\]

Each smaller cube has whole-number side length. Since the smaller cubes are not all the same size, we cannot just cut the cube into all unit cubes.

The possible smaller cube side lengths are 1 cm or 2 cm. A 3 cm cube would be the entire original cube.

Let \(x\) be the number of \(2\times2\times2\) cubes, and let \(y\) be the number of \(1\times1\times1\) cubes.

Then

\[
8x+y=27
\]

The total number of cubes is

\[
N=x+y
\]

From \(y=27-8x\),

\[
N=x+27-8x=27-7x
\]

Since the cubes are not all the same size, we need at least one 2 cm cube, so \(x\ge 1\).

Among the answer choices, only \(20\) fits:

\[
20=27-7\cdot 1
\]

This is possible: put one \(2\times2\times2\) cube in a corner, and fill the remaining volume with 19 unit cubes.

So

\[
N=20
\]

ANSWER 7: E

Problem 8:

Each of the 1200 students takes 5 classes per day, so the total number of student-class enrollments is

\[
1200\cdot 5=6000
\]

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

\[
6000\div 30=200
\]

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

\[
200\div 4=50
\]

The tempting answer \(40\) would come from just doing \(1200\div 30\), but that ignores the fact that each student takes 5 classes.

ANSWER 8: E

Problem 9:

We count times in a 12-hour period whose digits form a palindrome when the colon is ignored.

For one-digit hours \(1,2,\dots,9\), a time looks like

\[
h:ab
\]

Ignoring the colon, this is \(hab\). To be a palindrome, we need

\[
b=h
\]

The tens digit \(a\) of the minutes can be \(0,1,2,3,4,

Problem 1:
We are asked to find the number of ways to distribute 5 different awards to 3 students such that each student receives at least one award. 
This is equivalent to finding the number of surjective (onto) functions from a set of 5 distinct elements to a set of 3 distinct elements. 

First, let's find the total number of ways to distribute the 5 awards to the 3 students without any restrictions. Since each of the 5 awards can go to any of the 3 students, there are $3^5 = 243$ ways.

Next, we use the Principle of Inclusion-Exclusion to subtract the distributions where at least one student receives no awards:
- Number of ways where 1 specific student gets no awards: The awards are distributed among the remaining 2 students. This can be done in $2^5 = 32$ ways. Since there are 3 ways to choose which student gets no awards, this gives $3 \times 32 = 96$ ways.
- Number of ways where 2 specific students get no awards: All awards go to the 1 remaining student. This can be done in $1^5 = 1$ way. There are 3 ways to choose which two students get no awards, giving $3 \times 1 = 3$ ways.

Applying inclusion-exclusion, the number of invalid distributions (where at least one student gets nothing) is:
$96 - 3 = 93$

Subtracting the invalid distributions from the total gives the number of valid distributions:
$243 - 93 = 150$

ANSWER 1: B

Problem 2:
Alina wants to divide the numbers 1 through 9 into 3 groups of 3 cards each, such that the sum of the numbers in each group is the same.
The sum of the numbers from 1 to 9 is:
$1 + 2 + 3 + \dots + 9 = \frac{9 \times 10}{2} = 45$
For the 3 groups to have the same sum, each group must sum to $45 / 3 = 15$.

Let's find all possible groups of 3 numbers from 1 to 9 that sum to 15:
1. $9 + 5 + 1$
2. $9 + 4 + 2$
3. $8 + 6 + 1$
4. $8 + 5 + 2$
5. $8 + 4 + 3$
6. $7 + 6 + 2$
7. $7 + 5 + 3$
8. $6 + 5 + 4$

Notice that the number 1 is only present in two possible groups: $\{9, 5, 1\}$ and $\{8, 6, 1\}$. Thus, any valid division must use one of these two groups.

Case 1: We use the group $\{9, 5, 1\}$.
The remaining numbers are 2, 3, 4, 6, 7, 8. We need to form two more groups of 3 that sum to 15.
Looking at the remaining numbers, the only valid groups that sum to 15 are $\{8, 4, 3\}$ and $\{7, 6, 2\}$. 
This gives our first valid division: $\{9, 5, 1\}, \{8, 4, 3\}, \{7, 6, 2\}$.

Case 2: We use the group $\{8, 6, 1\}$.
The remaining numbers are 2, 3, 4, 5, 7, 9. We need to form two more groups of 3 that sum to 15.
From the remaining numbers, the only valid groups that sum to 15 are $\{9, 4, 2\}$ and $\{7, 5, 3\}$.
This gives our second valid division: $\{8, 6, 1\}, \{9, 4, 2\}, \{7, 5, 3\}$.

Since the number 1 must be in exactly one of these configurations, there are exactly 2 ways to divide the cards.

ANSWER 2: C

Problem 3:
We are asked to find the sum of the digits of the whole number $10^{93} - 93$.
Let's look at a smaller pattern to understand the subtraction:
$10^2 - 93 = 100 - 93 = 07$ (sum of digits = 7)
$10^3 - 93 = 1000 - 93 = 907$ (sum of digits = 16)
$10^4 - 93 = 10000 - 93 = 9907$ (sum of digits = 25)

Notice that $10^n - 93$ results in a number with $(n - 2)$ nines, followed by the digits 0 and 7.
For $10^{93} - 93$, the number of nines will be $93 - 2 = 91$.
The number looks like: $\underbrace{999\dots99}_{91 \text{ times}}07$

The sum of the digits is:
$91 \times 9 + 0 + 7 = 819 + 7 = 826$

ANSWER 3: D

Problem 4:
Let $P$ be the total number of points scored by the team.
Alexa scored $\frac{1}{4}P$ points, and Brittany scored $\frac{2}{7}P$ points.
Together, they scored:
$\frac{1}{4}P + \frac{2}{7}P = \frac{7P + 8P}{28} = \frac{15}{28}P$

This means the rest of the team scored the remaining $1 - \frac{15}{28} = \frac{13}{28}$ of the total points.
The rest of the team consists of Chelsea (who scored 15 points) and 7 other members. Let $X$ be the total points scored by the other 7 members.
So, $\frac{13}{28}P = 15 + X$.

For $P$ to be a whole number, $15 + X$ must be a multiple of 13.
We are given that none of the other 7 team members scored more than 2 points. Therefore, the maximum points they could have scored together is $7 \times 2 = 14$.
So, $0 \le X \le 14$.
This means $15 \le 15 + X \le 29$.

The only multiple of 13 in the range from 15 to 29 is 26.
Thus, $15 + X = 26$, which gives $X = 11$.
(This also perfectly yields $P = 56$ total points).

ANSWER 4: B

Problem 5:
We are given that $a, b,$ and $c$ are nonzero real numbers and $a + b + c = 0$. We want to find the possible values of:
$\frac{a}{|a|} + \frac{b}{|b|} + \frac{c}{|c|} + \frac{abc}{|abc|}$

The expression $\frac{x}{|x|}$ evaluates to $1$ if $x > 0$ and $-1$ if $x < 0$.
Since $a + b + c = 0$ and none of them are zero, they cannot all be positive or all be negative. This leaves two possible cases for their signs:

Case 1: Two are positive and one is negative (e.g., $a>0, b>0, c<0$).
The first three terms will be $1 + 1 + (-1) = 1$.
The product $abc$ will be positive $\times$ positive $\times$ negative = negative, so the fourth term is $-1$.
Total sum = $1 - 1 = 0$.

Case 2: One is positive and two are negative (e.g., $a>0, b<0, c<0$).
The first three terms will be $1 + (-1) + (-1) = -1$.
The product $abc$ will be positive $\times$ negative $\times$ negative = positive, so the fourth term is $1$.
Total sum = $-1 + 1 = 0$.

In all possible cases, the expression evaluates to 0.

ANSWER 5: A

Problem 6:
Let the 15 integers be an arithmetic sequence with a common difference $d$. 
We are given:
$1 \le a_1 \le 10$
$13 \le a_2 \le 20$
$241 \le a_{15} \le 250$

The terms can be written as $a_n = a_1 + (n-1)d$.
For $a_{15}$, we have $a_{15} = a_1 + 14d$.
Substituting bounds for $a_1$ and $a_{15}$:
$241 \le a_1 + 14d \le 250$
Since $1 \le a_1 \le 10$, we can estimate $14d$:
$241 - 10 \le 14d \le 250 - 1 \implies 231 \le 14d \le 249$
Dividing by 14 gives $16.5 \le d \le 17.78$. Since $d$ must be an integer, $d = 17$.

Now we substitute $d = 17$ into the inequalities:
$a_2 = a_1 + 17$. Sinc

## Step 1: Problem 1 - Restate the problem
We need to find the number of different ways to distribute 5 different awards to 3 students, with each student receiving at least one award.

## Step 2: Problem 1 - Calculate the total number of ways to distribute the awards without restrictions
First, we consider distributing the awards without any restrictions. For each award, there are 3 choices of students, so the total number of ways to distribute 5 awards is $3^5 = 243$.

## Step 3: Problem 1 - Apply the Principle of Inclusion-Exclusion to find the number of ways where at least one student receives no award
To ensure each student gets at least one award, we'll use the Principle of Inclusion-Exclusion. We calculate the number of ways to distribute the awards where at least one student gets no award. There are $3$ ways to choose a student to not receive an award, and for each such choice, there are $2^5$ ways to distribute the awards among the remaining 2 students. So, there are $3 \cdot 2^5 = 96$ ways. However, this counts cases where 2 students receive no awards twice, so we must adjust for that.

## Step 4: Problem 1 - Continue applying the Principle of Inclusion-Exclusion
There are $3$ ways to choose 2 students to not receive an award, and for each such choice, there's $1^5 = 1$ way for the remaining student to receive all awards. So, there are $3 \cdot 1 = 3$ such cases. The total number of ways where at least one student gets no award is $3 \cdot 2^5 - 3 \cdot 1 = 96 - 3 = 93$.

## Step 5: Problem 1 - Calculate the final number of ways to distribute the awards
The number of ways to distribute the awards such that each student gets at least one is $3^5 - 93 = 243 - 93 = 150$.

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

## Step 7: Problem 2 - Restate the problem
We are to determine the number of ways Alina can divide the numbers 1 through 9 into 3 groups of 3 such that the sum of the numbers in each group is the same.

## Step 8: Problem 2 - Calculate the total sum of the numbers and the required sum for each group
The sum of the numbers 1 through 9 is $1+2+3+...+9 = \frac{9(9+1)}{2} = 45$. For 3 groups to have the same sum, each group's sum must be $45 / 3 = 15$.

## Step 9: Problem 2 - Analyze the possible combinations for the groups
To achieve a sum of 15 with 3 numbers, we need to consider the combinations. The highest number is 9, and to reach 15, the other two numbers must sum to 6. Possible pairs with 9 are (9,5,1), (9,4,2), and (9,3,3), but since we can't repeat numbers, (9,3,3) is not valid. We continue this process for other high numbers.

## Step 10: Problem 2 - Enumerate valid combinations
Valid sets that sum to 15 include (9,5,1), (9,4,2), (8,6,1), (8,5,2), (8,4,3), (7,6,2), (7,5,3), (7,4,4) - which is not valid due to repetition, and (6,5,4). We need to find combinations of 3 such numbers that use each number from 1 to 9 exactly once.

## Step 11: Problem 2 - Check for valid groupings
After examining possible combinations and considering the restrictions, we find that there are limited ways to form 3 groups. One valid grouping is (9,2,4), (8,3,4) - which doesn't work because it repeats 4, and (7,6,2), (7,5,3), (8,1,6), (9,5,1), (7,4,4) - again, not valid. A valid example is (9,5,1), (8,6,1) - not valid, (8,4,3), (7,6,2), (7,5,3), and (9,4,2), (8,6,1), (7,5,3). Upon careful examination, one valid set is (9,4,2), (8,7,0) - not valid because 0 is not in the list, but (8,6,1), (7,5,3), and (9,4,2) is a valid grouping: {9,4,2}, {8,6,1}, {7,5,3}. Another is {9,5,1}, {8,4,3}, {7,6,2}.

## Step 12: Problem 2 - Count the distinct ways to form the groups
There are 2 distinct ways to form the groups as shown by the examples.

## Step 13: Problem 2 - State the answer
ANSWER 2: C

## Step 14: Problem 3 - Restate the problem
We need to find the sum of the digits when $10^{93} - 93$ is expressed as a whole number.

## Step 15: Problem 3 - Analyze the number $10^{93}$
$10^{93}$ is a 1 followed by 93 zeros.

## Step 16: Problem 3 - Subtract 93 from $10^{93}$
Subtracting 93 from $10^{93}$ gives us a number that is 93 less than $10^{93}$. This results in a number with 93 digits, where the last two digits are 07, and the rest are 9s except for the third last digit which is 9 - 1 = 8 (due to borrowing), but since we're subtracting from a number ending in 00, it actually results in ...9907.

## Step 17: Problem 3 - Calculate the sum of the digits
The number $10^{93} - 93$ is ...9999907. It has 91 nines, an 8 (from the subtraction affecting the hundreds place due to borrowing), a 0, and a 7. So, the sum is $91*9 + 8 + 0 + 7$.

## Step 18: Problem 3 - Perform the sum of the digits calculation
$91*9 = 819$, $819 + 8 + 7 = 834$.

## Step 19: Problem 3 - State the answer
ANSWER 3: E

## Step 20: Problem 4 - Restate the problem
We need to find the total number of points scored by the other 7 team members in a basketball game given that Alexa scored 1/4 of the points, Brittany scored 2/7, and Chelsea scored 15 points.

## Step 21: Problem 4 - Set up an equation for the total points
Let $T$ be the total points. Then, $\frac{1}{4}T + \frac{2}{7}T + 15 + x = T$, where $x$ is the total points scored by the other 7 team members.

## Step 22: Problem 4 - Simplify the equation
$\frac{7T + 8T}{28} + 15 + x = T$, so $\frac{15T}{28} + 15 + x = T$.

## Step 23: Problem 4 - Solve for $x$
$T - \frac{15T}{28} = 15 + x$, which simplifies to $\frac{13T}{28} = 15 + x$. Thus, $T = \frac{28}{13}(15 + x)$.

## Step 24: Problem 4 - Use the information that none of the other 7 team members scored more than 2 points
$x \leq 14$ because $7 \times 2 = 14$.

## Step 25: Problem 4 - Test possible values of $T$ and $x$
Since $T$ must be a whole number, and given that $\frac{1}{4}T$ and $\frac{2}{7}T$ are also whole numbers, $T$ must be divisible by both 4 and 7, hence divisible by 28.

## Step 26: Problem 4 - Find the appropriate $T$
Let's start with the smallest $T$ that is divisible by 28 and satisfies the given condi

Problem 1:

We are distributing 5 **different** awards to 3 students, with each student receiving at least one award.

First count all possible distributions with no restriction:

- Each of 5 awards can go to any of 3 students.
- Total: \(3^5=243\)

Now subtract distributions where at least one student gets no award.

If a particular student gets no award, then all 5 awards go to the other 2 students:

\[
2^5=32
\]

There are 3 choices for which student gets no award, so subtract:

\[
3\cdot 32=96
\]

But distributions where two students get no awards were subtracted twice, so add them back. If only one student gets all awards, there are 3 such distributions:

\[
243-96+3=150
\]

So the number of ways is \(150\).

ANSWER 1: B

Problem 2:

The numbers are \(1,2,\dots,9\), whose total sum is

\[
1+2+\cdots+9=45
\]

They are divided into 3 groups with equal sums, so each group must sum to

\[
45\div 3=15
\]

Now list triples from \(1\) to \(9\) that sum to 15:

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

Now form complete partitions into 3 disjoint triples.

If the group containing 1 is \(\{1,5,9\}\), the remaining numbers are

\[
2,3,4,6,7,8
\]

These form

\[
\{2,6,7\},\{3,4,8\}
\]

So one partition is

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

If the group containing 1 is \(\{1,6,8\}\), the remaining numbers are

\[
2,3,4,5,7,9
\]

These form

\[
\{2,4,9\},\{3,5,7\}
\]

So another partition is

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

There are exactly 2 ways.

ANSWER 2: C

Problem 3:

We need the sum of the digits of

\[
10^{93}-93
\]

Think of subtracting 93 from a 1 followed by 93 zeros.

A useful way:

\[
10^{93}-93=(10^{93}-100)+7
\]

Now \(10^{93}-100\) is a number with 91 nines followed by two zeros:

\[
999\cdots 99900
\]

with 91 nines. Adding 7 gives

\[
999\cdots 99907
\]

So the digit sum is

\[
91\cdot 9+0+7=819+7=826
\]

ANSWER 3: D

Problem 4:

Let the total number of points be \(T\).

Alexa scored

\[
\frac14T
\]

Brittany scored

\[
\frac27T
\]

Chelsea scored 15 points.

The fraction of the total scored by Alexa and Brittany together is

\[
\frac14+\frac27=\frac7{28}+\frac8{28}=\frac{15}{28}
\]

So the remaining fraction is

\[
1-\frac{15}{28}=\frac{13}{28}
\]

Chelsea plus the other 7 players scored

\[
\frac{13}{28}T
\]

Since \(\frac14T\) and \(\frac27T\) are whole numbers, \(T\) must be a multiple of 28. Let

\[
T=28k
\]

Then Chelsea plus the other players scored

\[
\frac{13}{28}\cdot 28k=13k
\]

So the other 7 players scored

\[
13k-15
\]

Each of the other 7 players scored at most 2 points, so together they scored at most

\[
7\cdot 2=14
\]

Thus

\[
0\le 13k-15\le 14
\]

Testing positive integers \(k\):

- \(k=1\): \(13-15=-2\), impossible.
- \(k=2\): \(26-15=11\), possible.
- \(k=3\): \(39-15=24\), too large.

So the other 7 players scored 11 points.

ANSWER 4: B

Problem 5:

The expression is

\[
\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}
\]

Each fraction is just the sign of the number:

\[
\frac{x}{|x|}=
\begin{cases}
1, & x>0\\
-1, & x<0
\end{cases}
\]

Since \(a+b+c=0\) and none of \(a,b,c\) is zero, they cannot all be positive or all be negative.

So there are two cases.

Case 1: Two are positive and one is negative.

Then

\[
\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}=1+1-1=1
\]

The product \(abc\) is negative, so

\[
\frac{abc}{|abc|}=-1
\]

Total:

\[
1+(-1)=0
\]

Case 2: One is positive and two are negative.

Then

\[
1-1-1=-1
\]

The product \(abc\) is positive, so

\[
\frac{abc}{|abc|}=1
\]

Total:

\[
-1+1=0
\]

So the only possible value is \(0\).

ANSWER 5: A

Problem 6:

The numbers

\[
a_1,a_2,\dots,a_{15}
\]

are equally spaced integers, so they form an arithmetic sequence with integer common difference \(d\).

Thus

\[
a_{15}=a_1+14d
\]

We know

\[
1\le a_1\le 10
\]

and

\[
241\le a_{15}\le 250
\]

So

\[
a_{15}-a_1
\]

is between

\[
241-10=231
\]

and

\[
250-1=249
\]

But

\[
a_{15}-a_1=14d
\]

So \(14d\) must be a multiple of 14 between 231 and 249. The only such multiple is

\[
238=14\cdot 17
\]

Thus

\[
d=17
\]

Now

\[
a_2=a_1+17
\]

and we are told

\[
13\le a_2\le 20
\]

So

\[
13\le a_1+17\le 20
\]

which gives

\[
-4\le a_1\le 3
\]

Combining with \(1\le a_1\le 10\), we get

\[
a_1=3
\]

Now

\[
a_{14}=a_1+13d=3+13\cdot 17=3+221=224
\]

The digit sum of 224 is

\[
2+2+4=8
\]

ANSWER 6: A

Problem 7:

The original cube has edge length 3 cm, so its volume is

\[
3^3=27
\]

Each smaller cube has whole-number side length. Since the smaller cubes are not all the same size, we cannot just cut the cube into all unit cubes.

The possible smaller cube side lengths are 1 cm or 2 cm. A 3 cm cube would be the entire original cube.

Let \(x\) be the number of \(2\times2\times2\) cubes, and let \(y\) be the number of \(1\times1\times1\) cubes.

Then

\[
8x+y=27
\]

The total number of cubes is

\[
N=x+y
\]

From \(y=27-8x\),

\[
N=x+27-8x=27-7x
\]

Since the cubes are not all the same size, we need at least one 2 cm cube, so \(x\ge 1\).

Among the answer choices, only \(20\) fits:

\[
20=27-7\cdot 1
\]

This is possible: put one \(2\times2\times2\) cube in a corner, and fill the remaining volume with 19 unit cubes.

So

\[
N=20
\]

ANSWER 7: E

Problem 8:

Each of the 1200 students takes 5 classes per day, so the total number of student-class enrollments is

\[
1200\cdot 5=6000
\]

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

\[
6000\div 30=200
\]

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

\[
200\div 4=50
\]

The tempting answer \(40\) would come from just doing \(1200\div 30\), but that ignores the fact that each student takes 5 classes.

ANSWER 8: E

Problem 9:

We count times in a 12-hour period whose digits form a palindrome when the colon is ignored.

For one-digit hours \(1,2,\dots,9\), a time looks like

\[
h:ab
\]

Ignoring the colon, this is \(hab\). To be a palindrome, we need

\[
b=h
\]

The tens digit \(a\) of the minutes can be \(0,1,2,3,4,

Problem 1:
We are asked to find the number of ways to distribute 5 different awards to 3 students such that each student receives at least one award. 
This is equivalent to finding the number of surjective (onto) functions from a set of 5 distinct elements to a set of 3 distinct elements. 

First, let's find the total number of ways to distribute the 5 awards to the 3 students without any restrictions. Since each of the 5 awards can go to any of the 3 students, there are $3^5 = 243$ ways.

Next, we use the Principle of Inclusion-Exclusion to subtract the distributions where at least one student receives no awards:
- Number of ways where 1 specific student gets no awards: The awards are distributed among the remaining 2 students. This can be done in $2^5 = 32$ ways. Since there are 3 ways to choose which student gets no awards, this gives $3 \times 32 = 96$ ways.
- Number of ways where 2 specific students get no awards: All awards go to the 1 remaining student. This can be done in $1^5 = 1$ way. There are 3 ways to choose which two students get no awards, giving $3 \times 1 = 3$ ways.

Applying inclusion-exclusion, the number of invalid distributions (where at least one student gets nothing) is:
$96 - 3 = 93$

Subtracting the invalid distributions from the total gives the number of valid distributions:
$243 - 93 = 150$

ANSWER 1: B

Problem 2:
Alina wants to divide the numbers 1 through 9 into 3 groups of 3 cards each, such that the sum of the numbers in each group is the same.
The sum of the numbers from 1 to 9 is:
$1 + 2 + 3 + \dots + 9 = \frac{9 \times 10}{2} = 45$
For the 3 groups to have the same sum, each group must sum to $45 / 3 = 15$.

Let's find all possible groups of 3 numbers from 1 to 9 that sum to 15:
1. $9 + 5 + 1$
2. $9 + 4 + 2$
3. $8 + 6 + 1$
4. $8 + 5 + 2$
5. $8 + 4 + 3$
6. $7 + 6 + 2$
7. $7 + 5 + 3$
8. $6 + 5 + 4$

Notice that the number 1 is only present in two possible groups: $\{9, 5, 1\}$ and $\{8, 6, 1\}$. Thus, any valid division must use one of these two groups.

Case 1: We use the group $\{9, 5, 1\}$.
The remaining numbers are 2, 3, 4, 6, 7, 8. We need to form two more groups of 3 that sum to 15.
Looking at the remaining numbers, the only valid groups that sum to 15 are $\{8, 4, 3\}$ and $\{7, 6, 2\}$. 
This gives our first valid division: $\{9, 5, 1\}, \{8, 4, 3\}, \{7, 6, 2\}$.

Case 2: We use the group $\{8, 6, 1\}$.
The remaining numbers are 2, 3, 4, 5, 7, 9. We need to form two more groups of 3 that sum to 15.
From the remaining numbers, the only valid groups that sum to 15 are $\{9, 4, 2\}$ and $\{7, 5, 3\}$.
This gives our second valid division: $\{8, 6, 1\}, \{9, 4, 2\}, \{7, 5, 3\}$.

Since the number 1 must be in exactly one of these configurations, there are exactly 2 ways to divide the cards.

ANSWER 2: C

Problem 3:
We are asked to find the sum of the digits of the whole number $10^{93} - 93$.
Let's look at a smaller pattern to understand the subtraction:
$10^2 - 93 = 100 - 93 = 07$ (sum of digits = 7)
$10^3 - 93 = 1000 - 93 = 907$ (sum of digits = 16)
$10^4 - 93 = 10000 - 93 = 9907$ (sum of digits = 25)

Notice that $10^n - 93$ results in a number with $(n - 2)$ nines, followed by the digits 0 and 7.
For $10^{93} - 93$, the number of nines will be $93 - 2 = 91$.
The number looks like: $\underbrace{999\dots99}_{91 \text{ times}}07$

The sum of the digits is:
$91 \times 9 + 0 + 7 = 819 + 7 = 826$

ANSWER 3: D

Problem 4:
Let $P$ be the total number of points scored by the team.
Alexa scored $\frac{1}{4}P$ points, and Brittany scored $\frac{2}{7}P$ points.
Together, they scored:
$\frac{1}{4}P + \frac{2}{7}P = \frac{7P + 8P}{28} = \frac{15}{28}P$

This means the rest of the team scored the remaining $1 - \frac{15}{28} = \frac{13}{28}$ of the total points.
The rest of the team consists of Chelsea (who scored 15 points) and 7 other members. Let $X$ be the total points scored by the other 7 members.
So, $\frac{13}{28}P = 15 + X$.

For $P$ to be a whole number, $15 + X$ must be a multiple of 13.
We are given that none of the other 7 team members scored more than 2 points. Therefore, the maximum points they could have scored together is $7 \times 2 = 14$.
So, $0 \le X \le 14$.
This means $15 \le 15 + X \le 29$.

The only multiple of 13 in the range from 15 to 29 is 26.
Thus, $15 + X = 26$, which gives $X = 11$.
(This also perfectly yields $P = 56$ total points).

ANSWER 4: B

Problem 5:
We are given that $a, b,$ and $c$ are nonzero real numbers and $a + b + c = 0$. We want to find the possible values of:
$\frac{a}{|a|} + \frac{b}{|b|} + \frac{c}{|c|} + \frac{abc}{|abc|}$

The expression $\frac{x}{|x|}$ evaluates to $1$ if $x > 0$ and $-1$ if $x < 0$.
Since $a + b + c = 0$ and none of them are zero, they cannot all be positive or all be negative. This leaves two possible cases for their signs:

Case 1: Two are positive and one is negative (e.g., $a>0, b>0, c<0$).
The first three terms will be $1 + 1 + (-1) = 1$.
The product $abc$ will be positive $\times$ positive $\times$ negative = negative, so the fourth term is $-1$.
Total sum = $1 - 1 = 0$.

Case 2: One is positive and two are negative (e.g., $a>0, b<0, c<0$).
The first three terms will be $1 + (-1) + (-1) = -1$.
The product $abc$ will be positive $\times$ negative $\times$ negative = positive, so the fourth term is $1$.
Total sum = $-1 + 1 = 0$.

In all possible cases, the expression evaluates to 0.

ANSWER 5: A

Problem 6:
Let the 15 integers be an arithmetic sequence with a common difference $d$. 
We are given:
$1 \le a_1 \le 10$
$13 \le a_2 \le 20$
$241 \le a_{15} \le 250$

The terms can be written as $a_n = a_1 + (n-1)d$.
For $a_{15}$, we have $a_{15} = a_1 + 14d$.
Substituting bounds for $a_1$ and $a_{15}$:
$241 \le a_1 + 14d \le 250$
Since $1 \le a_1 \le 10$, we can estimate $14d$:
$241 - 10 \le 14d \le 250 - 1 \implies 231 \le 14d \le 249$
Dividing by 14 gives $16.5 \le d \le 17.78$. Since $d$ must be an integer, $d = 17$.

Now we substitute $d = 17$ into the inequalities:
$a_2 = a_1 + 17$. Sinc

## Step 1: Problem 1 - Restate the problem
We need to find the number of different ways to distribute 5 different awards to 3 students, with each student receiving at least one award.

