Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: The sum \[\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{2021}{2022!}\] can be expressed as $a-\frac{1}{b!}$ , where $a$ and $b$ are positive integers. What is $a+b$
$\textbf{(A)}\ 2020 \qquad\textbf{(B)}\ 2021 \qquad\textbf{(C)}\ 2022 \qquad\textbf{(D)}\ 2023 \qquad\textbf{(E)}\ 2024$

Solution: By looking for a pattern, we see that $\tfrac{1}{2!} = 1 - \tfrac{1}{2!}$ and $\tfrac{1}{2!} + \tfrac{2}{3!} = \tfrac{5}{6} = 1 - \tfrac{1}{3!}$ , so we can conclude by engineer's induction that the sum in the problem is equal to $1 - \tfrac{1}{2022!}$ , for an answer of $\boxed{2023}$ , completing the proof.
~eibc

Answer: D 2023
Problem: The sum \[\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{2021}{2022!}\] can be expressed as $a-\frac{1}{b!}$ , where $a$ and $b$ are positive integers. What is $a+b$
$\textbf{(A)}\ 2020 \qquad\textbf{(B)}\ 2021 \qquad\textbf{(C)}\ 2022 \qquad\textbf{(D)}\ 2023 \qquad\textbf{(E)}\ 2024$

Solution: Let $x=\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\dots+\frac{1}{2022!}.$
Note that \begin{align*} \left(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\dots+\frac{2021}{2022!}\right)+\left(\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\dots+\frac{1}{2022!}\right)&=\frac{1}{1!}+\frac{2}{2!}+\frac{3}{3!}+\dots+\frac{2022}{2022!}\\ \left(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\dots+\frac{2021}{2022!}\right)+x&=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{2021!}\\ \left(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\dots+\frac{2021}{2022!}\right)+x&=x+1-\frac{1}{2022!}\\ \left(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\dots+\frac{2021}{2022!}\right)&=1-\frac{1}{2022!}. \end{align*} Therefore, the answer is $1+2022=\boxed{2023}.$

Answer: D 2023
Problem: The sum \[\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{2021}{2022!}\] can be expressed as $a-\frac{1}{b!}$ , where $a$ and $b$ are positive integers. What is $a+b$
$\textbf{(A)}\ 2020 \qquad\textbf{(B)}\ 2021 \qquad\textbf{(C)}\ 2022 \qquad\textbf{(D)}\ 2023 \qquad\textbf{(E)}\ 2024$

Solution: Because the fractions get smaller, it is obvious that the answer is less than $1$ , so we can safely assume that $a=1$ (this can also be guessed by intuition using similar math problems). Looking at the answer choices, $2018<b<2024$ . Because the last term consists of $2022!$ (and the year is $2022$ ) we can guess that $b=2022$ . Adding them yields $1+2022=\boxed{2023}$

Answer: D 2023
Problem: The sum \[\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{2021}{2022!}\] can be expressed as $a-\frac{1}{b!}$ , where $a$ and $b$ are positive integers. What is $a+b$
$\textbf{(A)}\ 2020 \qquad\textbf{(B)}\ 2021 \qquad\textbf{(C)}\ 2022 \qquad\textbf{(D)}\ 2023 \qquad\textbf{(E)}\ 2024$

Solution: Knowing that the answer will be in the form $a-\frac{1}{b!}$ , we can guess that the sum telescopes. Using partial fractions, we can hope to rewrite $\frac{n-1}{n!}$ as $\frac{A}{(n-1)!}-\frac{B}{n}$ . Setting these equal and multiplying by $n!$ , we get $n-1=An-B(n-1)!$ . Since $An$ is the only term with $n$ with degree $1$ , we can conclude that $A=1$ . This means that $B=\frac{1}{(n-1)!}$ . Substituting, we find that $\frac{n-1}{n!}=\frac{1}{(n-1)!}-\frac{1}{n!}$ . This sum clearly telescopes and we obtain $1-\frac{1}{2022!}$ . This means that our desired answer is $1+2022=\boxed{2023}.$