## Step 2: Problem 1 - Calculate the total number of ways to distribute the awards without restrictions
First, we consider distributing the awards without any restrictions. For each award, there are 3 choices of students, so the total number of ways to distribute 5 awards is $3^5 = 243$.

## Step 3: Problem 1 - Apply the Principle of Inclusion-Exclusion to find the number of ways where at least one student receives no award
To ensure each student gets at least one award, we'll use the Principle of Inclusion-Exclusion. We calculate the number of ways to distribute the awards where at least one student gets no award. There are $3$ ways to choose a student to not receive an award, and for each such choice, there are $2^5$ ways to distribute the awards among the remaining 2 students. So, there are $3 \cdot 2^5 = 96$ ways. However, this counts cases where 2 students receive no awards twice, so we must adjust for that.

## Step 4: Problem 1 - Continue applying the Principle of Inclusion-Exclusion
There are $3$ ways to choose 2 students to not receive an award, and for each such choice, there's $1^5 = 1$ way for the remaining student to receive all awards. So, there are $3 \cdot 1 = 3$ such cases. The total number of ways where at least one student gets no award is $3 \cdot 2^5 - 3 \cdot 1 = 96 - 3 = 93$.

## Step 5: Problem 1 - Calculate the final number of ways to distribute the awards
The number of ways to distribute the awards such that each student gets at least one is $3^5 - 93 = 243 - 93 = 150$.

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

## Step 7: Problem 2 - Restate the problem
We are to determine the number of ways Alina can divide the numbers 1 through 9 into 3 groups of 3 such that the sum of the numbers in each group is the same.

## Step 8: Problem 2 - Calculate the total sum of the numbers and the required sum for each group
The sum of the numbers 1 through 9 is $1+2+3+...+9 = \frac{9(9+1)}{2} = 45$. For 3 groups to have the same sum, each group's sum must be $45 / 3 = 15$.

## Step 9: Problem 2 - Analyze the possible combinations for the groups
To achieve a sum of 15 with 3 numbers, we need to consider the combinations. The highest number is 9, and to reach 15, the other two numbers must sum to 6. Possible pairs with 9 are (9,5,1), (9,4,2), and (9,3,3), but since we can't repeat numbers, (9,3,3) is not valid. We continue this process for other high numbers.

## Step 10: Problem 2 - Enumerate valid combinations
Valid sets that sum to 15 include (9,5,1), (9,4,2), (8,6,1), (8,5,2), (8,4,3), (7,6,2), (7,5,3), (7,4,4) - which is not valid due to repetition, and (6,5,4). We need to find combinations of 3 such numbers that use each number from 1 to 9 exactly once.

## Step 11: Problem 2 - Check for valid groupings
After examining possible combinations and considering the restrictions, we find that there are limited ways to form 3 groups. One valid grouping is (9,2,4), (8,3,4) - which doesn't work because it repeats 4, and (7,6,2), (7,5,3), (8,1,6), (9,5,1), (7,4,4) - again, not valid. A valid example is (9,5,1), (8,6,1) - not valid, (8,4,3), (7,6,2), (7,5,3), and (9,4,2), (8,6,1), (7,5,3). Upon careful examination, one valid set is (9,4,2), (8,7,0) - not valid because 0 is not in the list, but (8,6,1), (7,5,3), and (9,4,2) is a valid grouping: {9,4,2}, {8,6,1}, {7,5,3}. Another is {9,5,1}, {8,4,3}, {7,6,2}.

## Step 12: Problem 2 - Count the distinct ways to form the groups
There are 2 distinct ways to form the groups as shown by the examples.

## Step 13: Problem 2 - State the answer
ANSWER 2: C

## Step 14: Problem 3 - Restate the problem
We need to find the sum of the digits when $10^{93} - 93$ is expressed as a whole number.

## Step 15: Problem 3 - Analyze the number $10^{93}$
$10^{93}$ is a 1 followed by 93 zeros.

## Step 16: Problem 3 - Subtract 93 from $10^{93}$
Subtracting 93 from $10^{93}$ gives us a number that is 93 less than $10^{93}$. This results in a number with 93 digits, where the last two digits are 07, and the rest are 9s except for the third last digit which is 9 - 1 = 8 (due to borrowing), but since we're subtracting from a number ending in 00, it actually results in ...9907.

## Step 17: Problem 3 - Calculate the sum of the digits
The number $10^{93} - 93$ is ...9999907. It has 91 nines, an 8 (from the subtraction affecting the hundreds place due to borrowing), a 0, and a 7. So, the sum is $91*9 + 8 + 0 + 7$.

## Step 18: Problem 3 - Perform the sum of the digits calculation
$91*9 = 819$, $819 + 8 + 7 = 834$.

## Step 19: Problem 3 - State the answer
ANSWER 3: E

## Step 20: Problem 4 - Restate the problem
We need to find the total number of points scored by the other 7 team members in a basketball game given that Alexa scored 1/4 of the points, Brittany scored 2/7, and Chelsea scored 15 points.

## Step 21: Problem 4 - Set up an equation for the total points
Let $T$ be the total points. Then, $\frac{1}{4}T + \frac{2}{7}T + 15 + x = T$, where $x$ is the total points scored by the other 7 team members.

## Step 22: Problem 4 - Simplify the equation
$\frac{7T + 8T}{28} + 15 + x = T$, so $\frac{15T}{28} + 15 + x = T$.

## Step 23: Problem 4 - Solve for $x$
$T - \frac{15T}{28} = 15 + x$, which simplifies to $\frac{13T}{28} = 15 + x$. Thus, $T = \frac{28}{13}(15 + x)$.

## Step 24: Problem 4 - Use the information that none of the other 7 team members scored more than 2 points
$x \leq 14$ because $7 \times 2 = 14$.

## Step 25: Problem 4 - Test possible values of $T$ and $x$
Since $T$ must be a whole number, and given that $\frac{1}{4}T$ and $\frac{2}{7}T$ are also whole numbers, $T$ must be divisible by both 4 and 7, hence divisible by 28.

## Step 26: Problem 4 - Find the appropriate $T$
Let's start with the smallest $T$ that is divisible by 28 and satisfies the given condi

Problem 1:

We are distributing 5 **different** awards to 3 students, with each student receiving at least one award.

First count all possible distributions with no restriction:

- Each of 5 awards can go to any of 3 students.
- Total: \(3^5=243\)

Now subtract distributions where at least one student gets no award.

If a particular student gets no award, then all 5 awards go to the other 2 students:

\[
2^5=32
\]

There are 3 choices for which student gets no award, so subtract:

\[
3\cdot 32=96
\]

But distributions where two students get no awards were subtracted twice, so add them back. If only one student gets all awards, there are 3 such distributions:

\[
243-96+3=150
\]

So the number of ways is \(150\).

ANSWER 1: B

Problem 2:

The numbers are \(1,2,\dots,9\), whose total sum is

\[
1+2+\cdots+9=45
\]

They are divided into 3 groups with equal sums, so each group must sum to

\[
45\div 3=15
\]

Now list triples from \(1\) to \(9\) that sum to 15:

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

Now form complete partitions into 3 disjoint triples.

If the group containing 1 is \(\{1,5,9\}\), the remaining numbers are

\[
2,3,4,6,7,8
\]

These form

\[
\{2,6,7\},\{3,4,8\}
\]

So one partition is

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

If the group containing 1 is \(\{1,6,8\}\), the remaining numbers are

\[
2,3,4,5,7,9
\]

These form

\[
\{2,4,9\},\{3,5,7\}
\]

So another partition is

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

There are exactly 2 ways.

ANSWER 2: C

Problem 3:

We need the sum of the digits of

\[
10^{93}-93
\]

Think of subtracting 93 from a 1 followed by 93 zeros.

A useful way:

\[
10^{93}-93=(10^{93}-100)+7
\]

Now \(10^{93}-100\) is a number with 91 nines followed by two zeros:

\[
999\cdots 99900
\]

with 91 nines. Adding 7 gives

\[
999\cdots 99907
\]

So the digit sum is

\[
91\cdot 9+0+7=819+7=826
\]

ANSWER 3: D

Problem 4:

Let the total number of points be \(T\).

Alexa scored

\[
\frac14T
\]

Brittany scored

\[
\frac27T
\]

Chelsea scored 15 points.

The fraction of the total scored by Alexa and Brittany together is

\[
\frac14+\frac27=\frac7{28}+\frac8{28}=\frac{15}{28}
\]

So the remaining fraction is

\[
1-\frac{15}{28}=\frac{13}{28}
\]

Chelsea plus the other 7 players scored

\[
\frac{13}{28}T
\]

Since \(\frac14T\) and \(\frac27T\) are whole numbers, \(T\) must be a multiple of 28. Let

\[
T=28k
\]

Then Chelsea plus the other players scored

\[
\frac{13}{28}\cdot 28k=13k
\]

So the other 7 players scored

\[
13k-15
\]

Each of the other 7 players scored at most 2 points, so together they scored at most

\[
7\cdot 2=14
\]

Thus

\[
0\le 13k-15\le 14
\]

Testing positive integers \(k\):

- \(k=1\): \(13-15=-2\), impossible.
- \(k=2\): \(26-15=11\), possible.
- \(k=3\): \(39-15=24\), too large.

So the other 7 players scored 11 points.

ANSWER 4: B

Problem 5:

The expression is

\[
\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}
\]

Each fraction is just the sign of the number:

\[
\frac{x}{|x|}=
\begin{cases}
1, & x>0\\
-1, & x<0
\end{cases}
\]

Since \(a+b+c=0\) and none of \(a,b,c\) is zero, they cannot all be positive or all be negative.

So there are two cases.

Case 1: Two are positive and one is negative.

Then

\[
\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}=1+1-1=1
\]

The product \(abc\) is negative, so

\[
\frac{abc}{|abc|}=-1
\]

Total:

\[
1+(-1)=0
\]

Case 2: One is positive and two are negative.

Then

\[
1-1-1=-1
\]

The product \(abc\) is positive, so

\[
\frac{abc}{|abc|}=1
\]

Total:

\[
-1+1=0
\]

So the only possible value is \(0\).

ANSWER 5: A

Problem 6:

The numbers

\[
a_1,a_2,\dots,a_{15}
\]

are equally spaced integers, so they form an arithmetic sequence with integer common difference \(d\).

Thus

\[
a_{15}=a_1+14d
\]

We know

\[
1\le a_1\le 10
\]

and

\[
241\le a_{15}\le 250
\]

So

\[
a_{15}-a_1
\]

is between

\[
241-10=231
\]

and

\[
250-1=249
\]

But

\[
a_{15}-a_1=14d
\]

So \(14d\) must be a multiple of 14 between 231 and 249. The only such multiple is

\[
238=14\cdot 17
\]

Thus

\[
d=17
\]

Now

\[
a_2=a_1+17
\]

and we are told

\[
13\le a_2\le 20
\]

So

\[
13\le a_1+17\le 20
\]

which gives

\[
-4\le a_1\le 3
\]

Combining with \(1\le a_1\le 10\), we get

\[
a_1=3
\]

Now

\[
a_{14}=a_1+13d=3+13\cdot 17=3+221=224
\]

The digit sum of 224 is

\[
2+2+4=8
\]

ANSWER 6: A

Problem 7:

The original cube has edge length 3 cm, so its volume is

\[
3^3=27
\]

Each smaller cube has whole-number side length. Since the smaller cubes are not all the same size, we cannot just cut the cube into all unit cubes.

The possible smaller cube side lengths are 1 cm or 2 cm. A 3 cm cube would be the entire original cube.

Let \(x\) be the number of \(2\times2\times2\) cubes, and let \(y\) be the number of \(1\times1\times1\) cubes.

Then

\[
8x+y=27
\]

The total number of cubes is

\[
N=x+y
\]

From \(y=27-8x\),

\[
N=x+27-8x=27-7x
\]

Since the cubes are not all the same size, we need at least one 2 cm cube, so \(x\ge 1\).

Among the answer choices, only \(20\) fits:

\[
20=27-7\cdot 1
\]

This is possible: put one \(2\times2\times2\) cube in a corner, and fill the remaining volume with 19 unit cubes.

So

\[
N=20
\]

ANSWER 7: E

Problem 8:

Each of the 1200 students takes 5 classes per day, so the total number of student-class enrollments is

\[
1200\cdot 5=6000
\]

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

\[
6000\div 30=200
\]

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

\[
200\div 4=50
\]

The tempting answer \(40\) would come from just doing \(1200\div 30\), but that ignores the fact that each student takes 5 classes.

ANSWER 8: E

Problem 9:

We count times in a 12-hour period whose digits form a palindrome when the colon is ignored.

For one-digit hours \(1,2,\dots,9\), a time looks like

\[
h:ab
\]

Ignoring the colon, this is \(hab\). To be a palindrome, we need

\[
b=h
\]

The tens digit \(a\) of the minutes can be \(0,1,2,3,4,

Problem 1:
We are asked to find the number of ways to distribute 5 different awards to 3 students such that each student receives at least one award. 
This is equivalent to finding the number of surjective (onto) functions from a set of 5 distinct elements to a set of 3 distinct elements. 

First, let's find the total number of ways to distribute the 5 awards to the 3 students without any restrictions. Since each of the 5 awards can go to any of the 3 students, there are $3^5 = 243$ ways.

Next, we use the Principle of Inclusion-Exclusion to subtract the distributions where at least one student receives no awards:
- Number of ways where 1 specific student gets no awards: The awards are distributed among the remaining 2 students. This can be done in $2^5 = 32$ ways. Since there are 3 ways to choose which student gets no awards, this gives $3 \times 32 = 96$ ways.
- Number of ways where 2 specific students get no awards: All awards go to the 1 remaining student. This can be done in $1^5 = 1$ way. There are 3 ways to choose which two students get no awards, giving $3 \times 1 = 3$ ways.

Applying inclusion-exclusion, the number of invalid distributions (where at least one student gets nothing) is:
$96 - 3 = 93$

Subtracting the invalid distributions from the total gives the number of valid distributions:
$243 - 93 = 150$

ANSWER 1: B

Problem 2:
Alina wants to divide the numbers 1 through 9 into 3 groups of 3 cards each, such that the sum of the numbers in each group is the same.
The sum of the numbers from 1 to 9 is:
$1 + 2 + 3 + \dots + 9 = \frac{9 \times 10}{2} = 45$
For the 3 groups to have the same sum, each group must sum to $45 / 3 = 15$.

Let's find all possible groups of 3 numbers from 1 to 9 that sum to 15:
1. $9 + 5 + 1$
2. $9 + 4 + 2$
3. $8 + 6 + 1$
4. $8 + 5 + 2$
5. $8 + 4 + 3$
6. $7 + 6 + 2$
7. $7 + 5 + 3$
8. $6 + 5 + 4$

Notice that the number 1 is only present in two possible groups: $\{9, 5, 1\}$ and $\{8, 6, 1\}$. Thus, any valid division must use one of these two groups.

Case 1: We use the group $\{9, 5, 1\}$.
The remaining numbers are 2, 3, 4, 6, 7, 8. We need to form two more groups of 3 that sum to 15.
Looking at the remaining numbers, the only valid groups that sum to 15 are $\{8, 4, 3\}$ and $\{7, 6, 2\}$. 
This gives our first valid division: $\{9, 5, 1\}, \{8, 4, 3\}, \{7, 6, 2\}$.

Case 2: We use the group $\{8, 6, 1\}$.
The remaining numbers are 2, 3, 4, 5, 7, 9. We need to form two more groups of 3 that sum to 15.
From the remaining numbers, the only valid groups that sum to 15 are $\{9, 4, 2\}$ and $\{7, 5, 3\}$.
This gives our second valid division: $\{8, 6, 1\}, \{9, 4, 2\}, \{7, 5, 3\}$.

Since the number 1 must be in exactly one of these configurations, there are exactly 2 ways to divide the cards.

ANSWER 2: C

Problem 3:
We are asked to find the sum of the digits of the whole number $10^{93} - 93$.
Let's look at a smaller pattern to understand the subtraction:
$10^2 - 93 = 100 - 93 = 07$ (sum of digits = 7)
$10^3 - 93 = 1000 - 93 = 907$ (sum of digits = 16)
$10^4 - 93 = 10000 - 93 = 9907$ (sum of digits = 25)

Notice that $10^n - 93$ results in a number with $(n - 2)$ nines, followed by the digits 0 and 7.
For $10^{93} - 93$, the number of nines will be $93 - 2 = 91$.
The number looks like: $\underbrace{999\dots99}_{91 \text{ times}}07$

The sum of the digits is:
$91 \times 9 + 0 + 7 = 819 + 7 = 826$

ANSWER 3: D

Problem 4:
Let $P$ be the total number of points scored by the team.
Alexa scored $\frac{1}{4}P$ points, and Brittany scored $\frac{2}{7}P$ points.
Together, they scored:
$\frac{1}{4}P + \frac{2}{7}P = \frac{7P + 8P}{28} = \frac{15}{28}P$

This means the rest of the team scored the remaining $1 - \frac{15}{28} = \frac{13}{28}$ of the total points.
The rest of the team consists of Chelsea (who scored 15 points) and 7 other members. Let $X$ be the total points scored by the other 7 members.
So, $\frac{13}{28}P = 15 + X$.

For $P$ to be a whole number, $15 + X$ must be a multiple of 13.
We are given that none of the other 7 team members scored more than 2 points. Therefore, the maximum points they could have scored together is $7 \times 2 = 14$.
So, $0 \le X \le 14$.
This means $15 \le 15 + X \le 29$.

The only multiple of 13 in the range from 15 to 29 is 26.
Thus, $15 + X = 26$, which gives $X = 11$.
(This also perfectly yields $P = 56$ total points).

ANSWER 4: B

Problem 5:
We are given that $a, b,$ and $c$ are nonzero real numbers and $a + b + c = 0$. We want to find the possible values of:
$\frac{a}{|a|} + \frac{b}{|b|} + \frac{c}{|c|} + \frac{abc}{|abc|}$

The expression $\frac{x}{|x|}$ evaluates to $1$ if $x > 0$ and $-1$ if $x < 0$.
Since $a + b + c = 0$ and none of them are zero, they cannot all be positive or all be negative. This leaves two possible cases for their signs:

Case 1: Two are positive and one is negative (e.g., $a>0, b>0, c<0$).
The first three terms will be $1 + 1 + (-1) = 1$.
The product $abc$ will be positive $\times$ positive $\times$ negative = negative, so the fourth term is $-1$.
Total sum = $1 - 1 = 0$.

Case 2: One is positive and two are negative (e.g., $a>0, b<0, c<0$).
The first three terms will be $1 + (-1) + (-1) = -1$.
The product $abc$ will be positive $\times$ negative $\times$ negative = positive, so the fourth term is $1$.
Total sum = $-1 + 1 = 0$.

In all possible cases, the expression evaluates to 0.

ANSWER 5: A

Problem 6:
Let the 15 integers be an arithmetic sequence with a common difference $d$. 
We are given:
$1 \le a_1 \le 10$
$13 \le a_2 \le 20$
$241 \le a_{15} \le 250$

The terms can be written as $a_n = a_1 + (n-1)d$.
For $a_{15}$, we have $a_{15} = a_1 + 14d$.
Substituting bounds for $a_1$ and $a_{15}$:
$241 \le a_1 + 14d \le 250$
Since $1 \le a_1 \le 10$, we can estimate $14d$:
$241 - 10 \le 14d \le 250 - 1 \implies 231 \le 14d \le 249$
Dividing by 14 gives $16.5 \le d \le 17.78$. Since $d$ must be an integer, $d = 17$.

Now we substitute $d = 17$ into the inequalities:
$a_2 = a_1 + 17$. Sinc

## Step 1: Problem 1 - Restate the problem
We need to find the number of different ways to distribute 5 different awards to 3 students, with each student receiving at least one award.

## Step 2: Problem 1 - Calculate the total number of ways to distribute the awards without restrictions
First, we consider distributing the awards without any restrictions. For each award, there are 3 choices of students, so the total number of ways to distribute 5 awards is $3^5 = 243$.

## Step 3: Problem 1 - Apply the Principle of Inclusion-Exclusion to find the number of ways where at least one student receives no award
To ensure each student gets at least one award, we'll use the Principle of Inclusion-Exclusion. We calculate the number of ways to distribute the awards where at least one student gets no award. There are $3$ ways to choose a student to not receive an award, and for each such choice, there are $2^5$ ways to distribute the awards among the remaining 2 students. So, there are $3 \cdot 2^5 = 96$ ways. However, this counts cases where 2 students receive no awards twice, so we must adjust for that.

## Step 4: Problem 1 - Continue applying the Principle of Inclusion-Exclusion
There are $3$ ways to choose 2 students to not receive an award, and for each such choice, there's $1^5 = 1$ way for the remaining student to receive all awards. So, there are $3 \cdot 1 = 3$ such cases. The total number of ways where at least one student gets no award is $3 \cdot 2^5 - 3 \cdot 1 = 96 - 3 = 93$.

## Step 5: Problem 1 - Calculate the final number of ways to distribute the awards
The number of ways to distribute the awards such that each student gets at least one is $3^5 - 93 = 243 - 93 = 150$.

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

## Step 7: Problem 2 - Restate the problem
We are to determine the number of ways Alina can divide the numbers 1 through 9 into 3 groups of 3 such that the sum of the numbers in each group is the same.

## Step 8: Problem 2 - Calculate the total sum of the numbers and the required sum for each group
The sum of the numbers 1 through 9 is $1+2+3+...+9 = \frac{9(9+1)}{2} = 45$. For 3 groups to have the same sum, each group's sum must be $45 / 3 = 15$.

## Step 9: Problem 2 - Analyze the possible combinations for the groups
To achieve a sum of 15 with 3 numbers, we need to consider the combinations. The highest number is 9, and to reach 15, the other two numbers must sum to 6. Possible pairs with 9 are (9,5,1), (9,4,2), and (9,3,3), but since we can't repeat numbers, (9,3,3) is not valid. We continue this process for other high numbers.

## Step 10: Problem 2 - Enumerate valid combinations
Valid sets that sum to 15 include (9,5,1), (9,4,2), (8,6,1), (8,5,2), (8,4,3), (7,6,2), (7,5,3), (7,4,4) - which is not valid due to repetition, and (6,5,4). We need to find combinations of 3 such numbers that use each number from 1 to 9 exactly once.

## Step 11: Problem 2 - Check for valid groupings
After examining possible combinations and considering the restrictions, we find that there are limited ways to form 3 groups. One valid grouping is (9,2,4), (8,3,4) - which doesn't work because it repeats 4, and (7,6,2), (7,5,3), (8,1,6), (9,5,1), (7,4,4) - again, not valid. A valid example is (9,5,1), (8,6,1) - not valid, (8,4,3), (7,6,2), (7,5,3), and (9,4,2), (8,6,1), (7,5,3). Upon careful examination, one valid set is (9,4,2), (8,7,0) - not valid because 0 is not in the list, but (8,6,1), (7,5,3), and (9,4,2) is a valid grouping: {9,4,2}, {8,6,1}, {7,5,3}. Another is {9,5,1}, {8,4,3}, {7,6,2}.

## Step 12: Problem 2 - Count the distinct ways to form the groups
There are 2 distinct ways to form the groups as shown by the examples.

## Step 13: Problem 2 - State the answer
ANSWER 2: C

## Step 14: Problem 3 - Restate the problem
We need to find the sum of the digits when $10^{93} - 93$ is expressed as a whole number.

## Step 15: Problem 3 - Analyze the number $10^{93}$
$10^{93}$ is a 1 followed by 93 zeros.

## Step 16: Problem 3 - Subtract 93 from $10^{93}$
Subtracting 93 from $10^{93}$ gives us a number that is 93 less than $10^{93}$. This results in a number with 93 digits, where the last two digits are 07, and the rest are 9s except for the third last digit which is 9 - 1 = 8 (due to borrowing), but since we're subtracting from a number ending in 00, it actually results in ...9907.

## Step 17: Problem 3 - Calculate the sum of the digits
The number $10^{93} - 93$ is ...9999907. It has 91 nines, an 8 (from the subtraction affecting the hundreds place due to borrowing), a 0, and a 7. So, the sum is $91*9 + 8 + 0 + 7$.

## Step 18: Problem 3 - Perform the sum of the digits calculation
$91*9 = 819$, $819 + 8 + 7 = 834$.

## Step 19: Problem 3 - State the answer
ANSWER 3: E

## Step 20: Problem 4 - Restate the problem
We need to find the total number of points scored by the other 7 team members in a basketball game given that Alexa scored 1/4 of the points, Brittany scored 2/7, and Chelsea scored 15 points.

## Step 21: Problem 4 - Set up an equation for the total points
Let $T$ be the total points. Then, $\frac{1}{4}T + \frac{2}{7}T + 15 + x = T$, where $x$ is the total points scored by the other 7 team members.

## Step 22: Problem 4 - Simplify the equation
$\frac{7T + 8T}{28} + 15 + x = T$, so $\frac{15T}{28} + 15 + x = T$.

## Step 23: Problem 4 - Solve for $x$
$T - \frac{15T}{28} = 15 + x$, which simplifies to $\frac{13T}{28} = 15 + x$. Thus, $T = \frac{28}{13}(15 + x)$.

## Step 24: Problem 4 - Use the information that none of the other 7 team members scored more than 2 points
$x \leq 14$ because $7 \times 2 = 14$.

## Step 25: Problem 4 - Test possible values of $T$ and $x$
Since $T$ must be a whole number, and given that $\frac{1}{4}T$ and $\frac{2}{7}T$ are also whole numbers, $T$ must be divisible by both 4 and 7, hence divisible by 28.

## Step 26: Problem 4 - Find the appropriate $T$
Let's start with the smallest $T$ that is divisible by 28 and satisfies the given condi

Problem 1:

We are distributing 5 **different** awards to 3 students, with each student receiving at least one award.

First count all possible distributions with no restriction:

- Each of 5 awards can go to any of 3 students.
- Total: \(3^5=243\)

Now subtract distributions where at least one student gets no award.

If a particular student gets no award, then all 5 awards go to the other 2 students:

\[
2^5=32
\]

There are 3 choices for which student gets no award, so subtract:

\[
3\cdot 32=96
\]

But distributions where two students get no awards were subtracted twice, so add them back. If only one student gets all awards, there are 3 such distributions:

\[
243-96+3=150
\]

So the number of ways is \(150\).

ANSWER 1: B

Problem 2:

The numbers are \(1,2,\dots,9\), whose total sum is

\[
1+2+\cdots+9=45
\]

They are divided into 3 groups with equal sums, so each group must sum to

\[
45\div 3=15
\]

Now list triples from \(1\) to \(9\) that sum to 15:

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

Now form complete partitions into 3 disjoint triples.

If the group containing 1 is \(\{1,5,9\}\), the remaining numbers are

\[
2,3,4,6,7,8
\]

These form

\[
\{2,6,7\},\{3,4,8\}
\]

So one partition is

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

If the group containing 1 is \(\{1,6,8\}\), the remaining numbers are

\[
2,3,4,5,7,9
\]

These form

\[
\{2,4,9\},\{3,5,7\}
\]

So another partition is

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

There are exactly 2 ways.

ANSWER 2: C

Problem 3:

We need the sum of the digits of

\[
10^{93}-93
\]

Think of subtracting 93 from a 1 followed by 93 zeros.

A useful way:

\[
10^{93}-93=(10^{93}-100)+7
\]

Now \(10^{93}-100\) is a number with 91 nines followed by two zeros:

\[
999\cdots 99900
\]

with 91 nines. Adding 7 gives

\[
999\cdots 99907
\]

So the digit sum is

\[
91\cdot 9+0+7=819+7=826
\]

ANSWER 3: D

Problem 4:

Let the total number of points be \(T\).

Alexa scored

\[
\frac14T
\]

Brittany scored

\[
\frac27T
\]

Chelsea scored 15 points.

The fraction of the total scored by Alexa and Brittany together is

\[
\frac14+\frac27=\frac7{28}+\frac8{28}=\frac{15}{28}
\]

So the remaining fraction is

\[
1-\frac{15}{28}=\frac{13}{28}
\]

Chelsea plus the other 7 players scored

\[
\frac{13}{28}T
\]

Since \(\frac14T\) and \(\frac27T\) are whole numbers, \(T\) must be a multiple of 28. Let

\[
T=28k
\]

Then Chelsea plus the other players scored

\[
\frac{13}{28}\cdot 28k=13k
\]

So the other 7 players scored

\[
13k-15
\]

Each of the other 7 players scored at most 2 points, so together they scored at most

\[
7\cdot 2=14
\]

Thus

\[
0\le 13k-15\le 14
\]

Testing positive integers \(k\):

- \(k=1\): \(13-15=-2\), impossible.
- \(k=2\): \(26-15=11\), possible.
- \(k=3\): \(39-15=24\), too large.

So the other 7 players scored 11 points.

ANSWER 4: B

Problem 5:

The expression is

\[
\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}
\]

Each fraction is just the sign of the number:

\[
\frac{x}{|x|}=
\begin{cases}
1, & x>0\\
-1, & x<0
\end{cases}
\]

Since \(a+b+c=0\) and none of \(a,b,c\) is zero, they cannot all be positive or all be negative.

So there are two cases.

Case 1: Two are positive and one is negative.

Then

\[
\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}=1+1-1=1
\]

The product \(abc\) is negative, so

\[
\frac{abc}{|abc|}=-1
\]

Total:

\[
1+(-1)=0
\]

Case 2: One is positive and two are negative.

Then

\[
1-1-1=-1
\]

The product \(abc\) is positive, so

\[
\frac{abc}{|abc|}=1
\]

Total:

\[
-1+1=0
\]

So the only possible value is \(0\).

ANSWER 5: A

Problem 6:

The numbers

\[
a_1,a_2,\dots,a_{15}
\]

are equally spaced integers, so they form an arithmetic sequence with integer common difference \(d\).

Thus

\[
a_{15}=a_1+14d
\]

We know

\[
1\le a_1\le 10
\]

and

\[
241\le a_{15}\le 250
\]

So

\[
a_{15}-a_1
\]

is between

\[
241-10=231
\]

and

\[
250-1=249
\]

But

\[
a_{15}-a_1=14d
\]

So \(14d\) must be a multiple of 14 between 231 and 249. The only such multiple is

\[
238=14\cdot 17
\]

Thus

\[
d=17
\]

Now

\[
a_2=a_1+17
\]

and we are told

\[
13\le a_2\le 20
\]

So

\[
13\le a_1+17\le 20
\]

which gives

\[
-4\le a_1\le 3
\]

Combining with \(1\le a_1\le 10\), we get

\[
a_1=3
\]

Now

\[
a_{14}=a_1+13d=3+13\cdot 17=3+221=224
\]

The digit sum of 224 is

\[
2+2+4=8
\]

ANSWER 6: A

Problem 7:

The original cube has edge length 3 cm, so its volume is

\[
3^3=27
\]

Each smaller cube has whole-number side length. Since the smaller cubes are not all the same size, we cannot just cut the cube into all unit cubes.

The possible smaller cube side lengths are 1 cm or 2 cm. A 3 cm cube would be the entire original cube.

Let \(x\) be the number of \(2\times2\times2\) cubes, and let \(y\) be the number of \(1\times1\times1\) cubes.

Then

\[
8x+y=27
\]

The total number of cubes is

\[
N=x+y
\]

From \(y=27-8x\),

\[
N=x+27-8x=27-7x
\]

Since the cubes are not all the same size, we need at least one 2 cm cube, so \(x\ge 1\).

Among the answer choices, only \(20\) fits:

\[
20=27-7\cdot 1
\]

This is possible: put one \(2\times2\times2\) cube in a corner, and fill the remaining volume with 19 unit cubes.

So

\[
N=20
\]

ANSWER 7: E

Problem 8:

Each of the 1200 students takes 5 classes per day, so the total number of student-class enrollments is

\[
1200\cdot 5=6000
\]

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

\[
6000\div 30=200
\]

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

\[
200\div 4=50
\]

The tempting answer \(40\) would come from just doing \(1200\div 30\), but that ignores the fact that each student takes 5 classes.

ANSWER 8: E

Problem 9:

We count times in a 12-hour period whose digits form a palindrome when the colon is ignored.

For one-digit hours \(1,2,\dots,9\), a time looks like

\[
h:ab
\]

Ignoring the colon, this is \(hab\). To be a palindrome, we need

\[
b=h
\]

The tens digit \(a\) of the minutes can be \(0,1,2,3,4,

Problem 1:
We are asked to find the number of ways to distribute 5 different awards to 3 students such that each student receives at least one award. 
This is equivalent to finding the number of surjective (onto) functions from a set of 5 distinct elements to a set of 3 distinct elements. 

First, let's find the total number of ways to distribute the 5 awards to the 3 students without any restrictions. Since each of the 5 awards can go to any of the 3 students, there are $3^5 = 243$ ways.

Next, we use the Principle of Inclusion-Exclusion to subtract the distributions where at least one student receives no awards:
- Number of ways where 1 specific student gets no awards: The awards are distributed among the remaining 2 students. This can be done in $2^5 = 32$ ways. Since there are 3 ways to choose which student gets no awards, this gives $3 \times 32 = 96$ ways.
- Number of ways where 2 specific students get no awards: All awards go to the 1 remaining student. This can be done in $1^5 = 1$ way. There are 3 ways to choose which two students get no awards, giving $3 \times 1 = 3$ ways.

Applying inclusion-exclusion, the number of invalid distributions (where at least one student gets nothing) is:
$96 - 3 = 93$

Subtracting the invalid distributions from the total gives the number of valid distributions:
$243 - 93 = 150$

ANSWER 1: B

Problem 2:
Alina wants to divide the numbers 1 through 9 into 3 groups of 3 cards each, such that the sum of the numbers in each group is the same.
The sum of the numbers from 1 to 9 is:
$1 + 2 + 3 + \dots + 9 = \frac{9 \times 10}{2} = 45$
For the 3 groups to have the same sum, each group must sum to $45 / 3 = 15$.

Let's find all possible groups of 3 numbers from 1 to 9 that sum to 15:
1. $9 + 5 + 1$
2. $9 + 4 + 2$
3. $8 + 6 + 1$
4. $8 + 5 + 2$
5. $8 + 4 + 3$
6. $7 + 6 + 2$
7. $7 + 5 + 3$
8. $6 + 5 + 4$

Notice that the number 1 is only present in two possible groups: $\{9, 5, 1\}$ and $\{8, 6, 1\}$. Thus, any valid division must use one of these two groups.

Case 1: We use the group $\{9, 5, 1\}$.
The remaining numbers are 2, 3, 4, 6, 7, 8. We need to form two more groups of 3 that sum to 15.
Looking at the remaining numbers, the only valid groups that sum to 15 are $\{8, 4, 3\}$ and $\{7, 6, 2\}$. 
This gives our first valid division: $\{9, 5, 1\}, \{8, 4, 3\}, \{7, 6, 2\}$.

Case 2: We use the group $\{8, 6, 1\}$.
The remaining numbers are 2, 3, 4, 5, 7, 9. We need to form two more groups of 3 that sum to 15.
From the remaining numbers, the only valid groups that sum to 15 are $\{9, 4, 2\}$ and $\{7, 5, 3\}$.
This gives our second valid division: $\{8, 6, 1\}, \{9, 4, 2\}, \{7, 5, 3\}$.

Since the number 1 must be in exactly one of these configurations, there are exactly 2 ways to divide the cards.

ANSWER 2: C

Problem 3:
We are asked to find the sum of the digits of the whole number $10^{93} - 93$.
Let's look at a smaller pattern to understand the subtraction:
$10^2 - 93 = 100 - 93 = 07$ (sum of digits = 7)
$10^3 - 93 = 1000 - 93 = 907$ (sum of digits = 16)
$10^4 - 93 = 10000 - 93 = 9907$ (sum of digits = 25)

Notice that $10^n - 93$ results in a number with $(n - 2)$ nines, followed by the digits 0 and 7.
For $10^{93} - 93$, the number of nines will be $93 - 2 = 91$.
The number looks like: $\underbrace{999\dots99}_{91 \text{ times}}07$

The sum of the digits is:
$91 \times 9 + 0 + 7 = 819 + 7 = 826$

ANSWER 3: D

Problem 4:
Let $P$ be the total number of points scored by the team.
Alexa scored $\frac{1}{4}P$ points, and Brittany scored $\frac{2}{7}P$ points.
Together, they scored:
$\frac{1}{4}P + \frac{2}{7}P = \frac{7P + 8P}{28} = \frac{15}{28}P$

This means the rest of the team scored the remaining $1 - \frac{15}{28} = \frac{13}{28}$ of the total points.
The rest of the team consists of Chelsea (who scored 15 points) and 7 other members. Let $X$ be the total points scored by the other 7 members.
So, $\frac{13}{28}P = 15 + X$.

For $P$ to be a whole number, $15 + X$ must be a multiple of 13.
We are given that none of the other 7 team members scored more than 2 points. Therefore, the maximum points they could have scored together is $7 \times 2 = 14$.
So, $0 \le X \le 14$.
This means $15 \le 15 + X \le 29$.

The only multiple of 13 in the range from 15 to 29 is 26.
Thus, $15 + X = 26$, which gives $X = 11$.
(This also perfectly yields $P = 56$ total points).

ANSWER 4: B

Problem 5:
We are given that $a, b,$ and $c$ are nonzero real numbers and $a + b + c = 0$. We want to find the possible values of:
$\frac{a}{|a|} + \frac{b}{|b|} + \frac{c}{|c|} + \frac{abc}{|abc|}$

The expression $\frac{x}{|x|}$ evaluates to $1$ if $x > 0$ and $-1$ if $x < 0$.
Since $a + b + c = 0$ and none of them are zero, they cannot all be positive or all be negative. This leaves two possible cases for their signs:

Case 1: Two are positive and one is negative (e.g., $a>0, b>0, c<0$).
The first three terms will be $1 + 1 + (-1) = 1$.
The product $abc$ will be positive $\times$ positive $\times$ negative = negative, so the fourth term is $-1$.
Total sum = $1 - 1 = 0$.

Case 2: One is positive and two are negative (e.g., $a>0, b<0, c<0$).
The first three terms will be $1 + (-1) + (-1) = -1$.
The product $abc$ will be positive $\times$ negative $\times$ negative = positive, so the fourth term is $1$.
Total sum = $-1 + 1 = 0$.

In all possible cases, the expression evaluates to 0.

ANSWER 5: A

Problem 6:
Let the 15 integers be an arithmetic sequence with a common difference $d$. 
We are given:
$1 \le a_1 \le 10$
$13 \le a_2 \le 20$
$241 \le a_{15} \le 250$

The terms can be written as $a_n = a_1 + (n-1)d$.
For $a_{15}$, we have $a_{15} = a_1 + 14d$.
Substituting bounds for $a_1$ and $a_{15}$:
$241 \le a_1 + 14d \le 250$
Since $1 \le a_1 \le 10$, we can estimate $14d$:
$241 - 10 \le 14d \le 250 - 1 \implies 231 \le 14d \le 249$
Dividing by 14 gives $16.5 \le d \le 17.78$. Since $d$ must be an integer, $d = 17$.

Now we substitute $d = 17$ into the inequalities:
$a_2 = a_1 + 17$. Sinc

## Step 1: Problem 1 - Restate the problem
We need to find the number of different ways to distribute 5 different awards to 3 students, with each student receiving at least one award.

## Step 2: Problem 1 - Calculate the total number of ways to distribute the awards without restrictions
First, we consider distributing the awards without any restrictions. For each award, there are 3 choices of students, so the total number of ways to distribute 5 awards is $3^5 = 243$.

## Step 3: Problem 1 - Apply the Principle of Inclusion-Exclusion to find the number of ways where at least one student receives no award
To ensure each student gets at least one award, we'll use the Principle of Inclusion-Exclusion. We calculate the number of ways to distribute the awards where at least one student gets no award. There are $3$ ways to choose a student to not receive an award, and for each such choice, there are $2^5$ ways to distribute the awards among the remaining 2 students. So, there are $3 \cdot 2^5 = 96$ ways. However, this counts cases where 2 students receive no awards twice, so we must adjust for that.

## Step 4: Problem 1 - Continue applying the Principle of Inclusion-Exclusion
There are $3$ ways to choose 2 students to not receive an award, and for each such choice, there's $1^5 = 1$ way for the remaining student to receive all awards. So, there are $3 \cdot 1 = 3$ such cases. The total number of ways where at least one student gets no award is $3 \cdot 2^5 - 3 \cdot 1 = 96 - 3 = 93$.

## Step 5: Problem 1 - Calculate the final number of ways to distribute the awards
The number of ways to distribute the awards such that each student gets at least one is $3^5 - 93 = 243 - 93 = 150$.

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

## Step 7: Problem 2 - Restate the problem
We are to determine the number of ways Alina can divide the numbers 1 through 9 into 3 groups of 3 such that the sum of the numbers in each group is the same.

## Step 8: Problem 2 - Calculate the total sum of the numbers and the required sum for each group
The sum of the numbers 1 through 9 is $1+2+3+...+9 = \frac{9(9+1)}{2} = 45$. For 3 groups to have the same sum, each group's sum must be $45 / 3 = 15$.

## Step 9: Problem 2 - Analyze the possible combinations for the groups
To achieve a sum of 15 with 3 numbers, we need to consider the combinations. The highest number is 9, and to reach 15, the other two numbers must sum to 6. Possible pairs with 9 are (9,5,1), (9,4,2), and (9,3,3), but since we can't repeat numbers, (9,3,3) is not valid. We continue this process for other high numbers.

## Step 10: Problem 2 - Enumerate valid combinations
Valid sets that sum to 15 include (9,5,1), (9,4,2), (8,6,1), (8,5,2), (8,4,3), (7,6,2), (7,5,3), (7,4,4) - which is not valid due to repetition, and (6,5,4). We need to find combinations of 3 such numbers that use each number from 1 to 9 exactly once.

## Step 11: Problem 2 - Check for valid groupings
After examining possible combinations and considering the restrictions, we find that there are limited ways to form 3 groups. One valid grouping is (9,2,4), (8,3,4) - which doesn't work because it repeats 4, and (7,6,2), (7,5,3), (8,1,6), (9,5,1), (7,4,4) - again, not valid. A valid example is (9,5,1), (8,6,1) - not valid, (8,4,3), (7,6,2), (7,5,3), and (9,4,2), (8,6,1), (7,5,3). Upon careful examination, one valid set is (9,4,2), (8,7,0) - not valid because 0 is not in the list, but (8,6,1), (7,5,3), and (9,4,2) is a valid grouping: {9,4,2}, {8,6,1}, {7,5,3}. Another is {9,5,1}, {8,4,3}, {7,6,2}.

## Step 12: Problem 2 - Count the distinct ways to form the groups
There are 2 distinct ways to form the groups as shown by the examples.

## Step 13: Problem 2 - State the answer
ANSWER 2: C

## Step 14: Problem 3 - Restate the problem
We need to find the sum of the digits when $10^{93} - 93$ is expressed as a whole number.

## Step 15: Problem 3 - Analyze the number $10^{93}$
$10^{93}$ is a 1 followed by 93 zeros.

## Step 16: Problem 3 - Subtract 93 from $10^{93}$
Subtracting 93 from $10^{93}$ gives us a number that is 93 less than $10^{93}$. This results in a number with 93 digits, where the last two digits are 07, and the rest are 9s except for the third last digit which is 9 - 1 = 8 (due to borrowing), but since we're subtracting from a number ending in 00, it actually results in ...9907.

## Step 17: Problem 3 - Calculate the sum of the digits
The number $10^{93} - 93$ is ...9999907. It has 91 nines, an 8 (from the subtraction affecting the hundreds place due to borrowing), a 0, and a 7. So, the sum is $91*9 + 8 + 0 + 7$.

## Step 18: Problem 3 - Perform the sum of the digits calculation
$91*9 = 819$, $819 + 8 + 7 = 834$.

## Step 19: Problem 3 - State the answer
ANSWER 3: E

## Step 20: Problem 4 - Restate the problem
We need to find the total number of points scored by the other 7 team members in a basketball game given that Alexa scored 1/4 of the points, Brittany scored 2/7, and Chelsea scored 15 points.

## Step 21: Problem 4 - Set up an equation for the total points
Let $T$ be the total points. Then, $\frac{1}{4}T + \frac{2}{7}T + 15 + x = T$, where $x$ is the total points scored by the other 7 team members.

## Step 22: Problem 4 - Simplify the equation
$\frac{7T + 8T}{28} + 15 + x = T$, so $\frac{15T}{28} + 15 + x = T$.

## Step 23: Problem 4 - Solve for $x$
$T - \frac{15T}{28} = 15 + x$, which simplifies to $\frac{13T}{28} = 15 + x$. Thus, $T = \frac{28}{13}(15 + x)$.

## Step 24: Problem 4 - Use the information that none of the other 7 team members scored more than 2 points
$x \leq 14$ because $7 \times 2 = 14$.

## Step 25: Problem 4 - Test possible values of $T$ and $x$
Since $T$ must be a whole number, and given that $\frac{1}{4}T$ and $\frac{2}{7}T$ are also whole numbers, $T$ must be divisible by both 4 and 7, hence divisible by 28.

## Step 26: Problem 4 - Find the appropriate $T$
Let's start with the smallest $T$ that is divisible by 28 and satisfies the given condi

Problem 1:

We are distributing 5 **different** awards to 3 students, with each student receiving at least one award.

First count all possible distributions with no restriction:

- Each of 5 awards can go to any of 3 students.
- Total: \(3^5=243\)

Now subtract distributions where at least one student gets no award.

If a particular student gets no award, then all 5 awards go to the other 2 students:

\[
2^5=32
\]

There are 3 choices for which student gets no award, so subtract:

\[
3\cdot 32=96
\]

But distributions where two students get no awards were subtracted twice, so add them back. If only one student gets all awards, there are 3 such distributions:

\[
243-96+3=150
\]

So the number of ways is \(150\).

ANSWER 1: B

Problem 2:

The numbers are \(1,2,\dots,9\), whose total sum is

\[
1+2+\cdots+9=45
\]

They are divided into 3 groups with equal sums, so each group must sum to

\[
45\div 3=15
\]

Now list triples from \(1\) to \(9\) that sum to 15:

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

Now form complete partitions into 3 disjoint triples.

If the group containing 1 is \(\{1,5,9\}\), the remaining numbers are

\[
2,3,4,6,7,8
\]

These form

\[
\{2,6,7\},\{3,4,8\}
\]

So one partition is

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

If the group containing 1 is \(\{1,6,8\}\), the remaining numbers are

\[
2,3,4,5,7,9
\]

These form

\[
\{2,4,9\},\{3,5,7\}
\]

So another partition is

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

There are exactly 2 ways.

ANSWER 2: C

Problem 3:

We need the sum of the digits of

\[
10^{93}-93
\]

Think of subtracting 93 from a 1 followed by 93 zeros.

A useful way:

\[
10^{93}-93=(10^{93}-100)+7
\]

Now \(10^{93}-100\) is a number with 91 nines followed by two zeros:

\[
999\cdots 99900
\]

with 91 nines. Adding 7 gives

\[
999\cdots 99907
\]

So the digit sum is

\[
91\cdot 9+0+7=819+7=826
\]

ANSWER 3: D

Problem 4:

Let the total number of points be \(T\).

Alexa scored

\[
\frac14T
\]

Brittany scored

\[
\frac27T
\]

Chelsea scored 15 points.

The fraction of the total scored by Alexa and Brittany together is

\[
\frac14+\frac27=\frac7{28}+\frac8{28}=\frac{15}{28}
\]

So the remaining fraction is

\[
1-\frac{15}{28}=\frac{13}{28}
\]

Chelsea plus the other 7 players scored

\[
\frac{13}{28}T
\]

Since \(\frac14T\) and \(\frac27T\) are whole numbers, \(T\) must be a multiple of 28. Let

\[
T=28k
\]

Then Chelsea plus the other players scored

\[
\frac{13}{28}\cdot 28k=13k
\]

So the other 7 players scored

\[
13k-15
\]

Each of the other 7 players scored at most 2 points, so together they scored at most

\[
7\cdot 2=14
\]

Thus

\[
0\le 13k-15\le 14
\]

Testing positive integers \(k\):

- \(k=1\): \(13-15=-2\), impossible.
- \(k=2\): \(26-15=11\), possible.
- \(k=3\): \(39-15=24\), too large.

So the other 7 players scored 11 points.

ANSWER 4: B

Problem 5:

The expression is

\[
\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}
\]

Each fraction is just the sign of the number:

\[
\frac{x}{|x|}=
\begin{cases}
1, & x>0\\
-1, & x<0
\end{cases}
\]

Since \(a+b+c=0\) and none of \(a,b,c\) is zero, they cannot all be positive or all be negative.

So there are two cases.

Case 1: Two are positive and one is negative.

Then

\[
\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}=1+1-1=1
\]

The product \(abc\) is negative, so

\[
\frac{abc}{|abc|}=-1
\]

Total:

\[
1+(-1)=0
\]

Case 2: One is positive and two are negative.

Then

\[
1-1-1=-1
\]

The product \(abc\) is positive, so

\[
\frac{abc}{|abc|}=1
\]

Total:

\[
-1+1=0
\]

So the only possible value is \(0\).

ANSWER 5: A

Problem 6:

The numbers

\[
a_1,a_2,\dots,a_{15}
\]

are equally spaced integers, so they form an arithmetic sequence with integer common difference \(d\).

Thus

\[
a_{15}=a_1+14d
\]

We know

\[
1\le a_1\le 10
\]

and

\[
241\le a_{15}\le 250
\]

So

\[
a_{15}-a_1
\]

is between

\[
241-10=231
\]

and

\[
250-1=249
\]

But

\[
a_{15}-a_1=14d
\]

So \(14d\) must be a multiple of 14 between 231 and 249. The only such multiple is

\[
238=14\cdot 17
\]

Thus

\[
d=17
\]

Now

\[
a_2=a_1+17
\]

and we are told

\[
13\le a_2\le 20
\]

So

\[
13\le a_1+17\le 20
\]

which gives

\[
-4\le a_1\le 3
\]

Combining with \(1\le a_1\le 10\), we get

\[
a_1=3
\]

Now

\[
a_{14}=a_1+13d=3+13\cdot 17=3+221=224
\]

The digit sum of 224 is

\[
2+2+4=8
\]

ANSWER 6: A

Problem 7:

The original cube has edge length 3 cm, so its volume is

\[
3^3=27
\]

Each smaller cube has whole-number side length. Since the smaller cubes are not all the same size, we cannot just cut the cube into all unit cubes.

The possible smaller cube side lengths are 1 cm or 2 cm. A 3 cm cube would be the entire original cube.

Let \(x\) be the number of \(2\times2\times2\) cubes, and let \(y\) be the number of \(1\times1\times1\) cubes.

Then

\[
8x+y=27
\]

The total number of cubes is

\[
N=x+y
\]

From \(y=27-8x\),

\[
N=x+27-8x=27-7x
\]

Since the cubes are not all the same size, we need at least one 2 cm cube, so \(x\ge 1\).

Among the answer choices, only \(20\) fits:

\[
20=27-7\cdot 1
\]

This is possible: put one \(2\times2\times2\) cube in a corner, and fill the remaining volume with 19 unit cubes.

So

\[
N=20
\]

ANSWER 7: E

Problem 8:

Each of the 1200 students takes 5 classes per day, so the total number of student-class enrollments is

\[
1200\cdot 5=6000
\]

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

\[
6000\div 30=200
\]

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

\[
200\div 4=50
\]

The tempting answer \(40\) would come from just doing \(1200\div 30\), but that ignores the fact that each student takes 5 classes.

ANSWER 8: E

Problem 9:

We count times in a 12-hour period whose digits form a palindrome when the colon is ignored.

For one-digit hours \(1,2,\dots,9\), a time looks like

\[
h:ab
\]

Ignoring the colon, this is \(hab\). To be a palindrome, we need

\[
b=h
\]

The tens digit \(a\) of the minutes can be \(0,1,2,3,4,

Problem 1:
We are asked to find the number of ways to distribute 5 different awards to 3 students such that each student receives at least one award. 
This is equivalent to finding the number of surjective (onto) functions from a set of 5 distinct elements to a set of 3 distinct elements. 

First, let's find the total number of ways to distribute the 5 awards to the 3 students without any restrictions. Since each of the 5 awards can go to any of the 3 students, there are $3^5 = 243$ ways.

Next, we use the Principle of Inclusion-Exclusion to subtract the distributions where at least one student receives no awards:
- Number of ways where 1 specific student gets no awards: The awards are distributed among the remaining 2 students. This can be done in $2^5 = 32$ ways. Since there are 3 ways to choose which student gets no awards, this gives $3 \times 32 = 96$ ways.
- Number of ways where 2 specific students get no awards: All awards go to the 1 remaining student. This can be done in $1^5 = 1$ way. There are 3 ways to choose which two students get no awards, giving $3 \times 1 = 3$ ways.

Applying inclusion-exclusion, the number of invalid distributions (where at least one student gets nothing) is:
$96 - 3 = 93$

Subtracting the invalid distributions from the total gives the number of valid distributions:
$243 - 93 = 150$

ANSWER 1: B

Problem 2:
Alina wants to divide the numbers 1 through 9 into 3 groups of 3 cards each, such that the sum of the numbers in each group is the same.
The sum of the numbers from 1 to 9 is:
$1 + 2 + 3 + \dots + 9 = \frac{9 \times 10}{2} = 45$
For the 3 groups to have the same sum, each group must sum to $45 / 3 = 15$.

Let's find all possible groups of 3 numbers from 1 to 9 that sum to 15:
1. $9 + 5 + 1$
2. $9 + 4 + 2$
3. $8 + 6 + 1$
4. $8 + 5 + 2$
5. $8 + 4 + 3$
6. $7 + 6 + 2$
7. $7 + 5 + 3$
8. $6 + 5 + 4$

Notice that the number 1 is only present in two possible groups: $\{9, 5, 1\}$ and $\{8, 6, 1\}$. Thus, any valid division must use one of these two groups.

Case 1: We use the group $\{9, 5, 1\}$.
The remaining numbers are 2, 3, 4, 6, 7, 8. We need to form two more groups of 3 that sum to 15.
Looking at the remaining numbers, the only valid groups that sum to 15 are $\{8, 4, 3\}$ and $\{7, 6, 2\}$. 
This gives our first valid division: $\{9, 5, 1\}, \{8, 4, 3\}, \{7, 6, 2\}$.

Case 2: We use the group $\{8, 6, 1\}$.
The remaining numbers are 2, 3, 4, 5, 7, 9. We need to form two more groups of 3 that sum to 15.
From the remaining numbers, the only valid groups that sum to 15 are $\{9, 4, 2\}$ and $\{7, 5, 3\}$.
This gives our second valid division: $\{8, 6, 1\}, \{9, 4, 2\}, \{7, 5, 3\}$.

Since the number 1 must be in exactly one of these configurations, there are exactly 2 ways to divide the cards.

ANSWER 2: C

Problem 3:
We are asked to find the sum of the digits of the whole number $10^{93} - 93$.
Let's look at a smaller pattern to understand the subtraction:
$10^2 - 93 = 100 - 93 = 07$ (sum of digits = 7)
$10^3 - 93 = 1000 - 93 = 907$ (sum of digits = 16)
$10^4 - 93 = 10000 - 93 = 9907$ (sum of digits = 25)

Notice that $10^n - 93$ results in a number with $(n - 2)$ nines, followed by the digits 0 and 7.
For $10^{93} - 93$, the number of nines will be $93 - 2 = 91$.
The number looks like: $\underbrace{999\dots99}_{91 \text{ times}}07$

The sum of the digits is:
$91 \times 9 + 0 + 7 = 819 + 7 = 826$

ANSWER 3: D

Problem 4:
Let $P$ be the total number of points scored by the team.
Alexa scored $\frac{1}{4}P$ points, and Brittany scored $\frac{2}{7}P$ points.
Together, they scored:
$\frac{1}{4}P + \frac{2}{7}P = \frac{7P + 8P}{28} = \frac{15}{28}P$

This means the rest of the team scored the remaining $1 - \frac{15}{28} = \frac{13}{28}$ of the total points.
The rest of the team consists of Chelsea (who scored 15 points) and 7 other members. Let $X$ be the total points scored by the other 7 members.
So, $\frac{13}{28}P = 15 + X$.