Answer: D 2023
Problem: Camila writes down five positive integers. The unique mode of these integers is $2$ greater than their median, and the median is $2$ greater than their arithmetic mean. What is the least possible value for the mode?
$\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 7 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)}\ 13$

Solution: Let $M$ be the median. It follows that the two largest integers are both $M+2.$
Let $a$ and $b$ be the two smallest integers such that $a<b.$ The sorted list is \[a,b,M,M+2,M+2.\] Since the median is $2$ greater than their arithmetic mean, we have $\frac{a+b+M+(M+2)+(M+2)}{5}+2=M,$ or \[a+b+14=2M.\] Note that $a+b$ must be even. We minimize this sum so that the arithmetic mean, the median, and the unique mode are minimized. Let $a=1$ and $b=3,$ from which $M=9$ and $M+2=\boxed{11}.$

Answer: D 11
Problem: Camila writes down five positive integers. The unique mode of these integers is $2$ greater than their median, and the median is $2$ greater than their arithmetic mean. What is the least possible value for the mode?
$\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 7 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)}\ 13$

Solution: We can also easily test all the answer choices. (This strategy is generally good to use for multiple-choice questions if you don't have a concrete method to proceed with!)
For answer choice $\textbf{(A)},$ the mode is $5,$ the median is $3,$ and the arithmetic mean is $1.$ However, we can quickly see this doesn't work, as there are five integers, and they can't have an arithmetic mean of $1$ while having a mode of $5.$
Trying answer choice $\textbf{(B)},$ the mode is $7,$ the median is $5,$ and the arithmetic mean is $3.$ From the arithmetic mean, we know that all the numbers have to sum to $15.$ We know three of the numbers: $\underline{\hspace{3mm}},\underline{\hspace{3mm}},5,7,7.$ This exceeds the sum of $15.$
Now we try answer choice $\textbf{(C)}.$ The mode is $9,$ the median is $7,$ and the arithmetic mean is $5.$ From the arithmetic mean, we know that the list sums to $25.$ Three of the numbers are $\underline{\hspace{3mm}},\underline{\hspace{3mm}},7,9,9,$ which is exactly $25.$ However, our list needs positive integers, so this won't work.
Since we were really close on answer choice $\textbf{(C)},$ we can intuitively feel that the answer is probably going to be $\textbf{(D)}.$ We can confirm this by creating a list that satisfies the problem and choose $\textbf{(D)}: 1,3,9,11,11.$
So, our answer is $\boxed{11}.$

Answer: D 11
Problem: A pair of fair $6$ -sided dice is rolled $n$ times. What is the least value of $n$ such that the probability that the sum of the numbers face up on a roll equals $7$ at least once is greater than $\frac{1}{2}$
$\textbf{(A) } 2 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 4 \qquad \textbf{(D) } 5 \qquad \textbf{(E) } 6$

Solution: Rolling a pair of fair $6$ -sided dice, the probability of getting a sum of $7$ is $\frac16:$ Regardless what the first die shows, the second die has exactly one outcome to make the sum $7.$ We consider the complement: The probability of not getting a sum of $7$ is $1-\frac16=\frac56.$ Rolling the pair of dice $n$ times, the probability of getting a sum of $7$ at least once is $1-\left(\frac56\right)^n.$
Therefore, we have $1-\left(\frac56\right)^n>\frac12,$ or \[\left(\frac56\right)^n<\frac12.\] Since $\left(\frac56\right)^4<\frac12<\left(\frac56\right)^3,$ the least integer $n$ satisfying the inequality is $\boxed{4}.$

Answer: C 4
Problem: A pair of fair $6$ -sided dice is rolled $n$ times. What is the least value of $n$ such that the probability that the sum of the numbers face up on a roll equals $7$ at least once is greater than $\frac{1}{2}$
$\textbf{(A) } 2 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 4 \qquad \textbf{(D) } 5 \qquad \textbf{(E) } 6$