For $P$ to be a whole number, $15 + X$ must be a multiple of 13.
We are given that none of the other 7 team members scored more than 2 points. Therefore, the maximum points they could have scored together is $7 \times 2 = 14$.
So, $0 \le X \le 14$.
This means $15 \le 15 + X \le 29$.

The only multiple of 13 in the range from 15 to 29 is 26.
Thus, $15 + X = 26$, which gives $X = 11$.
(This also perfectly yields $P = 56$ total points).

ANSWER 4: B

Problem 5:
We are given that $a, b,$ and $c$ are nonzero real numbers and $a + b + c = 0$. We want to find the possible values of:
$\frac{a}{|a|} + \frac{b}{|b|} + \frac{c}{|c|} + \frac{abc}{|abc|}$

The expression $\frac{x}{|x|}$ evaluates to $1$ if $x > 0$ and $-1$ if $x < 0$.
Since $a + b + c = 0$ and none of them are zero, they cannot all be positive or all be negative. This leaves two possible cases for their signs:

Case 1: Two are positive and one is negative (e.g., $a>0, b>0, c<0$).
The first three terms will be $1 + 1 + (-1) = 1$.
The product $abc$ will be positive $\times$ positive $\times$ negative = negative, so the fourth term is $-1$.
Total sum = $1 - 1 = 0$.

Case 2: One is positive and two are negative (e.g., $a>0, b<0, c<0$).
The first three terms will be $1 + (-1) + (-1) = -1$.
The product $abc$ will be positive $\times$ negative $\times$ negative = positive, so the fourth term is $1$.
Total sum = $-1 + 1 = 0$.

In all possible cases, the expression evaluates to 0.

ANSWER 5: A

Problem 6:
Let the 15 integers be an arithmetic sequence with a common difference $d$. 
We are given:
$1 \le a_1 \le 10$
$13 \le a_2 \le 20$
$241 \le a_{15} \le 250$

The terms can be written as $a_n = a_1 + (n-1)d$.
For $a_{15}$, we have $a_{15} = a_1 + 14d$.
Substituting bounds for $a_1$ and $a_{15}$:
$241 \le a_1 + 14d \le 250$
Since $1 \le a_1 \le 10$, we can estimate $14d$:
$241 - 10 \le 14d \le 250 - 1 \implies 231 \le 14d \le 249$
Dividing by 14 gives $16.5 \le d \le 17.78$. Since $d$ must be an integer, $d = 17$.

Now we substitute $d = 17$ into the inequalities:
$a_2 = a_1 + 17$. Sinc

## Step 1: Problem 1 - Restate the problem
We need to find the number of different ways to distribute 5 different awards to 3 students, with each student receiving at least one award.

## Step 2: Problem 1 - Calculate the total number of ways to distribute the awards without restrictions
First, we consider distributing the awards without any restrictions. For each award, there are 3 choices of students, so the total number of ways to distribute 5 awards is $3^5 = 243$.

## Step 3: Problem 1 - Apply the Principle of Inclusion-Exclusion to find the number of ways where at least one student receives no award
To ensure each student gets at least one award, we'll use the Principle of Inclusion-Exclusion. We calculate the number of ways to distribute the awards where at least one student gets no award. There are $3$ ways to choose a student to not receive an award, and for each such choice, there are $2^5$ ways to distribute the awards among the remaining 2 students. So, there are $3 \cdot 2^5 = 96$ ways. However, this counts cases where 2 students receive no awards twice, so we must adjust for that.

## Step 4: Problem 1 - Continue applying the Principle of Inclusion-Exclusion
There are $3$ ways to choose 2 students to not receive an award, and for each such choice, there's $1^5 = 1$ way for the remaining student to receive all awards. So, there are $3 \cdot 1 = 3$ such cases. The total number of ways where at least one student gets no award is $3 \cdot 2^5 - 3 \cdot 1 = 96 - 3 = 93$.

## Step 5: Problem 1 - Calculate the final number of ways to distribute the awards
The number of ways to distribute the awards such that each student gets at least one is $3^5 - 93 = 243 - 93 = 150$.

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

## Step 7: Problem 2 - Restate the problem
We are to determine the number of ways Alina can divide the numbers 1 through 9 into 3 groups of 3 such that the sum of the numbers in each group is the same.

## Step 8: Problem 2 - Calculate the total sum of the numbers and the required sum for each group
The sum of the numbers 1 through 9 is $1+2+3+...+9 = \frac{9(9+1)}{2} = 45$. For 3 groups to have the same sum, each group's sum must be $45 / 3 = 15$.

## Step 9: Problem 2 - Analyze the possible combinations for the groups
To achieve a sum of 15 with 3 numbers, we need to consider the combinations. The highest number is 9, and to reach 15, the other two numbers must sum to 6. Possible pairs with 9 are (9,5,1), (9,4,2), and (9,3,3), but since we can't repeat numbers, (9,3,3) is not valid. We continue this process for other high numbers.

## Step 10: Problem 2 - Enumerate valid combinations
Valid sets that sum to 15 include (9,5,1), (9,4,2), (8,6,1), (8,5,2), (8,4,3), (7,6,2), (7,5,3), (7,4,4) - which is not valid due to repetition, and (6,5,4). We need to find combinations of 3 such numbers that use each number from 1 to 9 exactly once.

## Step 11: Problem 2 - Check for valid groupings
After examining possible combinations and considering the restrictions, we find that there are limited ways to form 3 groups. One valid grouping is (9,2,4), (8,3,4) - which doesn't work because it repeats 4, and (7,6,2), (7,5,3), (8,1,6), (9,5,1), (7,4,4) - again, not valid. A valid example is (9,5,1), (8,6,1) - not valid, (8,4,3), (7,6,2), (7,5,3), and (9,4,2), (8,6,1), (7,5,3). Upon careful examination, one valid set is (9,4,2), (8,7,0) - not valid because 0 is not in the list, but (8,6,1), (7,5,3), and (9,4,2) is a valid grouping: {9,4,2}, {8,6,1}, {7,5,3}. Another is {9,5,1}, {8,4,3}, {7,6,2}.

## Step 12: Problem 2 - Count the distinct ways to form the groups
There are 2 distinct ways to form the groups as shown by the examples.

## Step 13: Problem 2 - State the answer
ANSWER 2: C

## Step 14: Problem 3 - Restate the problem
We need to find the sum of the digits when $10^{93} - 93$ is expressed as a whole number.

## Step 15: Problem 3 - Analyze the number $10^{93}$
$10^{93}$ is a 1 followed by 93 zeros.

## Step 16: Problem 3 - Subtract 93 from $10^{93}$
Subtracting 93 from $10^{93}$ gives us a number that is 93 less than $10^{93}$. This results in a number with 93 digits, where the last two digits are 07, and the rest are 9s except for the third last digit which is 9 - 1 = 8 (due to borrowing), but since we're subtracting from a number ending in 00, it actually results in ...9907.

## Step 17: Problem 3 - Calculate the sum of the digits
The number $10^{93} - 93$ is ...9999907. It has 91 nines, an 8 (from the subtraction affecting the hundreds place due to borrowing), a 0, and a 7. So, the sum is $91*9 + 8 + 0 + 7$.

## Step 18: Problem 3 - Perform the sum of the digits calculation
$91*9 = 819$, $819 + 8 + 7 = 834$.

## Step 19: Problem 3 - State the answer
ANSWER 3: E

## Step 20: Problem 4 - Restate the problem
We need to find the total number of points scored by the other 7 team members in a basketball game given that Alexa scored 1/4 of the points, Brittany scored 2/7, and Chelsea scored 15 points.

## Step 21: Problem 4 - Set up an equation for the total points
Let $T$ be the total points. Then, $\frac{1}{4}T + \frac{2}{7}T + 15 + x = T$, where $x$ is the total points scored by the other 7 team members.

## Step 22: Problem 4 - Simplify the equation
$\frac{7T + 8T}{28} + 15 + x = T$, so $\frac{15T}{28} + 15 + x = T$.

## Step 23: Problem 4 - Solve for $x$
$T - \frac{15T}{28} = 15 + x$, which simplifies to $\frac{13T}{28} = 15 + x$. Thus, $T = \frac{28}{13}(15 + x)$.

## Step 24: Problem 4 - Use the information that none of the other 7 team members scored more than 2 points
$x \leq 14$ because $7 \times 2 = 14$.

## Step 25: Problem 4 - Test possible values of $T$ and $x$
Since $T$ must be a whole number, and given that $\frac{1}{4}T$ and $\frac{2}{7}T$ are also whole numbers, $T$ must be divisible by both 4 and 7, hence divisible by 28.

## Step 26: Problem 4 - Find the appropriate $T$
Let's start with the smallest $T$ that is divisible by 28 and satisfies the given condi

Problem 1:

We are distributing 5 **different** awards to 3 students, with each student receiving at least one award.

First count all possible distributions with no restriction:

- Each of 5 awards can go to any of 3 students.
- Total: \(3^5=243\)

Now subtract distributions where at least one student gets no award.

If a particular student gets no award, then all 5 awards go to the other 2 students:

\[
2^5=32
\]

There are 3 choices for which student gets no award, so subtract:

\[
3\cdot 32=96
\]

But distributions where two students get no awards were subtracted twice, so add them back. If only one student gets all awards, there are 3 such distributions:

\[
243-96+3=150
\]

So the number of ways is \(150\).

ANSWER 1: B

Problem 2:

The numbers are \(1,2,\dots,9\), whose total sum is

\[
1+2+\cdots+9=45
\]

They are divided into 3 groups with equal sums, so each group must sum to

\[
45\div 3=15
\]

Now list triples from \(1\) to \(9\) that sum to 15:

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

Now form complete partitions into 3 disjoint triples.

If the group containing 1 is \(\{1,5,9\}\), the remaining numbers are

\[
2,3,4,6,7,8
\]

These form

\[
\{2,6,7\},\{3,4,8\}
\]

So one partition is

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

If the group containing 1 is \(\{1,6,8\}\), the remaining numbers are

\[
2,3,4,5,7,9
\]

These form

\[
\{2,4,9\},\{3,5,7\}
\]

So another partition is

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

There are exactly 2 ways.

ANSWER 2: C

Problem 3:

We need the sum of the digits of

\[
10^{93}-93
\]

Think of subtracting 93 from a 1 followed by 93 zeros.

A useful way:

\[
10^{93}-93=(10^{93}-100)+7
\]

Now \(10^{93}-100\) is a number with 91 nines followed by two zeros:

\[
999\cdots 99900
\]

with 91 nines. Adding 7 gives

\[
999\cdots 99907
\]

So the digit sum is

\[
91\cdot 9+0+7=819+7=826
\]

ANSWER 3: D

Problem 4:

Let the total number of points be \(T\).

Alexa scored

\[
\frac14T
\]

Brittany scored

\[
\frac27T
\]

Chelsea scored 15 points.

The fraction of the total scored by Alexa and Brittany together is

\[
\frac14+\frac27=\frac7{28}+\frac8{28}=\frac{15}{28}
\]

So the remaining fraction is

\[
1-\frac{15}{28}=\frac{13}{28}
\]

Chelsea plus the other 7 players scored

\[
\frac{13}{28}T
\]

Since \(\frac14T\) and \(\frac27T\) are whole numbers, \(T\) must be a multiple of 28. Let

\[
T=28k
\]

Then Chelsea plus the other players scored

\[
\frac{13}{28}\cdot 28k=13k
\]

So the other 7 players scored

\[
13k-15
\]

Each of the other 7 players scored at most 2 points, so together they scored at most

\[
7\cdot 2=14
\]

Thus

\[
0\le 13k-15\le 14
\]

Testing positive integers \(k\):

- \(k=1\): \(13-15=-2\), impossible.
- \(k=2\): \(26-15=11\), possible.
- \(k=3\): \(39-15=24\), too large.

So the other 7 players scored 11 points.

ANSWER 4: B

Problem 5:

The expression is

\[
\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}
\]

Each fraction is just the sign of the number:

\[
\frac{x}{|x|}=
\begin{cases}
1, & x>0\\
-1, & x<0
\end{cases}
\]

Since \(a+b+c=0\) and none of \(a,b,c\) is zero, they cannot all be positive or all be negative.

So there are two cases.

Case 1: Two are positive and one is negative.

Then

\[
\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}=1+1-1=1
\]

The product \(abc\) is negative, so

\[
\frac{abc}{|abc|}=-1
\]

Total:

\[
1+(-1)=0
\]

Case 2: One is positive and two are negative.

Then

\[
1-1-1=-1
\]

The product \(abc\) is positive, so

\[
\frac{abc}{|abc|}=1
\]

Total:

\[
-1+1=0
\]

So the only possible value is \(0\).

ANSWER 5: A

Problem 6:

The numbers

\[
a_1,a_2,\dots,a_{15}
\]

are equally spaced integers, so they form an arithmetic sequence with integer common difference \(d\).

Thus

\[
a_{15}=a_1+14d
\]

We know

\[
1\le a_1\le 10
\]

and

\[
241\le a_{15}\le 250
\]

So

\[
a_{15}-a_1
\]

is between

\[
241-10=231
\]

and

\[
250-1=249
\]

But

\[
a_{15}-a_1=14d
\]

So \(14d\) must be a multiple of 14 between 231 and 249. The only such multiple is

\[
238=14\cdot 17
\]

Thus

\[
d=17
\]

Now

\[
a_2=a_1+17
\]

and we are told

\[
13\le a_2\le 20
\]

So

\[
13\le a_1+17\le 20
\]

which gives

\[
-4\le a_1\le 3
\]

Combining with \(1\le a_1\le 10\), we get

\[
a_1=3
\]

Now

\[
a_{14}=a_1+13d=3+13\cdot 17=3+221=224
\]

The digit sum of 224 is

\[
2+2+4=8
\]

ANSWER 6: A

Problem 7:

The original cube has edge length 3 cm, so its volume is

\[
3^3=27
\]

Each smaller cube has whole-number side length. Since the smaller cubes are not all the same size, we cannot just cut the cube into all unit cubes.

The possible smaller cube side lengths are 1 cm or 2 cm. A 3 cm cube would be the entire original cube.

Let \(x\) be the number of \(2\times2\times2\) cubes, and let \(y\) be the number of \(1\times1\times1\) cubes.

Then

\[
8x+y=27
\]

The total number of cubes is

\[
N=x+y
\]

From \(y=27-8x\),

\[
N=x+27-8x=27-7x
\]

Since the cubes are not all the same size, we need at least one 2 cm cube, so \(x\ge 1\).

Among the answer choices, only \(20\) fits:

\[
20=27-7\cdot 1
\]

This is possible: put one \(2\times2\times2\) cube in a corner, and fill the remaining volume with 19 unit cubes.

So

\[
N=20
\]

ANSWER 7: E

Problem 8:

Each of the 1200 students takes 5 classes per day, so the total number of student-class enrollments is

\[
1200\cdot 5=6000
\]

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

\[
6000\div 30=200
\]

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

\[
200\div 4=50
\]

The tempting answer \(40\) would come from just doing \(1200\div 30\), but that ignores the fact that each student takes 5 classes.

ANSWER 8: E

Problem 9:

We count times in a 12-hour period whose digits form a palindrome when the colon is ignored.

For one-digit hours \(1,2,\dots,9\), a time looks like

\[
h:ab
\]

Ignoring the colon, this is \(hab\). To be a palindrome, we need

\[
b=h
\]

The tens digit \(a\) of the minutes can be \(0,1,2,3,4,

Problem 1:
We are asked to find the number of ways to distribute 5 different awards to 3 students such that each student receives at least one award. 
This is equivalent to finding the number of surjective (onto) functions from a set of 5 distinct elements to a set of 3 distinct elements. 

First, let's find the total number of ways to distribute the 5 awards to the 3 students without any restrictions. Since each of the 5 awards can go to any of the 3 students, there are $3^5 = 243$ ways.

Next, we use the Principle of Inclusion-Exclusion to subtract the distributions where at least one student receives no awards:
- Number of ways where 1 specific student gets no awards: The awards are distributed among the remaining 2 students. This can be done in $2^5 = 32$ ways. Since there are 3 ways to choose which student gets no awards, this gives $3 \times 32 = 96$ ways.
- Number of ways where 2 specific students get no awards: All awards go to the 1 remaining student. This can be done in $1^5 = 1$ way. There are 3 ways to choose which two students get no awards, giving $3 \times 1 = 3$ ways.

Applying inclusion-exclusion, the number of invalid distributions (where at least one student gets nothing) is:
$96 - 3 = 93$

Subtracting the invalid distributions from the total gives the number of valid distributions:
$243 - 93 = 150$

ANSWER 1: B

Problem 2:
Alina wants to divide the numbers 1 through 9 into 3 groups of 3 cards each, such that the sum of the numbers in each group is the same.
The sum of the numbers from 1 to 9 is:
$1 + 2 + 3 + \dots + 9 = \frac{9 \times 10}{2} = 45$
For the 3 groups to have the same sum, each group must sum to $45 / 3 = 15$.

Let's find all possible groups of 3 numbers from 1 to 9 that sum to 15:
1. $9 + 5 + 1$
2. $9 + 4 + 2$
3. $8 + 6 + 1$
4. $8 + 5 + 2$
5. $8 + 4 + 3$
6. $7 + 6 + 2$
7. $7 + 5 + 3$
8. $6 + 5 + 4$

Notice that the number 1 is only present in two possible groups: $\{9, 5, 1\}$ and $\{8, 6, 1\}$. Thus, any valid division must use one of these two groups.

Case 1: We use the group $\{9, 5, 1\}$.
The remaining numbers are 2, 3, 4, 6, 7, 8. We need to form two more groups of 3 that sum to 15.
Looking at the remaining numbers, the only valid groups that sum to 15 are $\{8, 4, 3\}$ and $\{7, 6, 2\}$. 
This gives our first valid division: $\{9, 5, 1\}, \{8, 4, 3\}, \{7, 6, 2\}$.

Case 2: We use the group $\{8, 6, 1\}$.
The remaining numbers are 2, 3, 4, 5, 7, 9. We need to form two more groups of 3 that sum to 15.
From the remaining numbers, the only valid groups that sum to 15 are $\{9, 4, 2\}$ and $\{7, 5, 3\}$.
This gives our second valid division: $\{8, 6, 1\}, \{9, 4, 2\}, \{7, 5, 3\}$.

Since the number 1 must be in exactly one of these configurations, there are exactly 2 ways to divide the cards.

ANSWER 2: C

Problem 3:
We are asked to find the sum of the digits of the whole number $10^{93} - 93$.
Let's look at a smaller pattern to understand the subtraction:
$10^2 - 93 = 100 - 93 = 07$ (sum of digits = 7)
$10^3 - 93 = 1000 - 93 = 907$ (sum of digits = 16)
$10^4 - 93 = 10000 - 93 = 9907$ (sum of digits = 25)

Notice that $10^n - 93$ results in a number with $(n - 2)$ nines, followed by the digits 0 and 7.
For $10^{93} - 93$, the number of nines will be $93 - 2 = 91$.
The number looks like: $\underbrace{999\dots99}_{91 \text{ times}}07$

The sum of the digits is:
$91 \times 9 + 0 + 7 = 819 + 7 = 826$

ANSWER 3: D

Problem 4:
Let $P$ be the total number of points scored by the team.
Alexa scored $\frac{1}{4}P$ points, and Brittany scored $\frac{2}{7}P$ points.
Together, they scored:
$\frac{1}{4}P + \frac{2}{7}P = \frac{7P + 8P}{28} = \frac{15}{28}P$

This means the rest of the team scored the remaining $1 - \frac{15}{28} = \frac{13}{28}$ of the total points.
The rest of the team consists of Chelsea (who scored 15 points) and 7 other members. Let $X$ be the total points scored by the other 7 members.
So, $\frac{13}{28}P = 15 + X$.

For $P$ to be a whole number, $15 + X$ must be a multiple of 13.
We are given that none of the other 7 team members scored more than 2 points. Therefore, the maximum points they could have scored together is $7 \times 2 = 14$.
So, $0 \le X \le 14$.
This means $15 \le 15 + X \le 29$.

The only multiple of 13 in the range from 15 to 29 is 26.
Thus, $15 + X = 26$, which gives $X = 11$.
(This also perfectly yields $P = 56$ total points).

ANSWER 4: B

Problem 5:
We are given that $a, b,$ and $c$ are nonzero real numbers and $a + b + c = 0$. We want to find the possible values of:
$\frac{a}{|a|} + \frac{b}{|b|} + \frac{c}{|c|} + \frac{abc}{|abc|}$

The expression $\frac{x}{|x|}$ evaluates to $1$ if $x > 0$ and $-1$ if $x < 0$.
Since $a + b + c = 0$ and none of them are zero, they cannot all be positive or all be negative. This leaves two possible cases for their signs:

Case 1: Two are positive and one is negative (e.g., $a>0, b>0, c<0$).
The first three terms will be $1 + 1 + (-1) = 1$.
The product $abc$ will be positive $\times$ positive $\times$ negative = negative, so the fourth term is $-1$.
Total sum = $1 - 1 = 0$.

Case 2: One is positive and two are negative (e.g., $a>0, b<0, c<0$).
The first three terms will be $1 + (-1) + (-1) = -1$.
The product $abc$ will be positive $\times$ negative $\times$ negative = positive, so the fourth term is $1$.
Total sum = $-1 + 1 = 0$.

In all possible cases, the expression evaluates to 0.

ANSWER 5: A

Problem 6:
Let the 15 integers be an arithmetic sequence with a common difference $d$. 
We are given:
$1 \le a_1 \le 10$
$13 \le a_2 \le 20$
$241 \le a_{15} \le 250$

The terms can be written as $a_n = a_1 + (n-1)d$.
For $a_{15}$, we have $a_{15} = a_1 + 14d$.
Substituting bounds for $a_1$ and $a_{15}$:
$241 \le a_1 + 14d \le 250$
Since $1 \le a_1 \le 10$, we can estimate $14d$:
$241 - 10 \le 14d \le 250 - 1 \implies 231 \le 14d \le 249$
Dividing by 14 gives $16.5 \le d \le 17.78$. Since $d$ must be an integer, $d = 17$.

Now we substitute $d = 17$ into the inequalities:
$a_2 = a_1 + 17$. Sinc

## Step 1: Problem 1 - Restate the problem
We need to find the number of different ways to distribute 5 different awards to 3 students, with each student receiving at least one award.

## Step 2: Problem 1 - Calculate the total number of ways to distribute the awards without restrictions
First, we consider distributing the awards without any restrictions. For each award, there are 3 choices of students, so the total number of ways to distribute 5 awards is $3^5 = 243$.

## Step 3: Problem 1 - Apply the Principle of Inclusion-Exclusion to find the number of ways where at least one student receives no award
To ensure each student gets at least one award, we'll use the Principle of Inclusion-Exclusion. We calculate the number of ways to distribute the awards where at least one student gets no award. There are $3$ ways to choose a student to not receive an award, and for each such choice, there are $2^5$ ways to distribute the awards among the remaining 2 students. So, there are $3 \cdot 2^5 = 96$ ways. However, this counts cases where 2 students receive no awards twice, so we must adjust for that.

## Step 4: Problem 1 - Continue applying the Principle of Inclusion-Exclusion
There are $3$ ways to choose 2 students to not receive an award, and for each such choice, there's $1^5 = 1$ way for the remaining student to receive all awards. So, there are $3 \cdot 1 = 3$ such cases. The total number of ways where at least one student gets no award is $3 \cdot 2^5 - 3 \cdot 1 = 96 - 3 = 93$.

## Step 5: Problem 1 - Calculate the final number of ways to distribute the awards
The number of ways to distribute the awards such that each student gets at least one is $3^5 - 93 = 243 - 93 = 150$.

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

## Step 7: Problem 2 - Restate the problem
We are to determine the number of ways Alina can divide the numbers 1 through 9 into 3 groups of 3 such that the sum of the numbers in each group is the same.

## Step 8: Problem 2 - Calculate the total sum of the numbers and the required sum for each group
The sum of the numbers 1 through 9 is $1+2+3+...+9 = \frac{9(9+1)}{2} = 45$. For 3 groups to have the same sum, each group's sum must be $45 / 3 = 15$.

## Step 9: Problem 2 - Analyze the possible combinations for the groups
To achieve a sum of 15 with 3 numbers, we need to consider the combinations. The highest number is 9, and to reach 15, the other two numbers must sum to 6. Possible pairs with 9 are (9,5,1), (9,4,2), and (9,3,3), but since we can't repeat numbers, (9,3,3) is not valid. We continue this process for other high numbers.

## Step 10: Problem 2 - Enumerate valid combinations
Valid sets that sum to 15 include (9,5,1), (9,4,2), (8,6,1), (8,5,2), (8,4,3), (7,6,2), (7,5,3), (7,4,4) - which is not valid due to repetition, and (6,5,4). We need to find combinations of 3 such numbers that use each number from 1 to 9 exactly once.

## Step 11: Problem 2 - Check for valid groupings
After examining possible combinations and considering the restrictions, we find that there are limited ways to form 3 groups. One valid grouping is (9,2,4), (8,3,4) - which doesn't work because it repeats 4, and (7,6,2), (7,5,3), (8,1,6), (9,5,1), (7,4,4) - again, not valid. A valid example is (9,5,1), (8,6,1) - not valid, (8,4,3), (7,6,2), (7,5,3), and (9,4,2), (8,6,1), (7,5,3). Upon careful examination, one valid set is (9,4,2), (8,7,0) - not valid because 0 is not in the list, but (8,6,1), (7,5,3), and (9,4,2) is a valid grouping: {9,4,2}, {8,6,1}, {7,5,3}. Another is {9,5,1}, {8,4,3}, {7,6,2}.

## Step 12: Problem 2 - Count the distinct ways to form the groups
There are 2 distinct ways to form the groups as shown by the examples.

## Step 13: Problem 2 - State the answer
ANSWER 2: C

## Step 14: Problem 3 - Restate the problem
We need to find the sum of the digits when $10^{93} - 93$ is expressed as a whole number.

## Step 15: Problem 3 - Analyze the number $10^{93}$
$10^{93}$ is a 1 followed by 93 zeros.

## Step 16: Problem 3 - Subtract 93 from $10^{93}$
Subtracting 93 from $10^{93}$ gives us a number that is 93 less than $10^{93}$. This results in a number with 93 digits, where the last two digits are 07, and the rest are 9s except for the third last digit which is 9 - 1 = 8 (due to borrowing), but since we're subtracting from a number ending in 00, it actually results in ...9907.

## Step 17: Problem 3 - Calculate the sum of the digits
The number $10^{93} - 93$ is ...9999907. It has 91 nines, an 8 (from the subtraction affecting the hundreds place due to borrowing), a 0, and a 7. So, the sum is $91*9 + 8 + 0 + 7$.

## Step 18: Problem 3 - Perform the sum of the digits calculation
$91*9 = 819$, $819 + 8 + 7 = 834$.

## Step 19: Problem 3 - State the answer
ANSWER 3: E

## Step 20: Problem 4 - Restate the problem
We need to find the total number of points scored by the other 7 team members in a basketball game given that Alexa scored 1/4 of the points, Brittany scored 2/7, and Chelsea scored 15 points.

## Step 21: Problem 4 - Set up an equation for the total points
Let $T$ be the total points. Then, $\frac{1}{4}T + \frac{2}{7}T + 15 + x = T$, where $x$ is the total points scored by the other 7 team members.

## Step 22: Problem 4 - Simplify the equation
$\frac{7T + 8T}{28} + 15 + x = T$, so $\frac{15T}{28} + 15 + x = T$.

## Step 23: Problem 4 - Solve for $x$
$T - \frac{15T}{28} = 15 + x$, which simplifies to $\frac{13T}{28} = 15 + x$. Thus, $T = \frac{28}{13}(15 + x)$.

## Step 24: Problem 4 - Use the information that none of the other 7 team members scored more than 2 points
$x \leq 14$ because $7 \times 2 = 14$.

## Step 25: Problem 4 - Test possible values of $T$ and $x$
Since $T$ must be a whole number, and given that $\frac{1}{4}T$ and $\frac{2}{7}T$ are also whole numbers, $T$ must be divisible by both 4 and 7, hence divisible by 28.

## Step 26: Problem 4 - Find the appropriate $T$
Let's start with the smallest $T$ that is divisible by 28 and satisfies the given condi

Problem 1:

We are distributing 5 **different** awards to 3 students, with each student receiving at least one award.

First count all possible distributions with no restriction:

- Each of 5 awards can go to any of 3 students.
- Total: \(3^5=243\)

Now subtract distributions where at least one student gets no award.

If a particular student gets no award, then all 5 awards go to the other 2 students:

\[
2^5=32
\]

There are 3 choices for which student gets no award, so subtract:

\[
3\cdot 32=96
\]

But distributions where two students get no awards were subtracted twice, so add them back. If only one student gets all awards, there are 3 such distributions:

\[
243-96+3=150
\]

So the number of ways is \(150\).

ANSWER 1: B

Problem 2:

The numbers are \(1,2,\dots,9\), whose total sum is

\[
1+2+\cdots+9=45
\]

They are divided into 3 groups with equal sums, so each group must sum to

\[
45\div 3=15
\]

Now list triples from \(1\) to \(9\) that sum to 15:

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

Now form complete partitions into 3 disjoint triples.

If the group containing 1 is \(\{1,5,9\}\), the remaining numbers are

\[
2,3,4,6,7,8
\]

These form

\[
\{2,6,7\},\{3,4,8\}
\]

So one partition is

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

If the group containing 1 is \(\{1,6,8\}\), the remaining numbers are

\[
2,3,4,5,7,9
\]

These form

\[
\{2,4,9\},\{3,5,7\}
\]

So another partition is

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

There are exactly 2 ways.

ANSWER 2: C

Problem 3:

We need the sum of the digits of

\[
10^{93}-93
\]

Think of subtracting 93 from a 1 followed by 93 zeros.

A useful way:

\[
10^{93}-93=(10^{93}-100)+7
\]

Now \(10^{93}-100\) is a number with 91 nines followed by two zeros:

\[
999\cdots 99900
\]

with 91 nines. Adding 7 gives

\[
999\cdots 99907
\]

So the digit sum is

\[
91\cdot 9+0+7=819+7=826
\]

ANSWER 3: D

Problem 4:

Let the total number of points be \(T\).

Alexa scored

\[
\frac14T
\]

Brittany scored

\[
\frac27T
\]

Chelsea scored 15 points.

The fraction of the total scored by Alexa and Brittany together is

\[
\frac14+\frac27=\frac7{28}+\frac8{28}=\frac{15}{28}
\]

So the remaining fraction is

\[
1-\frac{15}{28}=\frac{13}{28}
\]

Chelsea plus the other 7 players scored

\[
\frac{13}{28}T
\]

Since \(\frac14T\) and \(\frac27T\) are whole numbers, \(T\) must be a multiple of 28. Let

\[
T=28k
\]

Then Chelsea plus the other players scored

\[
\frac{13}{28}\cdot 28k=13k
\]

So the other 7 players scored

\[
13k-15
\]

Each of the other 7 players scored at most 2 points, so together they scored at most

\[
7\cdot 2=14
\]

Thus

\[
0\le 13k-15\le 14
\]

Testing positive integers \(k\):

- \(k=1\): \(13-15=-2\), impossible.
- \(k=2\): \(26-15=11\), possible.
- \(k=3\): \(39-15=24\), too large.

So the other 7 players scored 11 points.

ANSWER 4: B

Problem 5:

The expression is

\[
\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}
\]

Each fraction is just the sign of the number:

\[
\frac{x}{|x|}=
\begin{cases}
1, & x>0\\
-1, & x<0
\end{cases}
\]

Since \(a+b+c=0\) and none of \(a,b,c\) is zero, they cannot all be positive or all be negative.

So there are two cases.

Case 1: Two are positive and one is negative.

Then

\[
\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}=1+1-1=1
\]

The product \(abc\) is negative, so

\[
\frac{abc}{|abc|}=-1
\]

Total:

\[
1+(-1)=0
\]

Case 2: One is positive and two are negative.

Then

\[
1-1-1=-1
\]

The product \(abc\) is positive, so

\[
\frac{abc}{|abc|}=1
\]

Total:

\[
-1+1=0
\]

So the only possible value is \(0\).

ANSWER 5: A

Problem 6:

The numbers

\[
a_1,a_2,\dots,a_{15}
\]

are equally spaced integers, so they form an arithmetic sequence with integer common difference \(d\).

Thus

\[
a_{15}=a_1+14d
\]

We know

\[
1\le a_1\le 10
\]

and

\[
241\le a_{15}\le 250
\]

So

\[
a_{15}-a_1
\]

is between

\[
241-10=231
\]

and

\[
250-1=249
\]

But

\[
a_{15}-a_1=14d
\]

So \(14d\) must be a multiple of 14 between 231 and 249. The only such multiple is

\[
238=14\cdot 17
\]

Thus

\[
d=17
\]

Now

\[
a_2=a_1+17
\]

and we are told

\[
13\le a_2\le 20
\]

So

\[
13\le a_1+17\le 20
\]

which gives

\[
-4\le a_1\le 3
\]

Combining with \(1\le a_1\le 10\), we get

\[
a_1=3
\]

Now

\[
a_{14}=a_1+13d=3+13\cdot 17=3+221=224
\]

The digit sum of 224 is

\[
2+2+4=8
\]

ANSWER 6: A

Problem 7:

The original cube has edge length 3 cm, so its volume is

\[
3^3=27
\]

Each smaller cube has whole-number side length. Since the smaller cubes are not all the same size, we cannot just cut the cube into all unit cubes.

The possible smaller cube side lengths are 1 cm or 2 cm. A 3 cm cube would be the entire original cube.

Let \(x\) be the number of \(2\times2\times2\) cubes, and let \(y\) be the number of \(1\times1\times1\) cubes.

Then

\[
8x+y=27
\]

The total number of cubes is

\[
N=x+y
\]

From \(y=27-8x\),

\[
N=x+27-8x=27-7x
\]

Since the cubes are not all the same size, we need at least one 2 cm cube, so \(x\ge 1\).

Among the answer choices, only \(20\) fits:

\[
20=27-7\cdot 1
\]

This is possible: put one \(2\times2\times2\) cube in a corner, and fill the remaining volume with 19 unit cubes.

So

\[
N=20
\]

ANSWER 7: E

Problem 8:

Each of the 1200 students takes 5 classes per day, so the total number of student-class enrollments is

\[
1200\cdot 5=6000
\]

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

\[
6000\div 30=200
\]

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

\[
200\div 4=50
\]

The tempting answer \(40\) would come from just doing \(1200\div 30\), but that ignores the fact that each student takes 5 classes.

ANSWER 8: E

Problem 9:

We count times in a 12-hour period whose digits form a palindrome when the colon is ignored.

For one-digit hours \(1,2,\dots,9\), a time looks like

\[
h:ab
\]

Ignoring the colon, this is \(hab\). To be a palindrome, we need

\[
b=h
\]

The tens digit \(a\) of the minutes can be \(0,1,2,3,4,

Problem 1:
We are asked to find the number of ways to distribute 5 different awards to 3 students such that each student receives at least one award. 
This is equivalent to finding the number of surjective (onto) functions from a set of 5 distinct elements to a set of 3 distinct elements. 

First, let's find the total number of ways to distribute the 5 awards to the 3 students without any restrictions. Since each of the 5 awards can go to any of the 3 students, there are $3^5 = 243$ ways.

Next, we use the Principle of Inclusion-Exclusion to subtract the distributions where at least one student receives no awards:
- Number of ways where 1 specific student gets no awards: The awards are distributed among the remaining 2 students. This can be done in $2^5 = 32$ ways. Since there are 3 ways to choose which student gets no awards, this gives $3 \times 32 = 96$ ways.
- Number of ways where 2 specific students get no awards: All awards go to the 1 remaining student. This can be done in $1^5 = 1$ way. There are 3 ways to choose which two students get no awards, giving $3 \times 1 = 3$ ways.

Applying inclusion-exclusion, the number of invalid distributions (where at least one student gets nothing) is:
$96 - 3 = 93$

Subtracting the invalid distributions from the total gives the number of valid distributions:
$243 - 93 = 150$

ANSWER 1: B

Problem 2:
Alina wants to divide the numbers 1 through 9 into 3 groups of 3 cards each, such that the sum of the numbers in each group is the same.
The sum of the numbers from 1 to 9 is:
$1 + 2 + 3 + \dots + 9 = \frac{9 \times 10}{2} = 45$
For the 3 groups to have the same sum, each group must sum to $45 / 3 = 15$.

Let's find all possible groups of 3 numbers from 1 to 9 that sum to 15:
1. $9 + 5 + 1$
2. $9 + 4 + 2$
3. $8 + 6 + 1$
4. $8 + 5 + 2$
5. $8 + 4 + 3$
6. $7 + 6 + 2$
7. $7 + 5 + 3$
8. $6 + 5 + 4$

Notice that the number 1 is only present in two possible groups: $\{9, 5, 1\}$ and $\{8, 6, 1\}$. Thus, any valid division must use one of these two groups.

Case 1: We use the group $\{9, 5, 1\}$.
The remaining numbers are 2, 3, 4, 6, 7, 8. We need to form two more groups of 3 that sum to 15.
Looking at the remaining numbers, the only valid groups that sum to 15 are $\{8, 4, 3\}$ and $\{7, 6, 2\}$. 
This gives our first valid division: $\{9, 5, 1\}, \{8, 4, 3\}, \{7, 6, 2\}$.

Case 2: We use the group $\{8, 6, 1\}$.
The remaining numbers are 2, 3, 4, 5, 7, 9. We need to form two more groups of 3 that sum to 15.
From the remaining numbers, the only valid groups that sum to 15 are $\{9, 4, 2\}$ and $\{7, 5, 3\}$.
This gives our second valid division: $\{8, 6, 1\}, \{9, 4, 2\}, \{7, 5, 3\}$.

Since the number 1 must be in exactly one of these configurations, there are exactly 2 ways to divide the cards.

ANSWER 2: C

Problem 3:
We are asked to find the sum of the digits of the whole number $10^{93} - 93$.
Let's look at a smaller pattern to understand the subtraction:
$10^2 - 93 = 100 - 93 = 07$ (sum of digits = 7)
$10^3 - 93 = 1000 - 93 = 907$ (sum of digits = 16)
$10^4 - 93 = 10000 - 93 = 9907$ (sum of digits = 25)

Notice that $10^n - 93$ results in a number with $(n - 2)$ nines, followed by the digits 0 and 7.
For $10^{93} - 93$, the number of nines will be $93 - 2 = 91$.
The number looks like: $\underbrace{999\dots99}_{91 \text{ times}}07$

The sum of the digits is:
$91 \times 9 + 0 + 7 = 819 + 7 = 826$

ANSWER 3: D

Problem 4:
Let $P$ be the total number of points scored by the team.
Alexa scored $\frac{1}{4}P$ points, and Brittany scored $\frac{2}{7}P$ points.
Together, they scored:
$\frac{1}{4}P + \frac{2}{7}P = \frac{7P + 8P}{28} = \frac{15}{28}P$

This means the rest of the team scored the remaining $1 - \frac{15}{28} = \frac{13}{28}$ of the total points.
The rest of the team consists of Chelsea (who scored 15 points) and 7 other members. Let $X$ be the total points scored by the other 7 members.
So, $\frac{13}{28}P = 15 + X$.

For $P$ to be a whole number, $15 + X$ must be a multiple of 13.
We are given that none of the other 7 team members scored more than 2 points. Therefore, the maximum points they could have scored together is $7 \times 2 = 14$.
So, $0 \le X \le 14$.
This means $15 \le 15 + X \le 29$.

The only multiple of 13 in the range from 15 to 29 is 26.
Thus, $15 + X = 26$, which gives $X = 11$.
(This also perfectly yields $P = 56$ total points).

ANSWER 4: B

Problem 5:
We are given that $a, b,$ and $c$ are nonzero real numbers and $a + b + c = 0$. We want to find the possible values of:
$\frac{a}{|a|} + \frac{b}{|b|} + \frac{c}{|c|} + \frac{abc}{|abc|}$

The expression $\frac{x}{|x|}$ evaluates to $1$ if $x > 0$ and $-1$ if $x < 0$.
Since $a + b + c = 0$ and none of them are zero, they cannot all be positive or all be negative. This leaves two possible cases for their signs:

Case 1: Two are positive and one is negative (e.g., $a>0, b>0, c<0$).
The first three terms will be $1 + 1 + (-1) = 1$.
The product $abc$ will be positive $\times$ positive $\times$ negative = negative, so the fourth term is $-1$.
Total sum = $1 - 1 = 0$.

Case 2: One is positive and two are negative (e.g., $a>0, b<0, c<0$).
The first three terms will be $1 + (-1) + (-1) = -1$.
The product $abc$ will be positive $\times$ negative $\times$ negative = positive, so the fourth term is $1$.
Total sum = $-1 + 1 = 0$.

In all possible cases, the expression evaluates to 0.

ANSWER 5: A

Problem 6:
Let the 15 integers be an arithmetic sequence with a common difference $d$. 
We are given:
$1 \le a_1 \le 10$
$13 \le a_2 \le 20$
$241 \le a_{15} \le 250$

The terms can be written as $a_n = a_1 + (n-1)d$.
For $a_{15}$, we have $a_{15} = a_1 + 14d$.
Substituting bounds for $a_1$ and $a_{15}$:
$241 \le a_1 + 14d \le 250$
Since $1 \le a_1 \le 10$, we can estimate $14d$:
$241 - 10 \le 14d \le 250 - 1 \implies 231 \le 14d \le 249$
Dividing by 14 gives $16.5 \le d \le 17.78$. Since $d$ must be an integer, $d = 17$.

Now we substitute $d = 17$ into the inequalities:
$a_2 = a_1 + 17$. Sinc

## Step 1: Problem 1 - Restate the problem
We need to find the number of different ways to distribute 5 different awards to 3 students, with each student receiving at least one award.

## Step 2: Problem 1 - Calculate the total number of ways to distribute the awards without restrictions
First, we consider distributing the awards without any restrictions. For each award, there are 3 choices of students, so the total number of ways to distribute 5 awards is $3^5 = 243$.

## Step 3: Problem 1 - Apply the Principle of Inclusion-Exclusion to find the number of ways where at least one student receives no award
To ensure each student gets at least one award, we'll use the Principle of Inclusion-Exclusion. We calculate the number of ways to distribute the awards where at least one student gets no award. There are $3$ ways to choose a student to not receive an award, and for each such choice, there are $2^5$ ways to distribute the awards among the remaining 2 students. So, there are $3 \cdot 2^5 = 96$ ways. However, this counts cases where 2 students receive no awards twice, so we must adjust for that.

## Step 4: Problem 1 - Continue applying the Principle of Inclusion-Exclusion
There are $3$ ways to choose 2 students to not receive an award, and for each such choice, there's $1^5 = 1$ way for the remaining student to receive all awards. So, there are $3 \cdot 1 = 3$ such cases. The total number of ways where at least one student gets no award is $3 \cdot 2^5 - 3 \cdot 1 = 96 - 3 = 93$.

## Step 5: Problem 1 - Calculate the final number of ways to distribute the awards
The number of ways to distribute the awards such that each student gets at least one is $3^5 - 93 = 243 - 93 = 150$.

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

## Step 7: Problem 2 - Restate the problem
We are to determine the number of ways Alina can divide the numbers 1 through 9 into 3 groups of 3 such that the sum of the numbers in each group is the same.

## Step 8: Problem 2 - Calculate the total sum of the numbers and the required sum for each group
The sum of the numbers 1 through 9 is $1+2+3+...+9 = \frac{9(9+1)}{2} = 45$. For 3 groups to have the same sum, each group's sum must be $45 / 3 = 15$.

## Step 9: Problem 2 - Analyze the possible combinations for the groups
To achieve a sum of 15 with 3 numbers, we need to consider the combinations. The highest number is 9, and to reach 15, the other two numbers must sum to 6. Possible pairs with 9 are (9,5,1), (9,4,2), and (9,3,3), but since we can't repeat numbers, (9,3,3) is not valid. We continue this process for other high numbers.

## Step 10: Problem 2 - Enumerate valid combinations
Valid sets that sum to 15 include (9,5,1), (9,4,2), (8,6,1), (8,5,2), (8,4,3), (7,6,2), (7,5,3), (7,4,4) - which is not valid due to repetition, and (6,5,4). We need to find combinations of 3 such numbers that use each number from 1 to 9 exactly once.

## Step 11: Problem 2 - Check for valid groupings
After examining possible combinations and considering the restrictions, we find that there are limited ways to form 3 groups. One valid grouping is (9,2,4), (8,3,4) - which doesn't work because it repeats 4, and (7,6,2), (7,5,3), (8,1,6), (9,5,1), (7,4,4) - again, not valid. A valid example is (9,5,1), (8,6,1) - not valid, (8,4,3), (7,6,2), (7,5,3), and (9,4,2), (8,6,1), (7,5,3). Upon careful examination, one valid set is (9,4,2), (8,7,0) - not valid because 0 is not in the list, but (8,6,1), (7,5,3), and (9,4,2) is a valid grouping: {9,4,2}, {8,6,1}, {7,5,3}. Another is {9,5,1}, {8,4,3}, {7,6,2}.

## Step 12: Problem 2 - Count the distinct ways to form the groups
There are 2 distinct ways to form the groups as shown by the examples.

## Step 13: Problem 2 - State the answer
ANSWER 2: C

## Step 14: Problem 3 - Restate the problem
We need to find the sum of the digits when $10^{93} - 93$ is expressed as a whole number.

## Step 15: Problem 3 - Analyze the number $10^{93}$
$10^{93}$ is a 1 followed by 93 zeros.

## Step 16: Problem 3 - Subtract 93 from $10^{93}$
Subtracting 93 from $10^{93}$ gives us a number that is 93 less than $10^{93}$. This results in a number with 93 digits, where the last two digits are 07, and the rest are 9s except for the third last digit which is 9 - 1 = 8 (due to borrowing), but since we're subtracting from a number ending in 00, it actually results in ...9907.

## Step 17: Problem 3 - Calculate the sum of the digits
The number $10^{93} - 93$ is ...9999907. It has 91 nines, an 8 (from the subtraction affecting the hundreds place due to borrowing), a 0, and a 7. So, the sum is $91*9 + 8 + 0 + 7$.

## Step 18: Problem 3 - Perform the sum of the digits calculation
$91*9 = 819$, $819 + 8 + 7 = 834$.

## Step 19: Problem 3 - State the answer
ANSWER 3: E

## Step 20: Problem 4 - Restate the problem
We need to find the total number of points scored by the other 7 team members in a basketball game given that Alexa scored 1/4 of the points, Brittany scored 2/7, and Chelsea scored 15 points.

## Step 21: Problem 4 - Set up an equation for the total points
Let $T$ be the total points. Then, $\frac{1}{4}T + \frac{2}{7}T + 15 + x = T$, where $x$ is the total points scored by the other 7 team members.

## Step 22: Problem 4 - Simplify the equation
$\frac{7T + 8T}{28} + 15 + x = T$, so $\frac{15T}{28} + 15 + x = T$.

## Step 23: Problem 4 - Solve for $x$
$T - \frac{15T}{28} = 15 + x$, which simplifies to $\frac{13T}{28} = 15 + x$. Thus, $T = \frac{28}{13}(15 + x)$.

## Step 24: Problem 4 - Use the information that none of the other 7 team members scored more than 2 points
$x \leq 14$ because $7 \times 2 = 14$.

## Step 25: Problem 4 - Test possible values of $T$ and $x$
Since $T$ must be a whole number, and given that $\frac{1}{4}T$ and $\frac{2}{7}T$ are also whole numbers, $T$ must be divisible by both 4 and 7, hence divisible by 28.

## Step 26: Problem 4 - Find the appropriate $T$
Let's start with the smallest $T$ that is divisible by 28 and satisfies the given condi

Problem 1:

We are distributing 5 **different** awards to 3 students, with each student receiving at least one award.

First count all possible distributions with no restriction:

- Each of 5 awards can go to any of 3 students.
- Total: \(3^5=243\)

Now subtract distributions where at least one student gets no award.

If a particular student gets no award, then all 5 awards go to the other 2 students:

\[
2^5=32
\]

There are 3 choices for which student gets no award, so subtract:

\[
3\cdot 32=96
\]

But distributions where two students get no awards were subtracted twice, so add them back. If only one student gets all awards, there are 3 such distributions:

\[
243-96+3=150
\]

So the number of ways is \(150\).

ANSWER 1: B

Problem 2:

The numbers are \(1,2,\dots,9\), whose total sum is

\[
1+2+\cdots+9=45
\]

They are divided into 3 groups with equal sums, so each group must sum to

\[
45\div 3=15
\]

Now list triples from \(1\) to \(9\) that sum to 15:

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

Now form complete partitions into 3 disjoint triples.

If the group containing 1 is \(\{1,5,9\}\), the remaining numbers are

\[
2,3,4,6,7,8
\]

These form

\[
\{2,6,7\},\{3,4,8\}
\]

So one partition is

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

If the group containing 1 is \(\{1,6,8\}\), the remaining numbers are

\[
2,3,4,5,7,9
\]

These form

\[
\{2,4,9\},\{3,5,7\}
\]

So another partition is

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

There are exactly 2 ways.

ANSWER 2: C

Problem 3:

We need the sum of the digits of

\[
10^{93}-93
\]

Think of subtracting 93 from a 1 followed by 93 zeros.

A useful way:

\[
10^{93}-93=(10^{93}-100)+7
\]

Now \(10^{93}-100\) is a number with 91 nines followed by two zeros:

\[
999\cdots 99900
\]

with 91 nines. Adding 7 gives

\[
999\cdots 99907
\]

So the digit sum is

\[
91\cdot 9+0+7=819+7=826
\]

ANSWER 3: D

Problem 4:

Let the total number of points be \(T\).

Alexa scored

\[
\frac14T
\]

Brittany scored

\[
\frac27T
\]

Chelsea scored 15 points.

The fraction of the total scored by Alexa and Brittany together is

\[
\frac14+\frac27=\frac7{28}+\frac8{28}=\frac{15}{28}
\]

So the remaining fraction is

\[
1-\frac{15}{28}=\frac{13}{28}
\]

Chelsea plus the other 7 players scored

\[
\frac{13}{28}T
\]

Since \(\frac14T\) and \(\frac27T\) are whole numbers, \(T\) must be a multiple of 28. Let

\[
T=28k
\]

Then Chelsea plus the other players scored

\[
\frac{13}{28}\cdot 28k=13k
\]

So the other 7 players scored

\[
13k-15
\]

Each of the other 7 players scored at most 2 points, so together they scored at most

\[
7\cdot 2=14
\]

Thus

\[
0\le 13k-15\le 14
\]

Testing positive integers \(k\):

- \(k=1\): \(13-15=-2\), impossible.
- \(k=2\): \(26-15=11\), possible.
- \(k=3\): \(39-15=24\), too large.

So the other 7 players scored 11 points.

ANSWER 4: B

Problem 5:

The expression is

\[
\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}
\]

Each fraction is just the sign of the number:

\[
\frac{x}{|x|}=
\begin{cases}
1, & x>0\\
-1, & x<0
\end{cases}
\]

Since \(a+b+c=0\) and none of \(a,b,c\) is zero, they cannot all be positive or all be negative.

So there are two cases.

Case 1: Two are positive and one is negative.

Then

\[
\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}=1+1-1=1
\]

The product \(abc\) is negative, so

\[
\frac{abc}{|abc|}=-1
\]

Total:

\[
1+(-1)=0
\]

Case 2: One is positive and two are negative.

Then

\[
1-1-1=-1
\]

The product \(abc\) is positive, so

\[
\frac{abc}{|abc|}=1
\]

Total:

\[
-1+1=0
\]

So the only possible value is \(0\).

ANSWER 5: A

Problem 6:

The numbers

\[
a_1,a_2,\dots,a_{15}
\]

are equally spaced integers, so they form an arithmetic sequence with integer common difference \(d\).

Thus

\[
a_{15}=a_1+14d
\]

We know

\[
1\le a_1\le 10
\]

and

\[
241\le a_{15}\le 250
\]

So

\[
a_{15}-a_1
\]

is between

\[
241-10=231
\]

and

\[
250-1=249
\]

But

\[
a_{15}-a_1=14d
\]

So \(14d\) must be a multiple of 14 between 231 and 249. The only such multiple is

\[
238=14\cdot 17
\]

Thus

\[
d=17
\]

Now

\[
a_2=a_1+17
\]

and we are told

\[
13\le a_2\le 20
\]

So

\[
13\le a_1+17\le 20
\]

which gives

\[
-4\le a_1\le 3
\]

Combining with \(1\le a_1\le 10\), we get

\[
a_1=3
\]

Now

\[
a_{14}=a_1+13d=3+13\cdot 17=3+221=224
\]

The digit sum of 224 is

\[
2+2+4=8
\]

ANSWER 6: A

Problem 7:

The original cube has edge length 3 cm, so its volume is

\[
3^3=27
\]

Each smaller cube has whole-number side length. Since the smaller cubes are not all the same size, we cannot just cut the cube into all unit cubes.

The possible smaller cube side lengths are 1 cm or 2 cm. A 3 cm cube would be the entire original cube.

Let \(x\) be the number of \(2\times2\times2\) cubes, and let \(y\) be the number of \(1\times1\times1\) cubes.

Then

\[
8x+y=27
\]

The total number of cubes is

\[
N=x+y
\]

From \(y=27-8x\),

\[
N=x+27-8x=27-7x
\]

Since the cubes are not all the same size, we need at least one 2 cm cube, so \(x\ge 1\).

Among the answer choices, only \(20\) fits:

\[
20=27-7\cdot 1
\]

This is possible: put one \(2\times2\times2\) cube in a corner, and fill the remaining volume with 19 unit cubes.

So

\[
N=20
\]

ANSWER 7: E

Problem 8:

Each of the 1200 students takes 5 classes per day, so the total number of student-class enrollments is

\[
1200\cdot 5=6000
\]

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

\[
6000\div 30=200
\]

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

\[
200\div 4=50
\]

The tempting answer \(40\) would come from just doing \(1200\div 30\), but that ignores the fact that each student takes 5 classes.

ANSWER 8: E

Problem 9:

We count times in a 12-hour period whose digits form a palindrome when the colon is ignored.

For one-digit hours \(1,2,\dots,9\), a time looks like

\[
h:ab
\]

Ignoring the colon, this is \(hab\). To be a palindrome, we need

\[
b=h
\]

The tens digit \(a\) of the minutes can be \(0,1,2,3,4,

Problem 1:
We are asked to find the number of ways to distribute 5 different awards to 3 students such that each student receives at least one award. 
This is equivalent to finding the number of surjective (onto) functions from a set of 5 distinct elements to a set of 3 distinct elements. 

First, let's find the total number of ways to distribute the 5 awards to the 3 students without any restrictions. Since each of the 5 awards can go to any of the 3 students, there are $3^5 = 243$ ways.

Next, we use the Principle of Inclusion-Exclusion to subtract the distributions where at least one student receives no awards:
- Number of ways where 1 specific student gets no awards: The awards are distributed among the remaining 2 students. This can be done in $2^5 = 32$ ways. Since there are 3 ways to choose which student gets no awards, this gives $3 \times 32 = 96$ ways.
- Number of ways where 2 specific students get no awards: All awards go to the 1 remaining student. This can be done in $1^5 = 1$ way. There are 3 ways to choose which two students get no awards, giving $3 \times 1 = 3$ ways.