Solution: Let's try the answer choices. We can quickly find that when we roll $3$ dice, either the first and second sum to $7$ , the first and third sum to $7$ , or the second and third sum to $7$ . There are $6$ ways for the first and second dice to sum to $7$ $6$ ways for the first and third to sum to $7$ , and $6$ ways for the second and third dice to sum to $7$ . However, we overcounted (but not by much) so we can assume that the answer is $\boxed{4}$

Answer: C 4
Problem: A pair of fair $6$ -sided dice is rolled $n$ times. What is the least value of $n$ such that the probability that the sum of the numbers face up on a roll equals $7$ at least once is greater than $\frac{1}{2}$
$\textbf{(A) } 2 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 4 \qquad \textbf{(D) } 5 \qquad \textbf{(E) } 6$

Solution: We can start by figuring out what the probability is for each die to add up to $7$ if there is only $1$ roll. We can quickly see that the probability is $\frac16$ , as there are $6$ ways to make $7$ from $2$ numbers on a die, and there are a total of $36$ ways to add $2$ numbers on a die. And since each time we roll the dice, we are adding to the probability, we can conclude that the total probability for $n$ rolls would be $\frac16$ $n$ . The smallest number that satisfies this is $\boxed{4}$

Answer: C 4
Problem: The positive difference between a pair of primes is equal to $2$ , and the positive difference between the cubes of the two primes is $31106$ . What is the sum of the digits of the least prime that is greater than those two primes?
$\textbf{(A)}\ 8 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 11 \qquad\textbf{(D)}\ 13 \qquad\textbf{(E)}\ 16$

Solution: Let the two primes be $a$ and $b$ . We would have $a-b=2$ and $a^{3}-b^{3}=31106$ . Using difference of cubes, we would have $(a-b)(a^{2}+ab+b^{2})=31106$ . Since we know $a-b$ is equal to $2$ $(a-b)(a^{2}+ab+b^{2})$ would become $2(a^{2}+ab+b^{2})=31106$ . Simplifying more, we would get $a^{2}+ab+b^{2}=15553$
Now let's introduce another variable. Instead of using $a$ and $b$ , we can express the primes as $x+2$ and $x$ where $a$ is $x+2$ and b is $x$ . Plugging $x$ and $x+2$ in, we would have $(x+2)^{2}+x(x+2)+x^{2}$ . When we expand the parenthesis, it would become $x^{2}+4x+4+x^{2}+2x+x^{2}$ . Then we combine like terms to get $3x^{2}+6x+4$ which equals $15553$ . Then we subtract 4 from both sides to get $3x^{2}+6x=15549$ . Since all three numbers are divisible by 3, we can divide by 3 to get $x^{2}+2x=5183$
Notice how if we add 1 to both sides, the left side would become a perfect square trinomial: $x^{2}+2x+1=5184$ which is $(x+1)^{2}=5184$ . Since $2$ is too small to be a valid number, the two primes must be odd, therefore $x+1$ is the number in the middle of them. Conveniently enough, $5184=72^{2}$ so the two numbers are $71$ and $73$ . The next prime number is $79$ , and $7+9=16$ so the answer is $\boxed{16}$

Answer: E 16
Problem: The positive difference between a pair of primes is equal to $2$ , and the positive difference between the cubes of the two primes is $31106$ . What is the sum of the digits of the least prime that is greater than those two primes?
$\textbf{(A)}\ 8 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 11 \qquad\textbf{(D)}\ 13 \qquad\textbf{(E)}\ 16$

Solution: Let the two primes be $a$ and $b$ , with $a$ being the larger prime. We have $a - b = 2$ , and $a^3 - b^3 = 31106$ . Using difference of cubes, we obtain $a^2 + ab + b^2 = 15553$ . Now, we use the equation $a - b = 2$ to obtain $a^2 - 2ab + b^2 = 4$ . Hence, \[a^2 + ab + b^2 - (a^2 - 2ab + b^2) = 3ab = 15553 - 4 = 15549\] \[ab = 5183.\] Because we have $b = a+2$ $ab = (a+1)^2 - (1)^2$ . Thus, $(a+1)^2 = 5183 + 1 = 5184$ , so $a+1 = 72$ . This implies $a = 71$ $b = 73$ , and thus the next biggest prime is $79$ , so our answer is $7 + 9 = \boxed{16}$