Applying inclusion-exclusion, the number of invalid distributions (where at least one student gets nothing) is:
$96 - 3 = 93$

Subtracting the invalid distributions from the total gives the number of valid distributions:
$243 - 93 = 150$

ANSWER 1: B

Problem 2:
Alina wants to divide the numbers 1 through 9 into 3 groups of 3 cards each, such that the sum of the numbers in each group is the same.
The sum of the numbers from 1 to 9 is:
$1 + 2 + 3 + \dots + 9 = \frac{9 \times 10}{2} = 45$
For the 3 groups to have the same sum, each group must sum to $45 / 3 = 15$.

Let's find all possible groups of 3 numbers from 1 to 9 that sum to 15:
1. $9 + 5 + 1$
2. $9 + 4 + 2$
3. $8 + 6 + 1$
4. $8 + 5 + 2$
5. $8 + 4 + 3$
6. $7 + 6 + 2$
7. $7 + 5 + 3$
8. $6 + 5 + 4$

Notice that the number 1 is only present in two possible groups: $\{9, 5, 1\}$ and $\{8, 6, 1\}$. Thus, any valid division must use one of these two groups.

Case 1: We use the group $\{9, 5, 1\}$.
The remaining numbers are 2, 3, 4, 6, 7, 8. We need to form two more groups of 3 that sum to 15.
Looking at the remaining numbers, the only valid groups that sum to 15 are $\{8, 4, 3\}$ and $\{7, 6, 2\}$. 
This gives our first valid division: $\{9, 5, 1\}, \{8, 4, 3\}, \{7, 6, 2\}$.

Case 2: We use the group $\{8, 6, 1\}$.
The remaining numbers are 2, 3, 4, 5, 7, 9. We need to form two more groups of 3 that sum to 15.
From the remaining numbers, the only valid groups that sum to 15 are $\{9, 4, 2\}$ and $\{7, 5, 3\}$.
This gives our second valid division: $\{8, 6, 1\}, \{9, 4, 2\}, \{7, 5, 3\}$.

Since the number 1 must be in exactly one of these configurations, there are exactly 2 ways to divide the cards.

ANSWER 2: C

Problem 3:
We are asked to find the sum of the digits of the whole number $10^{93} - 93$.
Let's look at a smaller pattern to understand the subtraction:
$10^2 - 93 = 100 - 93 = 07$ (sum of digits = 7)
$10^3 - 93 = 1000 - 93 = 907$ (sum of digits = 16)
$10^4 - 93 = 10000 - 93 = 9907$ (sum of digits = 25)

Notice that $10^n - 93$ results in a number with $(n - 2)$ nines, followed by the digits 0 and 7.
For $10^{93} - 93$, the number of nines will be $93 - 2 = 91$.
The number looks like: $\underbrace{999\dots99}_{91 \text{ times}}07$

The sum of the digits is:
$91 \times 9 + 0 + 7 = 819 + 7 = 826$

ANSWER 3: D

Problem 4:
Let $P$ be the total number of points scored by the team.
Alexa scored $\frac{1}{4}P$ points, and Brittany scored $\frac{2}{7}P$ points.
Together, they scored:
$\frac{1}{4}P + \frac{2}{7}P = \frac{7P + 8P}{28} = \frac{15}{28}P$

This means the rest of the team scored the remaining $1 - \frac{15}{28} = \frac{13}{28}$ of the total points.
The rest of the team consists of Chelsea (who scored 15 points) and 7 other members. Let $X$ be the total points scored by the other 7 members.
So, $\frac{13}{28}P = 15 + X$.

For $P$ to be a whole number, $15 + X$ must be a multiple of 13.
We are given that none of the other 7 team members scored more than 2 points. Therefore, the maximum points they could have scored together is $7 \times 2 = 14$.
So, $0 \le X \le 14$.
This means $15 \le 15 + X \le 29$.

The only multiple of 13 in the range from 15 to 29 is 26.
Thus, $15 + X = 26$, which gives $X = 11$.
(This also perfectly yields $P = 56$ total points).

ANSWER 4: B

Problem 5:
We are given that $a, b,$ and $c$ are nonzero real numbers and $a + b + c = 0$. We want to find the possible values of:
$\frac{a}{|a|} + \frac{b}{|b|} + \frac{c}{|c|} + \frac{abc}{|abc|}$

The expression $\frac{x}{|x|}$ evaluates to $1$ if $x > 0$ and $-1$ if $x < 0$.
Since $a + b + c = 0$ and none of them are zero, they cannot all be positive or all be negative. This leaves two possible cases for their signs:

Case 1: Two are positive and one is negative (e.g., $a>0, b>0, c<0$).
The first three terms will be $1 + 1 + (-1) = 1$.
The product $abc$ will be positive $\times$ positive $\times$ negative = negative, so the fourth term is $-1$.
Total sum = $1 - 1 = 0$.

Case 2: One is positive and two are negative (e.g., $a>0, b<0, c<0$).
The first three terms will be $1 + (-1) + (-1) = -1$.
The product $abc$ will be positive $\times$ negative $\times$ negative = positive, so the fourth term is $1$.
Total sum = $-1 + 1 = 0$.

In all possible cases, the expression evaluates to 0.

ANSWER 5: A

Problem 6:
Let the 15 integers be an arithmetic sequence with a common difference $d$. 
We are given:
$1 \le a_1 \le 10$
$13 \le a_2 \le 20$
$241 \le a_{15} \le 250$

The terms can be written as $a_n = a_1 + (n-1)d$.
For $a_{15}$, we have $a_{15} = a_1 + 14d$.
Substituting bounds for $a_1$ and $a_{15}$:
$241 \le a_1 + 14d \le 250$
Since $1 \le a_1 \le 10$, we can estimate $14d$:
$241 - 10 \le 14d \le 250 - 1 \implies 231 \le 14d \le 249$
Dividing by 14 gives $16.5 \le d \le 17.78$. Since $d$ must be an integer, $d = 17$.

Now we substitute $d = 17$ into the inequalities:
$a_2 = a_1 + 17$. Sinc

## Step 1: Problem 1 - Restate the problem
We need to find the number of different ways to distribute 5 different awards to 3 students, with each student receiving at least one award.

## Step 2: Problem 1 - Calculate the total number of ways to distribute the awards without restrictions
First, we consider distributing the awards without any restrictions. For each award, there are 3 choices of students, so the total number of ways to distribute 5 awards is $3^5 = 243$.

## Step 3: Problem 1 - Apply the Principle of Inclusion-Exclusion to find the number of ways where at least one student receives no award
To ensure each student gets at least one award, we'll use the Principle of Inclusion-Exclusion. We calculate the number of ways to distribute the awards where at least one student gets no award. There are $3$ ways to choose a student to not receive an award, and for each such choice, there are $2^5$ ways to distribute the awards among the remaining 2 students. So, there are $3 \cdot 2^5 = 96$ ways. However, this counts cases where 2 students receive no awards twice, so we must adjust for that.

## Step 4: Problem 1 - Continue applying the Principle of Inclusion-Exclusion
There are $3$ ways to choose 2 students to not receive an award, and for each such choice, there's $1^5 = 1$ way for the remaining student to receive all awards. So, there are $3 \cdot 1 = 3$ such cases. The total number of ways where at least one student gets no award is $3 \cdot 2^5 - 3 \cdot 1 = 96 - 3 = 93$.

## Step 5: Problem 1 - Calculate the final number of ways to distribute the awards
The number of ways to distribute the awards such that each student gets at least one is $3^5 - 93 = 243 - 93 = 150$.

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

## Step 7: Problem 2 - Restate the problem
We are to determine the number of ways Alina can divide the numbers 1 through 9 into 3 groups of 3 such that the sum of the numbers in each group is the same.

## Step 8: Problem 2 - Calculate the total sum of the numbers and the required sum for each group
The sum of the numbers 1 through 9 is $1+2+3+...+9 = \frac{9(9+1)}{2} = 45$. For 3 groups to have the same sum, each group's sum must be $45 / 3 = 15$.

## Step 9: Problem 2 - Analyze the possible combinations for the groups
To achieve a sum of 15 with 3 numbers, we need to consider the combinations. The highest number is 9, and to reach 15, the other two numbers must sum to 6. Possible pairs with 9 are (9,5,1), (9,4,2), and (9,3,3), but since we can't repeat numbers, (9,3,3) is not valid. We continue this process for other high numbers.

## Step 10: Problem 2 - Enumerate valid combinations
Valid sets that sum to 15 include (9,5,1), (9,4,2), (8,6,1), (8,5,2), (8,4,3), (7,6,2), (7,5,3), (7,4,4) - which is not valid due to repetition, and (6,5,4). We need to find combinations of 3 such numbers that use each number from 1 to 9 exactly once.

## Step 11: Problem 2 - Check for valid groupings
After examining possible combinations and considering the restrictions, we find that there are limited ways to form 3 groups. One valid grouping is (9,2,4), (8,3,4) - which doesn't work because it repeats 4, and (7,6,2), (7,5,3), (8,1,6), (9,5,1), (7,4,4) - again, not valid. A valid example is (9,5,1), (8,6,1) - not valid, (8,4,3), (7,6,2), (7,5,3), and (9,4,2), (8,6,1), (7,5,3). Upon careful examination, one valid set is (9,4,2), (8,7,0) - not valid because 0 is not in the list, but (8,6,1), (7,5,3), and (9,4,2) is a valid grouping: {9,4,2}, {8,6,1}, {7,5,3}. Another is {9,5,1}, {8,4,3}, {7,6,2}.

## Step 12: Problem 2 - Count the distinct ways to form the groups
There are 2 distinct ways to form the groups as shown by the examples.

## Step 13: Problem 2 - State the answer
ANSWER 2: C

## Step 14: Problem 3 - Restate the problem
We need to find the sum of the digits when $10^{93} - 93$ is expressed as a whole number.

## Step 15: Problem 3 - Analyze the number $10^{93}$
$10^{93}$ is a 1 followed by 93 zeros.

## Step 16: Problem 3 - Subtract 93 from $10^{93}$
Subtracting 93 from $10^{93}$ gives us a number that is 93 less than $10^{93}$. This results in a number with 93 digits, where the last two digits are 07, and the rest are 9s except for the third last digit which is 9 - 1 = 8 (due to borrowing), but since we're subtracting from a number ending in 00, it actually results in ...9907.

## Step 17: Problem 3 - Calculate the sum of the digits
The number $10^{93} - 93$ is ...9999907. It has 91 nines, an 8 (from the subtraction affecting the hundreds place due to borrowing), a 0, and a 7. So, the sum is $91*9 + 8 + 0 + 7$.

## Step 18: Problem 3 - Perform the sum of the digits calculation
$91*9 = 819$, $819 + 8 + 7 = 834$.

## Step 19: Problem 3 - State the answer
ANSWER 3: E

## Step 20: Problem 4 - Restate the problem
We need to find the total number of points scored by the other 7 team members in a basketball game given that Alexa scored 1/4 of the points, Brittany scored 2/7, and Chelsea scored 15 points.

## Step 21: Problem 4 - Set up an equation for the total points
Let $T$ be the total points. Then, $\frac{1}{4}T + \frac{2}{7}T + 15 + x = T$, where $x$ is the total points scored by the other 7 team members.

## Step 22: Problem 4 - Simplify the equation
$\frac{7T + 8T}{28} + 15 + x = T$, so $\frac{15T}{28} + 15 + x = T$.

## Step 23: Problem 4 - Solve for $x$
$T - \frac{15T}{28} = 15 + x$, which simplifies to $\frac{13T}{28} = 15 + x$. Thus, $T = \frac{28}{13}(15 + x)$.

## Step 24: Problem 4 - Use the information that none of the other 7 team members scored more than 2 points
$x \leq 14$ because $7 \times 2 = 14$.

## Step 25: Problem 4 - Test possible values of $T$ and $x$
Since $T$ must be a whole number, and given that $\frac{1}{4}T$ and $\frac{2}{7}T$ are also whole numbers, $T$ must be divisible by both 4 and 7, hence divisible by 28.

## Step 26: Problem 4 - Find the appropriate $T$
Let's start with the smallest $T$ that is divisible by 28 and satisfies the given condi

Problem 1:

We are distributing 5 **different** awards to 3 students, with each student receiving at least one award.

First count all possible distributions with no restriction:

- Each of 5 awards can go to any of 3 students.
- Total: \(3^5=243\)

Now subtract distributions where at least one student gets no award.

If a particular student gets no award, then all 5 awards go to the other 2 students:

\[
2^5=32
\]

There are 3 choices for which student gets no award, so subtract:

\[
3\cdot 32=96
\]

But distributions where two students get no awards were subtracted twice, so add them back. If only one student gets all awards, there are 3 such distributions:

\[
243-96+3=150
\]

So the number of ways is \(150\).

ANSWER 1: B

Problem 2:

The numbers are \(1,2,\dots,9\), whose total sum is

\[
1+2+\cdots+9=45
\]

They are divided into 3 groups with equal sums, so each group must sum to

\[
45\div 3=15
\]

Now list triples from \(1\) to \(9\) that sum to 15:

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

Now form complete partitions into 3 disjoint triples.

If the group containing 1 is \(\{1,5,9\}\), the remaining numbers are

\[
2,3,4,6,7,8
\]

These form

\[
\{2,6,7\},\{3,4,8\}
\]

So one partition is

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

If the group containing 1 is \(\{1,6,8\}\), the remaining numbers are

\[
2,3,4,5,7,9
\]

These form

\[
\{2,4,9\},\{3,5,7\}
\]

So another partition is

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

There are exactly 2 ways.

ANSWER 2: C

Problem 3:

We need the sum of the digits of

\[
10^{93}-93
\]

Think of subtracting 93 from a 1 followed by 93 zeros.

A useful way:

\[
10^{93}-93=(10^{93}-100)+7
\]

Now \(10^{93}-100\) is a number with 91 nines followed by two zeros:

\[
999\cdots 99900
\]

with 91 nines. Adding 7 gives

\[
999\cdots 99907
\]

So the digit sum is

\[
91\cdot 9+0+7=819+7=826
\]

ANSWER 3: D

Problem 4:

Let the total number of points be \(T\).

Alexa scored

\[
\frac14T
\]

Brittany scored

\[
\frac27T
\]

Chelsea scored 15 points.

The fraction of the total scored by Alexa and Brittany together is

\[
\frac14+\frac27=\frac7{28}+\frac8{28}=\frac{15}{28}
\]

So the remaining fraction is

\[
1-\frac{15}{28}=\frac{13}{28}
\]

Chelsea plus the other 7 players scored

\[
\frac{13}{28}T
\]

Since \(\frac14T\) and \(\frac27T\) are whole numbers, \(T\) must be a multiple of 28. Let

\[
T=28k
\]

Then Chelsea plus the other players scored

\[
\frac{13}{28}\cdot 28k=13k
\]

So the other 7 players scored

\[
13k-15
\]

Each of the other 7 players scored at most 2 points, so together they scored at most

\[
7\cdot 2=14
\]

Thus

\[
0\le 13k-15\le 14
\]

Testing positive integers \(k\):

- \(k=1\): \(13-15=-2\), impossible.
- \(k=2\): \(26-15=11\), possible.
- \(k=3\): \(39-15=24\), too large.

So the other 7 players scored 11 points.

ANSWER 4: B

Problem 5:

The expression is

\[
\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}
\]

Each fraction is just the sign of the number:

\[
\frac{x}{|x|}=
\begin{cases}
1, & x>0\\
-1, & x<0
\end{cases}
\]

Since \(a+b+c=0\) and none of \(a,b,c\) is zero, they cannot all be positive or all be negative.

So there are two cases.

Case 1: Two are positive and one is negative.

Then

\[
\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}=1+1-1=1
\]

The product \(abc\) is negative, so

\[
\frac{abc}{|abc|}=-1
\]

Total:

\[
1+(-1)=0
\]

Case 2: One is positive and two are negative.

Then

\[
1-1-1=-1
\]

The product \(abc\) is positive, so

\[
\frac{abc}{|abc|}=1
\]

Total:

\[
-1+1=0
\]

So the only possible value is \(0\).

ANSWER 5: A

Problem 6:

The numbers

\[
a_1,a_2,\dots,a_{15}
\]

are equally spaced integers, so they form an arithmetic sequence with integer common difference \(d\).

Thus

\[
a_{15}=a_1+14d
\]

We know

\[
1\le a_1\le 10
\]

and

\[
241\le a_{15}\le 250
\]

So

\[
a_{15}-a_1
\]

is between

\[
241-10=231
\]

and

\[
250-1=249
\]

But

\[
a_{15}-a_1=14d
\]

So \(14d\) must be a multiple of 14 between 231 and 249. The only such multiple is

\[
238=14\cdot 17
\]

Thus

\[
d=17
\]

Now

\[
a_2=a_1+17
\]

and we are told

\[
13\le a_2\le 20
\]

So

\[
13\le a_1+17\le 20
\]

which gives

\[
-4\le a_1\le 3
\]

Combining with \(1\le a_1\le 10\), we get

\[
a_1=3
\]

Now

\[
a_{14}=a_1+13d=3+13\cdot 17=3+221=224
\]

The digit sum of 224 is

\[
2+2+4=8
\]

ANSWER 6: A

Problem 7:

The original cube has edge length 3 cm, so its volume is

\[
3^3=27
\]

Each smaller cube has whole-number side length. Since the smaller cubes are not all the same size, we cannot just cut the cube into all unit cubes.

The possible smaller cube side lengths are 1 cm or 2 cm. A 3 cm cube would be the entire original cube.

Let \(x\) be the number of \(2\times2\times2\) cubes, and let \(y\) be the number of \(1\times1\times1\) cubes.

Then

\[
8x+y=27
\]

The total number of cubes is

\[
N=x+y
\]

From \(y=27-8x\),

\[
N=x+27-8x=27-7x
\]

Since the cubes are not all the same size, we need at least one 2 cm cube, so \(x\ge 1\).

Among the answer choices, only \(20\) fits:

\[
20=27-7\cdot 1
\]

This is possible: put one \(2\times2\times2\) cube in a corner, and fill the remaining volume with 19 unit cubes.

So

\[
N=20
\]

ANSWER 7: E

Problem 8:

Each of the 1200 students takes 5 classes per day, so the total number of student-class enrollments is

\[
1200\cdot 5=6000
\]

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

\[
6000\div 30=200
\]

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

\[
200\div 4=50
\]

The tempting answer \(40\) would come from just doing \(1200\div 30\), but that ignores the fact that each student takes 5 classes.

ANSWER 8: E

Problem 9:

We count times in a 12-hour period whose digits form a palindrome when the colon is ignored.

For one-digit hours \(1,2,\dots,9\), a time looks like

\[
h:ab
\]

Ignoring the colon, this is \(hab\). To be a palindrome, we need

\[
b=h
\]

The tens digit \(a\) of the minutes can be \(0,1,2,3,4,

Problem 1:
We are asked to find the number of ways to distribute 5 different awards to 3 students such that each student receives at least one award. 
This is equivalent to finding the number of surjective (onto) functions from a set of 5 distinct elements to a set of 3 distinct elements. 

First, let's find the total number of ways to distribute the 5 awards to the 3 students without any restrictions. Since each of the 5 awards can go to any of the 3 students, there are $3^5 = 243$ ways.

Next, we use the Principle of Inclusion-Exclusion to subtract the distributions where at least one student receives no awards:
- Number of ways where 1 specific student gets no awards: The awards are distributed among the remaining 2 students. This can be done in $2^5 = 32$ ways. Since there are 3 ways to choose which student gets no awards, this gives $3 \times 32 = 96$ ways.
- Number of ways where 2 specific students get no awards: All awards go to the 1 remaining student. This can be done in $1^5 = 1$ way. There are 3 ways to choose which two students get no awards, giving $3 \times 1 = 3$ ways.

Applying inclusion-exclusion, the number of invalid distributions (where at least one student gets nothing) is:
$96 - 3 = 93$

Subtracting the invalid distributions from the total gives the number of valid distributions:
$243 - 93 = 150$

ANSWER 1: B

Problem 2:
Alina wants to divide the numbers 1 through 9 into 3 groups of 3 cards each, such that the sum of the numbers in each group is the same.
The sum of the numbers from 1 to 9 is:
$1 + 2 + 3 + \dots + 9 = \frac{9 \times 10}{2} = 45$
For the 3 groups to have the same sum, each group must sum to $45 / 3 = 15$.

Let's find all possible groups of 3 numbers from 1 to 9 that sum to 15:
1. $9 + 5 + 1$
2. $9 + 4 + 2$
3. $8 + 6 + 1$
4. $8 + 5 + 2$
5. $8 + 4 + 3$
6. $7 + 6 + 2$
7. $7 + 5 + 3$
8. $6 + 5 + 4$

Notice that the number 1 is only present in two possible groups: $\{9, 5, 1\}$ and $\{8, 6, 1\}$. Thus, any valid division must use one of these two groups.

Case 1: We use the group $\{9, 5, 1\}$.
The remaining numbers are 2, 3, 4, 6, 7, 8. We need to form two more groups of 3 that sum to 15.
Looking at the remaining numbers, the only valid groups that sum to 15 are $\{8, 4, 3\}$ and $\{7, 6, 2\}$. 
This gives our first valid division: $\{9, 5, 1\}, \{8, 4, 3\}, \{7, 6, 2\}$.

Case 2: We use the group $\{8, 6, 1\}$.
The remaining numbers are 2, 3, 4, 5, 7, 9. We need to form two more groups of 3 that sum to 15.
From the remaining numbers, the only valid groups that sum to 15 are $\{9, 4, 2\}$ and $\{7, 5, 3\}$.
This gives our second valid division: $\{8, 6, 1\}, \{9, 4, 2\}, \{7, 5, 3\}$.

Since the number 1 must be in exactly one of these configurations, there are exactly 2 ways to divide the cards.

ANSWER 2: C

Problem 3:
We are asked to find the sum of the digits of the whole number $10^{93} - 93$.
Let's look at a smaller pattern to understand the subtraction:
$10^2 - 93 = 100 - 93 = 07$ (sum of digits = 7)
$10^3 - 93 = 1000 - 93 = 907$ (sum of digits = 16)
$10^4 - 93 = 10000 - 93 = 9907$ (sum of digits = 25)

Notice that $10^n - 93$ results in a number with $(n - 2)$ nines, followed by the digits 0 and 7.
For $10^{93} - 93$, the number of nines will be $93 - 2 = 91$.
The number looks like: $\underbrace{999\dots99}_{91 \text{ times}}07$

The sum of the digits is:
$91 \times 9 + 0 + 7 = 819 + 7 = 826$

ANSWER 3: D

Problem 4:
Let $P$ be the total number of points scored by the team.
Alexa scored $\frac{1}{4}P$ points, and Brittany scored $\frac{2}{7}P$ points.
Together, they scored:
$\frac{1}{4}P + \frac{2}{7}P = \frac{7P + 8P}{28} = \frac{15}{28}P$

This means the rest of the team scored the remaining $1 - \frac{15}{28} = \frac{13}{28}$ of the total points.
The rest of the team consists of Chelsea (who scored 15 points) and 7 other members. Let $X$ be the total points scored by the other 7 members.
So, $\frac{13}{28}P = 15 + X$.

For $P$ to be a whole number, $15 + X$ must be a multiple of 13.
We are given that none of the other 7 team members scored more than 2 points. Therefore, the maximum points they could have scored together is $7 \times 2 = 14$.
So, $0 \le X \le 14$.
This means $15 \le 15 + X \le 29$.

The only multiple of 13 in the range from 15 to 29 is 26.
Thus, $15 + X = 26$, which gives $X = 11$.
(This also perfectly yields $P = 56$ total points).

ANSWER 4: B

Problem 5:
We are given that $a, b,$ and $c$ are nonzero real numbers and $a + b + c = 0$. We want to find the possible values of:
$\frac{a}{|a|} + \frac{b}{|b|} + \frac{c}{|c|} + \frac{abc}{|abc|}$

The expression $\frac{x}{|x|}$ evaluates to $1$ if $x > 0$ and $-1$ if $x < 0$.
Since $a + b + c = 0$ and none of them are zero, they cannot all be positive or all be negative. This leaves two possible cases for their signs:

Case 1: Two are positive and one is negative (e.g., $a>0, b>0, c<0$).
The first three terms will be $1 + 1 + (-1) = 1$.
The product $abc$ will be positive $\times$ positive $\times$ negative = negative, so the fourth term is $-1$.
Total sum = $1 - 1 = 0$.

Case 2: One is positive and two are negative (e.g., $a>0, b<0, c<0$).
The first three terms will be $1 + (-1) + (-1) = -1$.
The product $abc$ will be positive $\times$ negative $\times$ negative = positive, so the fourth term is $1$.
Total sum = $-1 + 1 = 0$.

In all possible cases, the expression evaluates to 0.

ANSWER 5: A

Problem 6:
Let the 15 integers be an arithmetic sequence with a common difference $d$. 
We are given:
$1 \le a_1 \le 10$
$13 \le a_2 \le 20$
$241 \le a_{15} \le 250$

The terms can be written as $a_n = a_1 + (n-1)d$.
For $a_{15}$, we have $a_{15} = a_1 + 14d$.
Substituting bounds for $a_1$ and $a_{15}$:
$241 \le a_1 + 14d \le 250$
Since $1 \le a_1 \le 10$, we can estimate $14d$:
$241 - 10 \le 14d \le 250 - 1 \implies 231 \le 14d \le 249$
Dividing by 14 gives $16.5 \le d \le 17.78$. Since $d$ must be an integer, $d = 17$.

Now we substitute $d = 17$ into the inequalities:
$a_2 = a_1 + 17$. Sinc

## Step 1: Problem 1 - Restate the problem
We need to find the number of different ways to distribute 5 different awards to 3 students, with each student receiving at least one award.

## Step 2: Problem 1 - Calculate the total number of ways to distribute the awards without restrictions
First, we consider distributing the awards without any restrictions. For each award, there are 3 choices of students, so the total number of ways to distribute 5 awards is $3^5 = 243$.

## Step 3: Problem 1 - Apply the Principle of Inclusion-Exclusion to find the number of ways where at least one student receives no award
To ensure each student gets at least one award, we'll use the Principle of Inclusion-Exclusion. We calculate the number of ways to distribute the awards where at least one student gets no award. There are $3$ ways to choose a student to not receive an award, and for each such choice, there are $2^5$ ways to distribute the awards among the remaining 2 students. So, there are $3 \cdot 2^5 = 96$ ways. However, this counts cases where 2 students receive no awards twice, so we must adjust for that.

## Step 4: Problem 1 - Continue applying the Principle of Inclusion-Exclusion
There are $3$ ways to choose 2 students to not receive an award, and for each such choice, there's $1^5 = 1$ way for the remaining student to receive all awards. So, there are $3 \cdot 1 = 3$ such cases. The total number of ways where at least one student gets no award is $3 \cdot 2^5 - 3 \cdot 1 = 96 - 3 = 93$.

## Step 5: Problem 1 - Calculate the final number of ways to distribute the awards
The number of ways to distribute the awards such that each student gets at least one is $3^5 - 93 = 243 - 93 = 150$.

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

## Step 7: Problem 2 - Restate the problem
We are to determine the number of ways Alina can divide the numbers 1 through 9 into 3 groups of 3 such that the sum of the numbers in each group is the same.

## Step 8: Problem 2 - Calculate the total sum of the numbers and the required sum for each group
The sum of the numbers 1 through 9 is $1+2+3+...+9 = \frac{9(9+1)}{2} = 45$. For 3 groups to have the same sum, each group's sum must be $45 / 3 = 15$.

## Step 9: Problem 2 - Analyze the possible combinations for the groups
To achieve a sum of 15 with 3 numbers, we need to consider the combinations. The highest number is 9, and to reach 15, the other two numbers must sum to 6. Possible pairs with 9 are (9,5,1), (9,4,2), and (9,3,3), but since we can't repeat numbers, (9,3,3) is not valid. We continue this process for other high numbers.

## Step 10: Problem 2 - Enumerate valid combinations
Valid sets that sum to 15 include (9,5,1), (9,4,2), (8,6,1), (8,5,2), (8,4,3), (7,6,2), (7,5,3), (7,4,4) - which is not valid due to repetition, and (6,5,4). We need to find combinations of 3 such numbers that use each number from 1 to 9 exactly once.

## Step 11: Problem 2 - Check for valid groupings
After examining possible combinations and considering the restrictions, we find that there are limited ways to form 3 groups. One valid grouping is (9,2,4), (8,3,4) - which doesn't work because it repeats 4, and (7,6,2), (7,5,3), (8,1,6), (9,5,1), (7,4,4) - again, not valid. A valid example is (9,5,1), (8,6,1) - not valid, (8,4,3), (7,6,2), (7,5,3), and (9,4,2), (8,6,1), (7,5,3). Upon careful examination, one valid set is (9,4,2), (8,7,0) - not valid because 0 is not in the list, but (8,6,1), (7,5,3), and (9,4,2) is a valid grouping: {9,4,2}, {8,6,1}, {7,5,3}. Another is {9,5,1}, {8,4,3}, {7,6,2}.

## Step 12: Problem 2 - Count the distinct ways to form the groups
There are 2 distinct ways to form the groups as shown by the examples.

## Step 13: Problem 2 - State the answer
ANSWER 2: C

## Step 14: Problem 3 - Restate the problem
We need to find the sum of the digits when $10^{93} - 93$ is expressed as a whole number.

## Step 15: Problem 3 - Analyze the number $10^{93}$
$10^{93}$ is a 1 followed by 93 zeros.

## Step 16: Problem 3 - Subtract 93 from $10^{93}$
Subtracting 93 from $10^{93}$ gives us a number that is 93 less than $10^{93}$. This results in a number with 93 digits, where the last two digits are 07, and the rest are 9s except for the third last digit which is 9 - 1 = 8 (due to borrowing), but since we're subtracting from a number ending in 00, it actually results in ...9907.

## Step 17: Problem 3 - Calculate the sum of the digits
The number $10^{93} - 93$ is ...9999907. It has 91 nines, an 8 (from the subtraction affecting the hundreds place due to borrowing), a 0, and a 7. So, the sum is $91*9 + 8 + 0 + 7$.

## Step 18: Problem 3 - Perform the sum of the digits calculation
$91*9 = 819$, $819 + 8 + 7 = 834$.

## Step 19: Problem 3 - State the answer
ANSWER 3: E

## Step 20: Problem 4 - Restate the problem
We need to find the total number of points scored by the other 7 team members in a basketball game given that Alexa scored 1/4 of the points, Brittany scored 2/7, and Chelsea scored 15 points.

## Step 21: Problem 4 - Set up an equation for the total points
Let $T$ be the total points. Then, $\frac{1}{4}T + \frac{2}{7}T + 15 + x = T$, where $x$ is the total points scored by the other 7 team members.

## Step 22: Problem 4 - Simplify the equation
$\frac{7T + 8T}{28} + 15 + x = T$, so $\frac{15T}{28} + 15 + x = T$.

## Step 23: Problem 4 - Solve for $x$
$T - \frac{15T}{28} = 15 + x$, which simplifies to $\frac{13T}{28} = 15 + x$. Thus, $T = \frac{28}{13}(15 + x)$.

## Step 24: Problem 4 - Use the information that none of the other 7 team members scored more than 2 points
$x \leq 14$ because $7 \times 2 = 14$.

## Step 25: Problem 4 - Test possible values of $T$ and $x$
Since $T$ must be a whole number, and given that $\frac{1}{4}T$ and $\frac{2}{7}T$ are also whole numbers, $T$ must be divisible by both 4 and 7, hence divisible by 28.

## Step 26: Problem 4 - Find the appropriate $T$
Let's start with the smallest $T$ that is divisible by 28 and satisfies the given condi

Problem 1:

We are distributing 5 **different** awards to 3 students, with each student receiving at least one award.

First count all possible distributions with no restriction:

- Each of 5 awards can go to any of 3 students.
- Total: \(3^5=243\)

Now subtract distributions where at least one student gets no award.

If a particular student gets no award, then all 5 awards go to the other 2 students:

\[
2^5=32
\]

There are 3 choices for which student gets no award, so subtract:

\[
3\cdot 32=96
\]

But distributions where two students get no awards were subtracted twice, so add them back. If only one student gets all awards, there are 3 such distributions:

\[
243-96+3=150
\]

So the number of ways is \(150\).

ANSWER 1: B

Problem 2:

The numbers are \(1,2,\dots,9\), whose total sum is

\[
1+2+\cdots+9=45
\]

They are divided into 3 groups with equal sums, so each group must sum to

\[
45\div 3=15
\]

Now list triples from \(1\) to \(9\) that sum to 15:

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

Now form complete partitions into 3 disjoint triples.

If the group containing 1 is \(\{1,5,9\}\), the remaining numbers are

\[
2,3,4,6,7,8
\]

These form

\[
\{2,6,7\},\{3,4,8\}
\]

So one partition is

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

If the group containing 1 is \(\{1,6,8\}\), the remaining numbers are

\[
2,3,4,5,7,9
\]

These form

\[
\{2,4,9\},\{3,5,7\}
\]

So another partition is

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

There are exactly 2 ways.

ANSWER 2: C

Problem 3:

We need the sum of the digits of

\[
10^{93}-93
\]

Think of subtracting 93 from a 1 followed by 93 zeros.

A useful way:

\[
10^{93}-93=(10^{93}-100)+7
\]

Now \(10^{93}-100\) is a number with 91 nines followed by two zeros:

\[
999\cdots 99900
\]

with 91 nines. Adding 7 gives

\[
999\cdots 99907
\]

So the digit sum is

\[
91\cdot 9+0+7=819+7=826
\]

ANSWER 3: D

Problem 4:

Let the total number of points be \(T\).

Alexa scored

\[
\frac14T
\]

Brittany scored

\[
\frac27T
\]

Chelsea scored 15 points.

The fraction of the total scored by Alexa and Brittany together is

\[
\frac14+\frac27=\frac7{28}+\frac8{28}=\frac{15}{28}
\]

So the remaining fraction is

\[
1-\frac{15}{28}=\frac{13}{28}
\]

Chelsea plus the other 7 players scored

\[
\frac{13}{28}T
\]

Since \(\frac14T\) and \(\frac27T\) are whole numbers, \(T\) must be a multiple of 28. Let

\[
T=28k
\]

Then Chelsea plus the other players scored

\[
\frac{13}{28}\cdot 28k=13k
\]

So the other 7 players scored

\[
13k-15
\]

Each of the other 7 players scored at most 2 points, so together they scored at most

\[
7\cdot 2=14
\]

Thus

\[
0\le 13k-15\le 14
\]

Testing positive integers \(k\):

- \(k=1\): \(13-15=-2\), impossible.
- \(k=2\): \(26-15=11\), possible.
- \(k=3\): \(39-15=24\), too large.

So the other 7 players scored 11 points.

ANSWER 4: B

Problem 5:

The expression is

\[
\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}
\]

Each fraction is just the sign of the number:

\[
\frac{x}{|x|}=
\begin{cases}
1, & x>0\\
-1, & x<0
\end{cases}
\]

Since \(a+b+c=0\) and none of \(a,b,c\) is zero, they cannot all be positive or all be negative.

So there are two cases.

Case 1: Two are positive and one is negative.

Then

\[
\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}=1+1-1=1
\]

The product \(abc\) is negative, so

\[
\frac{abc}{|abc|}=-1
\]

Total:

\[
1+(-1)=0
\]

Case 2: One is positive and two are negative.

Then

\[
1-1-1=-1
\]

The product \(abc\) is positive, so

\[
\frac{abc}{|abc|}=1
\]

Total:

\[
-1+1=0
\]

So the only possible value is \(0\).

ANSWER 5: A

Problem 6:

The numbers

\[
a_1,a_2,\dots,a_{15}
\]

are equally spaced integers, so they form an arithmetic sequence with integer common difference \(d\).

Thus

\[
a_{15}=a_1+14d
\]

We know

\[
1\le a_1\le 10
\]

and

\[
241\le a_{15}\le 250
\]

So

\[
a_{15}-a_1
\]

is between

\[
241-10=231
\]

and

\[
250-1=249
\]

But

\[
a_{15}-a_1=14d
\]

So \(14d\) must be a multiple of 14 between 231 and 249. The only such multiple is

\[
238=14\cdot 17
\]

Thus

\[
d=17
\]

Now

\[
a_2=a_1+17
\]

and we are told

\[
13\le a_2\le 20
\]

So

\[
13\le a_1+17\le 20
\]

which gives

\[
-4\le a_1\le 3
\]

Combining with \(1\le a_1\le 10\), we get

\[
a_1=3
\]

Now

\[
a_{14}=a_1+13d=3+13\cdot 17=3+221=224
\]

The digit sum of 224 is

\[
2+2+4=8
\]

ANSWER 6: A

Problem 7:

The original cube has edge length 3 cm, so its volume is

\[
3^3=27
\]

Each smaller cube has whole-number side length. Since the smaller cubes are not all the same size, we cannot just cut the cube into all unit cubes.

The possible smaller cube side lengths are 1 cm or 2 cm. A 3 cm cube would be the entire original cube.

Let \(x\) be the number of \(2\times2\times2\) cubes, and let \(y\) be the number of \(1\times1\times1\) cubes.

Then

\[
8x+y=27
\]

The total number of cubes is

\[
N=x+y
\]

From \(y=27-8x\),

\[
N=x+27-8x=27-7x
\]

Since the cubes are not all the same size, we need at least one 2 cm cube, so \(x\ge 1\).

Among the answer choices, only \(20\) fits:

\[
20=27-7\cdot 1
\]

This is possible: put one \(2\times2\times2\) cube in a corner, and fill the remaining volume with 19 unit cubes.

So

\[
N=20
\]

ANSWER 7: E

Problem 8:

Each of the 1200 students takes 5 classes per day, so the total number of student-class enrollments is

\[
1200\cdot 5=6000
\]

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

\[
6000\div 30=200
\]

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

\[
200\div 4=50
\]

The tempting answer \(40\) would come from just doing \(1200\div 30\), but that ignores the fact that each student takes 5 classes.

ANSWER 8: E

Problem 9:

We count times in a 12-hour period whose digits form a palindrome when the colon is ignored.

For one-digit hours \(1,2,\dots,9\), a time looks like

\[
h:ab
\]

Ignoring the colon, this is \(hab\). To be a palindrome, we need

\[
b=h
\]

The tens digit \(a\) of the minutes can be \(0,1,2,3,4,

Problem 1:
We are asked to find the number of ways to distribute 5 different awards to 3 students such that each student receives at least one award. 
This is equivalent to finding the number of surjective (onto) functions from a set of 5 distinct elements to a set of 3 distinct elements. 

First, let's find the total number of ways to distribute the 5 awards to the 3 students without any restrictions. Since each of the 5 awards can go to any of the 3 students, there are $3^5 = 243$ ways.

Next, we use the Principle of Inclusion-Exclusion to subtract the distributions where at least one student receives no awards:
- Number of ways where 1 specific student gets no awards: The awards are distributed among the remaining 2 students. This can be done in $2^5 = 32$ ways. Since there are 3 ways to choose which student gets no awards, this gives $3 \times 32 = 96$ ways.
- Number of ways where 2 specific students get no awards: All awards go to the 1 remaining student. This can be done in $1^5 = 1$ way. There are 3 ways to choose which two students get no awards, giving $3 \times 1 = 3$ ways.

Applying inclusion-exclusion, the number of invalid distributions (where at least one student gets nothing) is:
$96 - 3 = 93$

Subtracting the invalid distributions from the total gives the number of valid distributions:
$243 - 93 = 150$

ANSWER 1: B

Problem 2:
Alina wants to divide the numbers 1 through 9 into 3 groups of 3 cards each, such that the sum of the numbers in each group is the same.
The sum of the numbers from 1 to 9 is:
$1 + 2 + 3 + \dots + 9 = \frac{9 \times 10}{2} = 45$
For the 3 groups to have the same sum, each group must sum to $45 / 3 = 15$.

Let's find all possible groups of 3 numbers from 1 to 9 that sum to 15:
1. $9 + 5 + 1$
2. $9 + 4 + 2$
3. $8 + 6 + 1$
4. $8 + 5 + 2$
5. $8 + 4 + 3$
6. $7 + 6 + 2$
7. $7 + 5 + 3$
8. $6 + 5 + 4$

Notice that the number 1 is only present in two possible groups: $\{9, 5, 1\}$ and $\{8, 6, 1\}$. Thus, any valid division must use one of these two groups.

Case 1: We use the group $\{9, 5, 1\}$.
The remaining numbers are 2, 3, 4, 6, 7, 8. We need to form two more groups of 3 that sum to 15.
Looking at the remaining numbers, the only valid groups that sum to 15 are $\{8, 4, 3\}$ and $\{7, 6, 2\}$. 
This gives our first valid division: $\{9, 5, 1\}, \{8, 4, 3\}, \{7, 6, 2\}$.

Case 2: We use the group $\{8, 6, 1\}$.
The remaining numbers are 2, 3, 4, 5, 7, 9. We need to form two more groups of 3 that sum to 15.
From the remaining numbers, the only valid groups that sum to 15 are $\{9, 4, 2\}$ and $\{7, 5, 3\}$.
This gives our second valid division: $\{8, 6, 1\}, \{9, 4, 2\}, \{7, 5, 3\}$.

Since the number 1 must be in exactly one of these configurations, there are exactly 2 ways to divide the cards.

ANSWER 2: C

Problem 3:
We are asked to find the sum of the digits of the whole number $10^{93} - 93$.
Let's look at a smaller pattern to understand the subtraction:
$10^2 - 93 = 100 - 93 = 07$ (sum of digits = 7)
$10^3 - 93 = 1000 - 93 = 907$ (sum of digits = 16)
$10^4 - 93 = 10000 - 93 = 9907$ (sum of digits = 25)

Notice that $10^n - 93$ results in a number with $(n - 2)$ nines, followed by the digits 0 and 7.
For $10^{93} - 93$, the number of nines will be $93 - 2 = 91$.
The number looks like: $\underbrace{999\dots99}_{91 \text{ times}}07$

The sum of the digits is:
$91 \times 9 + 0 + 7 = 819 + 7 = 826$

ANSWER 3: D

Problem 4:
Let $P$ be the total number of points scored by the team.
Alexa scored $\frac{1}{4}P$ points, and Brittany scored $\frac{2}{7}P$ points.
Together, they scored:
$\frac{1}{4}P + \frac{2}{7}P = \frac{7P + 8P}{28} = \frac{15}{28}P$

This means the rest of the team scored the remaining $1 - \frac{15}{28} = \frac{13}{28}$ of the total points.
The rest of the team consists of Chelsea (who scored 15 points) and 7 other members. Let $X$ be the total points scored by the other 7 members.
So, $\frac{13}{28}P = 15 + X$.

For $P$ to be a whole number, $15 + X$ must be a multiple of 13.
We are given that none of the other 7 team members scored more than 2 points. Therefore, the maximum points they could have scored together is $7 \times 2 = 14$.
So, $0 \le X \le 14$.
This means $15 \le 15 + X \le 29$.

The only multiple of 13 in the range from 15 to 29 is 26.
Thus, $15 + X = 26$, which gives $X = 11$.
(This also perfectly yields $P = 56$ total points).

ANSWER 4: B

Problem 5:
We are given that $a, b,$ and $c$ are nonzero real numbers and $a + b + c = 0$. We want to find the possible values of:
$\frac{a}{|a|} + \frac{b}{|b|} + \frac{c}{|c|} + \frac{abc}{|abc|}$

The expression $\frac{x}{|x|}$ evaluates to $1$ if $x > 0$ and $-1$ if $x < 0$.
Since $a + b + c = 0$ and none of them are zero, they cannot all be positive or all be negative. This leaves two possible cases for their signs:

Case 1: Two are positive and one is negative (e.g., $a>0, b>0, c<0$).
The first three terms will be $1 + 1 + (-1) = 1$.
The product $abc$ will be positive $\times$ positive $\times$ negative = negative, so the fourth term is $-1$.
Total sum = $1 - 1 = 0$.

Case 2: One is positive and two are negative (e.g., $a>0, b<0, c<0$).
The first three terms will be $1 + (-1) + (-1) = -1$.
The product $abc$ will be positive $\times$ negative $\times$ negative = positive, so the fourth term is $1$.
Total sum = $-1 + 1 = 0$.

In all possible cases, the expression evaluates to 0.

ANSWER 5: A

Problem 6:
Let the 15 integers be an arithmetic sequence with a common difference $d$. 
We are given:
$1 \le a_1 \le 10$
$13 \le a_2 \le 20$
$241 \le a_{15} \le 250$

The terms can be written as $a_n = a_1 + (n-1)d$.
For $a_{15}$, we have $a_{15} = a_1 + 14d$.
Substituting bounds for $a_1$ and $a_{15}$:
$241 \le a_1 + 14d \le 250$
Since $1 \le a_1 \le 10$, we can estimate $14d$:
$241 - 10 \le 14d \le 250 - 1 \implies 231 \le 14d \le 249$
Dividing by 14 gives $16.5 \le d \le 17.78$. Since $d$ must be an integer, $d = 17$.

Now we substitute $d = 17$ into the inequalities:
$a_2 = a_1 + 17$. Sinc

## Step 1: Problem 1 - Restate the problem
We need to find the number of different ways to distribute 5 different awards to 3 students, with each student receiving at least one award.

## Step 2: Problem 1 - Calculate the total number of ways to distribute the awards without restrictions
First, we consider distributing the awards without any restrictions. For each award, there are 3 choices of students, so the total number of ways to distribute 5 awards is $3^5 = 243$.

## Step 3: Problem 1 - Apply the Principle of Inclusion-Exclusion to find the number of ways where at least one student receives no award
To ensure each student gets at least one award, we'll use the Principle of Inclusion-Exclusion. We calculate the number of ways to distribute the awards where at least one student gets no award. There are $3$ ways to choose a student to not receive an award, and for each such choice, there are $2^5$ ways to distribute the awards among the remaining 2 students. So, there are $3 \cdot 2^5 = 96$ ways. However, this counts cases where 2 students receive no awards twice, so we must adjust for that.

## Step 4: Problem 1 - Continue applying the Principle of Inclusion-Exclusion
There are $3$ ways to choose 2 students to not receive an award, and for each such choice, there's $1^5 = 1$ way for the remaining student to receive all awards. So, there are $3 \cdot 1 = 3$ such cases. The total number of ways where at least one student gets no award is $3 \cdot 2^5 - 3 \cdot 1 = 96 - 3 = 93$.

## Step 5: Problem 1 - Calculate the final number of ways to distribute the awards
The number of ways to distribute the awards such that each student gets at least one is $3^5 - 93 = 243 - 93 = 150$.

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

## Step 7: Problem 2 - Restate the problem
We are to determine the number of ways Alina can divide the numbers 1 through 9 into 3 groups of 3 such that the sum of the numbers in each group is the same.

## Step 8: Problem 2 - Calculate the total sum of the numbers and the required sum for each group
The sum of the numbers 1 through 9 is $1+2+3+...+9 = \frac{9(9+1)}{2} = 45$. For 3 groups to have the same sum, each group's sum must be $45 / 3 = 15$.

## Step 9: Problem 2 - Analyze the possible combinations for the groups
To achieve a sum of 15 with 3 numbers, we need to consider the combinations. The highest number is 9, and to reach 15, the other two numbers must sum to 6. Possible pairs with 9 are (9,5,1), (9,4,2), and (9,3,3), but since we can't repeat numbers, (9,3,3) is not valid. We continue this process for other high numbers.

## Step 10: Problem 2 - Enumerate valid combinations
Valid sets that sum to 15 include (9,5,1), (9,4,2), (8,6,1), (8,5,2), (8,4,3), (7,6,2), (7,5,3), (7,4,4) - which is not valid due to repetition, and (6,5,4). We need to find combinations of 3 such numbers that use each number from 1 to 9 exactly once.

## Step 11: Problem 2 - Check for valid groupings
After examining possible combinations and considering the restrictions, we find that there are limited ways to form 3 groups. One valid grouping is (9,2,4), (8,3,4) - which doesn't work because it repeats 4, and (7,6,2), (7,5,3), (8,1,6), (9,5,1), (7,4,4) - again, not valid. A valid example is (9,5,1), (8,6,1) - not valid, (8,4,3), (7,6,2), (7,5,3), and (9,4,2), (8,6,1), (7,5,3). Upon careful examination, one valid set is (9,4,2), (8,7,0) - not valid because 0 is not in the list, but (8,6,1), (7,5,3), and (9,4,2) is a valid grouping: {9,4,2}, {8,6,1}, {7,5,3}. Another is {9,5,1}, {8,4,3}, {7,6,2}.

## Step 12: Problem 2 - Count the distinct ways to form the groups
There are 2 distinct ways to form the groups as shown by the examples.

## Step 13: Problem 2 - State the answer
ANSWER 2: C

## Step 14: Problem 3 - Restate the problem
We need to find the sum of the digits when $10^{93} - 93$ is expressed as a whole number.

## Step 15: Problem 3 - Analyze the number $10^{93}$
$10^{93}$ is a 1 followed by 93 zeros.

## Step 16: Problem 3 - Subtract 93 from $10^{93}$
Subtracting 93 from $10^{93}$ gives us a number that is 93 less than $10^{93}$. This results in a number with 93 digits, where the last two digits are 07, and the rest are 9s except for the third last digit which is 9 - 1 = 8 (due to borrowing), but since we're subtracting from a number ending in 00, it actually results in ...9907.

## Step 17: Problem 3 - Calculate the sum of the digits
The number $10^{93} - 93$ is ...9999907. It has 91 nines, an 8 (from the subtraction affecting the hundreds place due to borrowing), a 0, and a 7. So, the sum is $91*9 + 8 + 0 + 7$.

## Step 18: Problem 3 - Perform the sum of the digits calculation
$91*9 = 819$, $819 + 8 + 7 = 834$.

## Step 19: Problem 3 - State the answer
ANSWER 3: E

## Step 20: Problem 4 - Restate the problem
We need to find the total number of points scored by the other 7 team members in a basketball game given that Alexa scored 1/4 of the points, Brittany scored 2/7, and Chelsea scored 15 points.

## Step 21: Problem 4 - Set up an equation for the total points
Let $T$ be the total points. Then, $\frac{1}{4}T + \frac{2}{7}T + 15 + x = T$, where $x$ is the total points scored by the other 7 team members.

## Step 22: Problem 4 - Simplify the equation
$\frac{7T + 8T}{28} + 15 + x = T$, so $\frac{15T}{28} + 15 + x = T$.

## Step 23: Problem 4 - Solve for $x$
$T - \frac{15T}{28} = 15 + x$, which simplifies to $\frac{13T}{28} = 15 + x$. Thus, $T = \frac{28}{13}(15 + x)$.

## Step 24: Problem 4 - Use the information that none of the other 7 team members scored more than 2 points
$x \leq 14$ because $7 \times 2 = 14$.

## Step 25: Problem 4 - Test possible values of $T$ and $x$
Since $T$ must be a whole number, and given that $\frac{1}{4}T$ and $\frac{2}{7}T$ are also whole numbers, $T$ must be divisible by both 4 and 7, hence divisible by 28.

## Step 26: Problem 4 - Find the appropriate $T$
Let's start with the smallest $T$ that is divisible by 28 and satisfies the given condi

Problem 1:

We are distributing 5 **different** awards to 3 students, with each student receiving at least one award.

First count all possible distributions with no restriction:

- Each of 5 awards can go to any of 3 students.
- Total: \(3^5=243\)

Now subtract distributions where at least one student gets no award.

If a particular student gets no award, then all 5 awards go to the other 2 students:

\[
2^5=32
\]

There are 3 choices for which student gets no award, so subtract:

\[
3\cdot 32=96
\]

But distributions where two students get no awards were subtracted twice, so add them back. If only one student gets all awards, there are 3 such distributions:

\[
243-96+3=150
\]

So the number of ways is \(150\).

ANSWER 1: B

Problem 2:

The numbers are \(1,2,\dots,9\), whose total sum is

\[
1+2+\cdots+9=45
\]

They are divided into 3 groups with equal sums, so each group must sum to

\[
45\div 3=15
\]

Now list triples from \(1\) to \(9\) that sum to 15:

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

Now form complete partitions into 3 disjoint triples.

If the group containing 1 is \(\{1,5,9\}\), the remaining numbers are

\[
2,3,4,6,7,8
\]

These form

\[
\{2,6,7\},\{3,4,8\}
\]

So one partition is

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

If the group containing 1 is \(\{1,6,8\}\), the remaining numbers are

\[
2,3,4,5,7,9
\]

These form

\[
\{2,4,9\},\{3,5,7\}
\]

So another partition is

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

There are exactly 2 ways.

ANSWER 2: C

Problem 3:

We need the sum of the digits of

\[
10^{93}-93
\]

Think of subtracting 93 from a 1 followed by 93 zeros.

A useful way:

\[
10^{93}-93=(10^{93}-100)+7
\]

Now \(10^{93}-100\) is a number with 91 nines followed by two zeros:

\[
999\cdots 99900
\]

with 91 nines. Adding 7 gives

\[
999\cdots 99907
\]

So the digit sum is

\[
91\cdot 9+0+7=819+7=826
\]

ANSWER 3: D

Problem 4:

Let the total number of points be \(T\).

Alexa scored

\[
\frac14T
\]

Brittany scored

\[
\frac27T
\]

Chelsea scored 15 points.

The fraction of the total scored by Alexa and Brittany together is

\[
\frac14+\frac27=\frac7{28}+\frac8{28}=\frac{15}{28}
\]

So the remaining fraction is

\[
1-\frac{15}{28}=\frac{13}{28}
\]

Chelsea plus the other 7 players scored

\[
\frac{13}{28}T
\]

Since \(\frac14T\) and \(\frac27T\) are whole numbers, \(T\) must be a multiple of 28. Let

\[
T=28k
\]

Then Chelsea plus the other players scored

\[
\frac{13}{28}\cdot 28k=13k
\]

So the other 7 players scored

\[
13k-15
\]

Each of the other 7 players scored at most 2 points, so together they scored at most

\[
7\cdot 2=14
\]

Thus

\[
0\le 13k-15\le 14
\]

Testing positive integers \(k\):

- \(k=1\): \(13-15=-2\), impossible.
- \(k=2\): \(26-15=11\), possible.
- \(k=3\): \(39-15=24\), too large.

So the other 7 players scored 11 points.

ANSWER 4: B

Problem 5:

The expression is

\[
\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}
\]

Each fraction is just the sign of the number:

\[
\frac{x}{|x|}=
\begin{cases}
1, & x>0\\
-1, & x<0
\end{cases}
\]

Since \(a+b+c=0\) and none of \(a,b,c\) is zero, they cannot all be positive or all be negative.

So there are two cases.

Case 1: Two are positive and one is negative.

Then

\[
\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}=1+1-1=1
\]

The product \(abc\) is negative, so

\[
\frac{abc}{|abc|}=-1
\]

Total:

\[
1+(-1)=0
\]

Case 2: One is positive and two are negative.

Then

\[
1-1-1=-1
\]

The product \(abc\) is positive, so

\[
\frac{abc}{|abc|}=1
\]

Total:

\[
-1+1=0
\]

So the only possible value is \(0\).

ANSWER 5: A

Problem 6:

The numbers

\[
a_1,a_2,\dots,a_{15}
\]

are equally spaced integers, so they form an arithmetic sequence with integer common difference \(d\).

Thus

\[
a_{15}=a_1+14d
\]

We know

\[
1\le a_1\le 10
\]

and

\[
241\le a_{15}\le 250
\]

So

\[
a_{15}-a_1
\]

is between

\[
241-10=231
\]

and

\[
250-1=249
\]

But

\[
a_{15}-a_1=14d
\]

So \(14d\) must be a multiple of 14 between 231 and 249. The only such multiple is

\[
238=14\cdot 17
\]

Thus

\[
d=17
\]

Now

\[
a_2=a_1+17
\]

and we are told

\[
13\le a_2\le 20
\]

So

\[
13\le a_1+17\le 20
\]

which gives

\[
-4\le a_1\le 3
\]

Combining with \(1\le a_1\le 10\), we get

\[
a_1=3
\]

Now

\[
a_{14}=a_1+13d=3+13\cdot 17=3+221=224
\]

The digit sum of 224 is

\[
2+2+4=8
\]

ANSWER 6: A

Problem 7:

The original cube has edge length 3 cm, so its volume is

\[
3^3=27
\]

Each smaller cube has whole-number side length. Since the smaller cubes are not all the same size, we cannot just cut the cube into all unit cubes.