Answer: E 16
Problem: The positive difference between a pair of primes is equal to $2$ , and the positive difference between the cubes of the two primes is $31106$ . What is the sum of the digits of the least prime that is greater than those two primes?
$\textbf{(A)}\ 8 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 11 \qquad\textbf{(D)}\ 13 \qquad\textbf{(E)}\ 16$

Solution: Let the two primes be $p$ and $q$ such that $p-q=2$ and $p^{3}-q^{3}=31106$
By the difference of cubes formula, $p^{3}-q^{3}=(p-q)(p^{2}+pq+q^{2})$
Plugging in $p-q=2$ and $p^{3}-q^{3}=31106$
$31106=2(p^{2}+pq+q^{2})$
Through the givens, we can see that $p \approx q$
Thus, $31106=2(p^{2}+pq+q^{2})\approx 6p^{2}\\p^2\approx \tfrac{31106}{6}\approx 5200$
Recall that $70^2=4900$ and $80^2=6400$ . It follows that our primes must be only marginally larger than $70$ , where we conveniently find $p=73, q=71$
The least prime greater than these two primes is $79 \implies 7 + 9 \implies \boxed{16}$

Answer: E 16
Problem: The positive difference between a pair of primes is equal to $2$ , and the positive difference between the cubes of the two primes is $31106$ . What is the sum of the digits of the least prime that is greater than those two primes?
$\textbf{(A)}\ 8 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 11 \qquad\textbf{(D)}\ 13 \qquad\textbf{(E)}\ 16$

Solution: Let the two primes be $x + 1$ and $x - 1$ . Then, plugging it into the second condition, we get $(x + 1)^3 - (x - 1)^3 = 31106.$ Expanding the left side, \[6x^2 + 2 = 31106 \implies x^2 = 5184.\] Taking the square root of both sides, we get that $x = 72$ and the larger prime is $73$ . The smallest prime larger than $73$ is $79$ , which has a digit sum of $7 + 9 = \boxed{16}.$

Answer:  16
Problem: Suppose that $S$ is a subset of $\left\{ 1, 2, 3, \cdots , 25 \right\}$ such that the sum of any two (not necessarily distinct) elements of $S$ is never an element of $S.$ What is the maximum number of elements $S$ may contain?
$\textbf{(A)}\ 12 \qquad\textbf{(B)}\ 13 \qquad\textbf{(C)}\ 14 \qquad\textbf{(D)}\ 15 \qquad\textbf{(E)}\ 16$

Solution: Let $M$ be the largest number in $S$ .
We categorize numbers $\left\{ 1, 2, \ldots , M-1 \right\}$ (except $\frac{M}{2}$ if $M$ is even) into $\left\lfloor \frac{M-1}{2} \right\rfloor$ groups, such that the $i$ th group contains two numbers $i$ and $M-i$
Recall that $M \in S$ and the sum of two numbers in $S$ cannot be equal to $M$ , and the sum of numbers in each group above is equal to $S$ . Thus, each of the above $\left\lfloor \frac{M-1}{2} \right\rfloor$ groups can have at most one number in $S$ .
Therefore, \begin{align*} |S| & \leq 1 + \left\lfloor \frac{M-1}{2} \right\rfloor \\ & \leq 1 + \left\lfloor \frac{25}{2} \right\rfloor \\ & = 13. \end{align*}
Next, we construct an instance of $S$ with $|S| = 13$ .
Let $S = \left\{ 13, 14, \ldots , 25 \right\}$ .
Thus, this set is feasible.
Therefore, the most number of elements in $S$ is $\boxed{13}$

Answer: B 13
Problem: Suppose that $S$ is a subset of $\left\{ 1, 2, 3, \cdots , 25 \right\}$ such that the sum of any two (not necessarily distinct) elements of $S$ is never an element of $S.$ What is the maximum number of elements $S$ may contain?
$\textbf{(A)}\ 12 \qquad\textbf{(B)}\ 13 \qquad\textbf{(C)}\ 14 \qquad\textbf{(D)}\ 15 \qquad\textbf{(E)}\ 16$