The possible smaller cube side lengths are 1 cm or 2 cm. A 3 cm cube would be the entire original cube.

Let \(x\) be the number of \(2\times2\times2\) cubes, and let \(y\) be the number of \(1\times1\times1\) cubes.

Then

\[
8x+y=27
\]

The total number of cubes is

\[
N=x+y
\]

From \(y=27-8x\),

\[
N=x+27-8x=27-7x
\]

Since the cubes are not all the same size, we need at least one 2 cm cube, so \(x\ge 1\).

Among the answer choices, only \(20\) fits:

\[
20=27-7\cdot 1
\]

This is possible: put one \(2\times2\times2\) cube in a corner, and fill the remaining volume with 19 unit cubes.

So

\[
N=20
\]

ANSWER 7: E

Problem 8:

Each of the 1200 students takes 5 classes per day, so the total number of student-class enrollments is

\[
1200\cdot 5=6000
\]

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

\[
6000\div 30=200
\]

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

\[
200\div 4=50
\]

The tempting answer \(40\) would come from just doing \(1200\div 30\), but that ignores the fact that each student takes 5 classes.

ANSWER 8: E

Problem 9:

We count times in a 12-hour period whose digits form a palindrome when the colon is ignored.

For one-digit hours \(1,2,\dots,9\), a time looks like

\[
h:ab
\]

Ignoring the colon, this is \(hab\). To be a palindrome, we need

\[
b=h
\]

The tens digit \(a\) of the minutes can be \(0,1,2,3,4,

Problem 1:
We are asked to find the number of ways to distribute 5 different awards to 3 students such that each student receives at least one award. 
This is equivalent to finding the number of surjective (onto) functions from a set of 5 distinct elements to a set of 3 distinct elements. 

First, let's find the total number of ways to distribute the 5 awards to the 3 students without any restrictions. Since each of the 5 awards can go to any of the 3 students, there are $3^5 = 243$ ways.

Next, we use the Principle of Inclusion-Exclusion to subtract the distributions where at least one student receives no awards:
- Number of ways where 1 specific student gets no awards: The awards are distributed among the remaining 2 students. This can be done in $2^5 = 32$ ways. Since there are 3 ways to choose which student gets no awards, this gives $3 \times 32 = 96$ ways.
- Number of ways where 2 specific students get no awards: All awards go to the 1 remaining student. This can be done in $1^5 = 1$ way. There are 3 ways to choose which two students get no awards, giving $3 \times 1 = 3$ ways.

Applying inclusion-exclusion, the number of invalid distributions (where at least one student gets nothing) is:
$96 - 3 = 93$

Subtracting the invalid distributions from the total gives the number of valid distributions:
$243 - 93 = 150$

ANSWER 1: B

Problem 2:
Alina wants to divide the numbers 1 through 9 into 3 groups of 3 cards each, such that the sum of the numbers in each group is the same.
The sum of the numbers from 1 to 9 is:
$1 + 2 + 3 + \dots + 9 = \frac{9 \times 10}{2} = 45$
For the 3 groups to have the same sum, each group must sum to $45 / 3 = 15$.

Let's find all possible groups of 3 numbers from 1 to 9 that sum to 15:
1. $9 + 5 + 1$
2. $9 + 4 + 2$
3. $8 + 6 + 1$
4. $8 + 5 + 2$
5. $8 + 4 + 3$
6. $7 + 6 + 2$
7. $7 + 5 + 3$
8. $6 + 5 + 4$

Notice that the number 1 is only present in two possible groups: $\{9, 5, 1\}$ and $\{8, 6, 1\}$. Thus, any valid division must use one of these two groups.

Case 1: We use the group $\{9, 5, 1\}$.
The remaining numbers are 2, 3, 4, 6, 7, 8. We need to form two more groups of 3 that sum to 15.
Looking at the remaining numbers, the only valid groups that sum to 15 are $\{8, 4, 3\}$ and $\{7, 6, 2\}$. 
This gives our first valid division: $\{9, 5, 1\}, \{8, 4, 3\}, \{7, 6, 2\}$.

Case 2: We use the group $\{8, 6, 1\}$.
The remaining numbers are 2, 3, 4, 5, 7, 9. We need to form two more groups of 3 that sum to 15.
From the remaining numbers, the only valid groups that sum to 15 are $\{9, 4, 2\}$ and $\{7, 5, 3\}$.
This gives our second valid division: $\{8, 6, 1\}, \{9, 4, 2\}, \{7, 5, 3\}$.

Since the number 1 must be in exactly one of these configurations, there are exactly 2 ways to divide the cards.

ANSWER 2: C

Problem 3:
We are asked to find the sum of the digits of the whole number $10^{93} - 93$.
Let's look at a smaller pattern to understand the subtraction:
$10^2 - 93 = 100 - 93 = 07$ (sum of digits = 7)
$10^3 - 93 = 1000 - 93 = 907$ (sum of digits = 16)
$10^4 - 93 = 10000 - 93 = 9907$ (sum of digits = 25)

Notice that $10^n - 93$ results in a number with $(n - 2)$ nines, followed by the digits 0 and 7.
For $10^{93} - 93$, the number of nines will be $93 - 2 = 91$.
The number looks like: $\underbrace{999\dots99}_{91 \text{ times}}07$

The sum of the digits is:
$91 \times 9 + 0 + 7 = 819 + 7 = 826$

ANSWER 3: D

Problem 4:
Let $P$ be the total number of points scored by the team.
Alexa scored $\frac{1}{4}P$ points, and Brittany scored $\frac{2}{7}P$ points.
Together, they scored:
$\frac{1}{4}P + \frac{2}{7}P = \frac{7P + 8P}{28} = \frac{15}{28}P$

This means the rest of the team scored the remaining $1 - \frac{15}{28} = \frac{13}{28}$ of the total points.
The rest of the team consists of Chelsea (who scored 15 points) and 7 other members. Let $X$ be the total points scored by the other 7 members.
So, $\frac{13}{28}P = 15 + X$.

For $P$ to be a whole number, $15 + X$ must be a multiple of 13.
We are given that none of the other 7 team members scored more than 2 points. Therefore, the maximum points they could have scored together is $7 \times 2 = 14$.
So, $0 \le X \le 14$.
This means $15 \le 15 + X \le 29$.

The only multiple of 13 in the range from 15 to 29 is 26.
Thus, $15 + X = 26$, which gives $X = 11$.
(This also perfectly yields $P = 56$ total points).

ANSWER 4: B

Problem 5:
We are given that $a, b,$ and $c$ are nonzero real numbers and $a + b + c = 0$. We want to find the possible values of:
$\frac{a}{|a|} + \frac{b}{|b|} + \frac{c}{|c|} + \frac{abc}{|abc|}$

The expression $\frac{x}{|x|}$ evaluates to $1$ if $x > 0$ and $-1$ if $x < 0$.
Since $a + b + c = 0$ and none of them are zero, they cannot all be positive or all be negative. This leaves two possible cases for their signs:

Case 1: Two are positive and one is negative (e.g., $a>0, b>0, c<0$).
The first three terms will be $1 + 1 + (-1) = 1$.
The product $abc$ will be positive $\times$ positive $\times$ negative = negative, so the fourth term is $-1$.
Total sum = $1 - 1 = 0$.

Case 2: One is positive and two are negative (e.g., $a>0, b<0, c<0$).
The first three terms will be $1 + (-1) + (-1) = -1$.
The product $abc$ will be positive $\times$ negative $\times$ negative = positive, so the fourth term is $1$.
Total sum = $-1 + 1 = 0$.

In all possible cases, the expression evaluates to 0.

ANSWER 5: A

Problem 6:
Let the 15 integers be an arithmetic sequence with a common difference $d$. 
We are given:
$1 \le a_1 \le 10$
$13 \le a_2 \le 20$
$241 \le a_{15} \le 250$

The terms can be written as $a_n = a_1 + (n-1)d$.
For $a_{15}$, we have $a_{15} = a_1 + 14d$.
Substituting bounds for $a_1$ and $a_{15}$:
$241 \le a_1 + 14d \le 250$
Since $1 \le a_1 \le 10$, we can estimate $14d$:
$241 - 10 \le 14d \le 250 - 1 \implies 231 \le 14d \le 249$
Dividing by 14 gives $16.5 \le d \le 17.78$. Since $d$ must be an integer, $d = 17$.

Now we substitute $d = 17$ into the inequalities:
$a_2 = a_1 + 17$. Sinc

## Step 1: Problem 1 - Restate the problem
We need to find the number of different ways to distribute 5 different awards to 3 students, with each student receiving at least one award.

## Step 2: Problem 1 - Calculate the total number of ways to distribute the awards without restrictions
First, we consider distributing the awards without any restrictions. For each award, there are 3 choices of students, so the total number of ways to distribute 5 awards is $3^5 = 243$.

## Step 3: Problem 1 - Apply the Principle of Inclusion-Exclusion to find the number of ways where at least one student receives no award
To ensure each student gets at least one award, we'll use the Principle of Inclusion-Exclusion. We calculate the number of ways to distribute the awards where at least one student gets no award. There are $3$ ways to choose a student to not receive an award, and for each such choice, there are $2^5$ ways to distribute the awards among the remaining 2 students. So, there are $3 \cdot 2^5 = 96$ ways. However, this counts cases where 2 students receive no awards twice, so we must adjust for that.

## Step 4: Problem 1 - Continue applying the Principle of Inclusion-Exclusion
There are $3$ ways to choose 2 students to not receive an award, and for each such choice, there's $1^5 = 1$ way for the remaining student to receive all awards. So, there are $3 \cdot 1 = 3$ such cases. The total number of ways where at least one student gets no award is $3 \cdot 2^5 - 3 \cdot 1 = 96 - 3 = 93$.

## Step 5: Problem 1 - Calculate the final number of ways to distribute the awards
The number of ways to distribute the awards such that each student gets at least one is $3^5 - 93 = 243 - 93 = 150$.

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

## Step 7: Problem 2 - Restate the problem
We are to determine the number of ways Alina can divide the numbers 1 through 9 into 3 groups of 3 such that the sum of the numbers in each group is the same.

## Step 8: Problem 2 - Calculate the total sum of the numbers and the required sum for each group
The sum of the numbers 1 through 9 is $1+2+3+...+9 = \frac{9(9+1)}{2} = 45$. For 3 groups to have the same sum, each group's sum must be $45 / 3 = 15$.

## Step 9: Problem 2 - Analyze the possible combinations for the groups
To achieve a sum of 15 with 3 numbers, we need to consider the combinations. The highest number is 9, and to reach 15, the other two numbers must sum to 6. Possible pairs with 9 are (9,5,1), (9,4,2), and (9,3,3), but since we can't repeat numbers, (9,3,3) is not valid. We continue this process for other high numbers.

## Step 10: Problem 2 - Enumerate valid combinations
Valid sets that sum to 15 include (9,5,1), (9,4,2), (8,6,1), (8,5,2), (8,4,3), (7,6,2), (7,5,3), (7,4,4) - which is not valid due to repetition, and (6,5,4). We need to find combinations of 3 such numbers that use each number from 1 to 9 exactly once.

## Step 11: Problem 2 - Check for valid groupings
After examining possible combinations and considering the restrictions, we find that there are limited ways to form 3 groups. One valid grouping is (9,2,4), (8,3,4) - which doesn't work because it repeats 4, and (7,6,2), (7,5,3), (8,1,6), (9,5,1), (7,4,4) - again, not valid. A valid example is (9,5,1), (8,6,1) - not valid, (8,4,3), (7,6,2), (7,5,3), and (9,4,2), (8,6,1), (7,5,3). Upon careful examination, one valid set is (9,4,2), (8,7,0) - not valid because 0 is not in the list, but (8,6,1), (7,5,3), and (9,4,2) is a valid grouping: {9,4,2}, {8,6,1}, {7,5,3}. Another is {9,5,1}, {8,4,3}, {7,6,2}.

## Step 12: Problem 2 - Count the distinct ways to form the groups
There are 2 distinct ways to form the groups as shown by the examples.

## Step 13: Problem 2 - State the answer
ANSWER 2: C

## Step 14: Problem 3 - Restate the problem
We need to find the sum of the digits when $10^{93} - 93$ is expressed as a whole number.

## Step 15: Problem 3 - Analyze the number $10^{93}$
$10^{93}$ is a 1 followed by 93 zeros.

## Step 16: Problem 3 - Subtract 93 from $10^{93}$
Subtracting 93 from $10^{93}$ gives us a number that is 93 less than $10^{93}$. This results in a number with 93 digits, where the last two digits are 07, and the rest are 9s except for the third last digit which is 9 - 1 = 8 (due to borrowing), but since we're subtracting from a number ending in 00, it actually results in ...9907.

## Step 17: Problem 3 - Calculate the sum of the digits
The number $10^{93} - 93$ is ...9999907. It has 91 nines, an 8 (from the subtraction affecting the hundreds place due to borrowing), a 0, and a 7. So, the sum is $91*9 + 8 + 0 + 7$.

## Step 18: Problem 3 - Perform the sum of the digits calculation
$91*9 = 819$, $819 + 8 + 7 = 834$.

## Step 19: Problem 3 - State the answer
ANSWER 3: E

## Step 20: Problem 4 - Restate the problem
We need to find the total number of points scored by the other 7 team members in a basketball game given that Alexa scored 1/4 of the points, Brittany scored 2/7, and Chelsea scored 15 points.

## Step 21: Problem 4 - Set up an equation for the total points
Let $T$ be the total points. Then, $\frac{1}{4}T + \frac{2}{7}T + 15 + x = T$, where $x$ is the total points scored by the other 7 team members.

## Step 22: Problem 4 - Simplify the equation
$\frac{7T + 8T}{28} + 15 + x = T$, so $\frac{15T}{28} + 15 + x = T$.

## Step 23: Problem 4 - Solve for $x$
$T - \frac{15T}{28} = 15 + x$, which simplifies to $\frac{13T}{28} = 15 + x$. Thus, $T = \frac{28}{13}(15 + x)$.

## Step 24: Problem 4 - Use the information that none of the other 7 team members scored more than 2 points
$x \leq 14$ because $7 \times 2 = 14$.

## Step 25: Problem 4 - Test possible values of $T$ and $x$
Since $T$ must be a whole number, and given that $\frac{1}{4}T$ and $\frac{2}{7}T$ are also whole numbers, $T$ must be divisible by both 4 and 7, hence divisible by 28.

## Step 26: Problem 4 - Find the appropriate $T$
Let's start with the smallest $T$ that is divisible by 28 and satisfies the given condi

Problem 1:

We are distributing 5 **different** awards to 3 students, with each student receiving at least one award.

First count all possible distributions with no restriction:

- Each of 5 awards can go to any of 3 students.
- Total: \(3^5=243\)

Now subtract distributions where at least one student gets no award.

If a particular student gets no award, then all 5 awards go to the other 2 students:

\[
2^5=32
\]

There are 3 choices for which student gets no award, so subtract:

\[
3\cdot 32=96
\]

But distributions where two students get no awards were subtracted twice, so add them back. If only one student gets all awards, there are 3 such distributions:

\[
243-96+3=150
\]

So the number of ways is \(150\).

ANSWER 1: B

Problem 2:

The numbers are \(1,2,\dots,9\), whose total sum is

\[
1+2+\cdots+9=45
\]

They are divided into 3 groups with equal sums, so each group must sum to

\[
45\div 3=15
\]

Now list triples from \(1\) to \(9\) that sum to 15:

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

Now form complete partitions into 3 disjoint triples.

If the group containing 1 is \(\{1,5,9\}\), the remaining numbers are

\[
2,3,4,6,7,8
\]

These form

\[
\{2,6,7\},\{3,4,8\}
\]

So one partition is

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

If the group containing 1 is \(\{1,6,8\}\), the remaining numbers are

\[
2,3,4,5,7,9
\]

These form

\[
\{2,4,9\},\{3,5,7\}
\]

So another partition is

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

There are exactly 2 ways.

ANSWER 2: C

Problem 3:

We need the sum of the digits of

\[
10^{93}-93
\]

Think of subtracting 93 from a 1 followed by 93 zeros.

A useful way:

\[
10^{93}-93=(10^{93}-100)+7
\]

Now \(10^{93}-100\) is a number with 91 nines followed by two zeros:

\[
999\cdots 99900
\]

with 91 nines. Adding 7 gives

\[
999\cdots 99907
\]

So the digit sum is

\[
91\cdot 9+0+7=819+7=826
\]

ANSWER 3: D

Problem 4:

Let the total number of points be \(T\).

Alexa scored

\[
\frac14T
\]

Brittany scored

\[
\frac27T
\]

Chelsea scored 15 points.

The fraction of the total scored by Alexa and Brittany together is

\[
\frac14+\frac27=\frac7{28}+\frac8{28}=\frac{15}{28}
\]

So the remaining fraction is

\[
1-\frac{15}{28}=\frac{13}{28}
\]

Chelsea plus the other 7 players scored

\[
\frac{13}{28}T
\]

Since \(\frac14T\) and \(\frac27T\) are whole numbers, \(T\) must be a multiple of 28. Let

\[
T=28k
\]

Then Chelsea plus the other players scored

\[
\frac{13}{28}\cdot 28k=13k
\]

So the other 7 players scored

\[
13k-15
\]

Each of the other 7 players scored at most 2 points, so together they scored at most

\[
7\cdot 2=14
\]

Thus

\[
0\le 13k-15\le 14
\]

Testing positive integers \(k\):

- \(k=1\): \(13-15=-2\), impossible.
- \(k=2\): \(26-15=11\), possible.
- \(k=3\): \(39-15=24\), too large.

So the other 7 players scored 11 points.

ANSWER 4: B

Problem 5:

The expression is

\[
\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}
\]

Each fraction is just the sign of the number:

\[
\frac{x}{|x|}=
\begin{cases}
1, & x>0\\
-1, & x<0
\end{cases}
\]

Since \(a+b+c=0\) and none of \(a,b,c\) is zero, they cannot all be positive or all be negative.

So there are two cases.

Case 1: Two are positive and one is negative.

Then

\[
\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}=1+1-1=1
\]

The product \(abc\) is negative, so

\[
\frac{abc}{|abc|}=-1
\]

Total:

\[
1+(-1)=0
\]

Case 2: One is positive and two are negative.

Then

\[
1-1-1=-1
\]

The product \(abc\) is positive, so

\[
\frac{abc}{|abc|}=1
\]

Total:

\[
-1+1=0
\]

So the only possible value is \(0\).

ANSWER 5: A

Problem 6:

The numbers

\[
a_1,a_2,\dots,a_{15}
\]

are equally spaced integers, so they form an arithmetic sequence with integer common difference \(d\).

Thus

\[
a_{15}=a_1+14d
\]

We know

\[
1\le a_1\le 10
\]

and

\[
241\le a_{15}\le 250
\]

So

\[
a_{15}-a_1
\]

is between

\[
241-10=231
\]

and

\[
250-1=249
\]

But

\[
a_{15}-a_1=14d
\]

So \(14d\) must be a multiple of 14 between 231 and 249. The only such multiple is

\[
238=14\cdot 17
\]

Thus

\[
d=17
\]

Now

\[
a_2=a_1+17
\]

and we are told

\[
13\le a_2\le 20
\]

So

\[
13\le a_1+17\le 20
\]

which gives

\[
-4\le a_1\le 3
\]

Combining with \(1\le a_1\le 10\), we get

\[
a_1=3
\]

Now

\[
a_{14}=a_1+13d=3+13\cdot 17=3+221=224
\]

The digit sum of 224 is

\[
2+2+4=8
\]

ANSWER 6: A

Problem 7:

The original cube has edge length 3 cm, so its volume is

\[
3^3=27
\]

Each smaller cube has whole-number side length. Since the smaller cubes are not all the same size, we cannot just cut the cube into all unit cubes.

The possible smaller cube side lengths are 1 cm or 2 cm. A 3 cm cube would be the entire original cube.

Let \(x\) be the number of \(2\times2\times2\) cubes, and let \(y\) be the number of \(1\times1\times1\) cubes.

Then

\[
8x+y=27
\]

The total number of cubes is

\[
N=x+y
\]

From \(y=27-8x\),

\[
N=x+27-8x=27-7x
\]

Since the cubes are not all the same size, we need at least one 2 cm cube, so \(x\ge 1\).

Among the answer choices, only \(20\) fits:

\[
20=27-7\cdot 1
\]

This is possible: put one \(2\times2\times2\) cube in a corner, and fill the remaining volume with 19 unit cubes.

So

\[
N=20
\]

ANSWER 7: E

Problem 8:

Each of the 1200 students takes 5 classes per day, so the total number of student-class enrollments is

\[
1200\cdot 5=6000
\]

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

\[
6000\div 30=200
\]

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

\[
200\div 4=50
\]

The tempting answer \(40\) would come from just doing \(1200\div 30\), but that ignores the fact that each student takes 5 classes.

ANSWER 8: E

Problem 9:

We count times in a 12-hour period whose digits form a palindrome when the colon is ignored.

For one-digit hours \(1,2,\dots,9\), a time looks like

\[
h:ab
\]

Ignoring the colon, this is \(hab\). To be a palindrome, we need

\[
b=h
\]

The tens digit \(a\) of the minutes can be \(0,1,2,3,4,

Problem 1:
We are asked to find the number of ways to distribute 5 different awards to 3 students such that each student receives at least one award. 
This is equivalent to finding the number of surjective (onto) functions from a set of 5 distinct elements to a set of 3 distinct elements. 

First, let's find the total number of ways to distribute the 5 awards to the 3 students without any restrictions. Since each of the 5 awards can go to any of the 3 students, there are $3^5 = 243$ ways.

Next, we use the Principle of Inclusion-Exclusion to subtract the distributions where at least one student receives no awards:
- Number of ways where 1 specific student gets no awards: The awards are distributed among the remaining 2 students. This can be done in $2^5 = 32$ ways. Since there are 3 ways to choose which student gets no awards, this gives $3 \times 32 = 96$ ways.
- Number of ways where 2 specific students get no awards: All awards go to the 1 remaining student. This can be done in $1^5 = 1$ way. There are 3 ways to choose which two students get no awards, giving $3 \times 1 = 3$ ways.

Applying inclusion-exclusion, the number of invalid distributions (where at least one student gets nothing) is:
$96 - 3 = 93$

Subtracting the invalid distributions from the total gives the number of valid distributions:
$243 - 93 = 150$

ANSWER 1: B

Problem 2:
Alina wants to divide the numbers 1 through 9 into 3 groups of 3 cards each, such that the sum of the numbers in each group is the same.
The sum of the numbers from 1 to 9 is:
$1 + 2 + 3 + \dots + 9 = \frac{9 \times 10}{2} = 45$
For the 3 groups to have the same sum, each group must sum to $45 / 3 = 15$.

Let's find all possible groups of 3 numbers from 1 to 9 that sum to 15:
1. $9 + 5 + 1$
2. $9 + 4 + 2$
3. $8 + 6 + 1$
4. $8 + 5 + 2$
5. $8 + 4 + 3$
6. $7 + 6 + 2$
7. $7 + 5 + 3$
8. $6 + 5 + 4$

Notice that the number 1 is only present in two possible groups: $\{9, 5, 1\}$ and $\{8, 6, 1\}$. Thus, any valid division must use one of these two groups.

Case 1: We use the group $\{9, 5, 1\}$.
The remaining numbers are 2, 3, 4, 6, 7, 8. We need to form two more groups of 3 that sum to 15.
Looking at the remaining numbers, the only valid groups that sum to 15 are $\{8, 4, 3\}$ and $\{7, 6, 2\}$. 
This gives our first valid division: $\{9, 5, 1\}, \{8, 4, 3\}, \{7, 6, 2\}$.

Case 2: We use the group $\{8, 6, 1\}$.
The remaining numbers are 2, 3, 4, 5, 7, 9. We need to form two more groups of 3 that sum to 15.
From the remaining numbers, the only valid groups that sum to 15 are $\{9, 4, 2\}$ and $\{7, 5, 3\}$.
This gives our second valid division: $\{8, 6, 1\}, \{9, 4, 2\}, \{7, 5, 3\}$.

Since the number 1 must be in exactly one of these configurations, there are exactly 2 ways to divide the cards.

ANSWER 2: C

Problem 3:
We are asked to find the sum of the digits of the whole number $10^{93} - 93$.
Let's look at a smaller pattern to understand the subtraction:
$10^2 - 93 = 100 - 93 = 07$ (sum of digits = 7)
$10^3 - 93 = 1000 - 93 = 907$ (sum of digits = 16)
$10^4 - 93 = 10000 - 93 = 9907$ (sum of digits = 25)

Notice that $10^n - 93$ results in a number with $(n - 2)$ nines, followed by the digits 0 and 7.
For $10^{93} - 93$, the number of nines will be $93 - 2 = 91$.
The number looks like: $\underbrace{999\dots99}_{91 \text{ times}}07$

The sum of the digits is:
$91 \times 9 + 0 + 7 = 819 + 7 = 826$

ANSWER 3: D

Problem 4:
Let $P$ be the total number of points scored by the team.
Alexa scored $\frac{1}{4}P$ points, and Brittany scored $\frac{2}{7}P$ points.
Together, they scored:
$\frac{1}{4}P + \frac{2}{7}P = \frac{7P + 8P}{28} = \frac{15}{28}P$

This means the rest of the team scored the remaining $1 - \frac{15}{28} = \frac{13}{28}$ of the total points.
The rest of the team consists of Chelsea (who scored 15 points) and 7 other members. Let $X$ be the total points scored by the other 7 members.
So, $\frac{13}{28}P = 15 + X$.

For $P$ to be a whole number, $15 + X$ must be a multiple of 13.
We are given that none of the other 7 team members scored more than 2 points. Therefore, the maximum points they could have scored together is $7 \times 2 = 14$.
So, $0 \le X \le 14$.
This means $15 \le 15 + X \le 29$.

The only multiple of 13 in the range from 15 to 29 is 26.
Thus, $15 + X = 26$, which gives $X = 11$.
(This also perfectly yields $P = 56$ total points).

ANSWER 4: B

Problem 5:
We are given that $a, b,$ and $c$ are nonzero real numbers and $a + b + c = 0$. We want to find the possible values of:
$\frac{a}{|a|} + \frac{b}{|b|} + \frac{c}{|c|} + \frac{abc}{|abc|}$

The expression $\frac{x}{|x|}$ evaluates to $1$ if $x > 0$ and $-1$ if $x < 0$.
Since $a + b + c = 0$ and none of them are zero, they cannot all be positive or all be negative. This leaves two possible cases for their signs:

Case 1: Two are positive and one is negative (e.g., $a>0, b>0, c<0$).
The first three terms will be $1 + 1 + (-1) = 1$.
The product $abc$ will be positive $\times$ positive $\times$ negative = negative, so the fourth term is $-1$.
Total sum = $1 - 1 = 0$.

Case 2: One is positive and two are negative (e.g., $a>0, b<0, c<0$).
The first three terms will be $1 + (-1) + (-1) = -1$.
The product $abc$ will be positive $\times$ negative $\times$ negative = positive, so the fourth term is $1$.
Total sum = $-1 + 1 = 0$.

In all possible cases, the expression evaluates to 0.

ANSWER 5: A

Problem 6:
Let the 15 integers be an arithmetic sequence with a common difference $d$. 
We are given:
$1 \le a_1 \le 10$
$13 \le a_2 \le 20$
$241 \le a_{15} \le 250$

The terms can be written as $a_n = a_1 + (n-1)d$.
For $a_{15}$, we have $a_{15} = a_1 + 14d$.
Substituting bounds for $a_1$ and $a_{15}$:
$241 \le a_1 + 14d \le 250$
Since $1 \le a_1 \le 10$, we can estimate $14d$:
$241 - 10 \le 14d \le 250 - 1 \implies 231 \le 14d \le 249$
Dividing by 14 gives $16.5 \le d \le 17.78$. Since $d$ must be an integer, $d = 17$.

Now we substitute $d = 17$ into the inequalities:
$a_2 = a_1 + 17$. Sinc