Solution: We know that two odd numbers sum to an even number, so we can easily say that odd numbers $1-25$ can be included in the list, making for $13$ elements. But, how do we know we can't include even numbers for a higher element value? Well, to get a higher element value than $13$ , odd numbers as well as even numbers would have to be included in the list (since there are only $12$ even numbers from $1-25$ , and many of those even numbers are the sum of even numbers). However, for every even value we add to our odd list, we have to take away an odd number because there are either two odd numbers that sum to that even value, or that even value and another odd number will sum to an odd number later in the list. So, $\boxed{13}$ elements is the highest we can go.

Answer: B 13
Problem: Suppose that $S$ is a subset of $\left\{ 1, 2, 3, \cdots , 25 \right\}$ such that the sum of any two (not necessarily distinct) elements of $S$ is never an element of $S.$ What is the maximum number of elements $S$ may contain?
$\textbf{(A)}\ 12 \qquad\textbf{(B)}\ 13 \qquad\textbf{(C)}\ 14 \qquad\textbf{(D)}\ 15 \qquad\textbf{(E)}\ 16$

Solution: The smallest sum of a number $a + b$ where $b \geq a$ is $a + a = 2a$ as we are using the smallest value of $b$ . Using this, we can say that if $12$ were an element of $S$ , then one of the sums (the smallest) would be $12 + 12 = 24 < 25$ . Thus $13$ must be the smallest element. So the largest amount of elements that could be in $S$ is the list of numbers from $13$ to $25$ as they all work. Because it is inclusive we have, $25 - 13 + 1 =   \boxed{13}$

Answer: B 13
Problem: Suppose that $S$ is a subset of $\left\{ 1, 2, 3, \cdots , 25 \right\}$ such that the sum of any two (not necessarily distinct) elements of $S$ is never an element of $S.$ What is the maximum number of elements $S$ may contain?
$\textbf{(A)}\ 12 \qquad\textbf{(B)}\ 13 \qquad\textbf{(C)}\ 14 \qquad\textbf{(D)}\ 15 \qquad\textbf{(E)}\ 16$

Solution: We construct a possible subset $S$ with $13$ elements by including all odd integers from $1$ to $25$ , inclusive. $S=\left\{ 1, 3, 5, \cdots , 25 \right\}$ . The sum of any $2$ elements is even, and thus cannot be an element of $S$
To show that $S$ cannot have more than $13$ elements, assume for sake of contradiction that $|S| \geq 14$ . Let $S=\left\{ x_1, x_2, \cdots , x_n \right\}$ where $n \geq 14$ and $x_1 < x_2 < \cdots < x_n$ . Because the sums of any $2$ (not necessarily distinct) elements do not appear in $S$ $x_1+x_i$ is not an element of $S$ for all $1 \leq i \leq n$ . So, $x_1, x_2, \cdots , x_n , x_1+x_1, x_1+x_2, \cdots , x_1+x_n$ are all distinct integers. Let these integers be elements of the set $T$ $|T|=2n$ , and because $n \geq 14$ $|T| \geq 28$ . But all elements of $T$ must be $\geq x_1$ and $\leq x_1+x_n \leq x_1+25$ , leaving only 26 possible values for the elements in $T$ . By the Pigeonhole Principle, the elements cannot be distinct, and we have a contradiction.
Thus, $\boxed{13}$ is the maximum possible size of $S$

Answer: B 13
Problem: Suppose that $S$ is a subset of $\left\{ 1, 2, 3, \cdots , 25 \right\}$ such that the sum of any two (not necessarily distinct) elements of $S$ is never an element of $S.$ What is the maximum number of elements $S$ may contain?
$\textbf{(A)}\ 12 \qquad\textbf{(B)}\ 13 \qquad\textbf{(C)}\ 14 \qquad\textbf{(D)}\ 15 \qquad\textbf{(E)}\ 16$

Solution: We can start by building a list of the elements of $S$ ,and see if we can find a pattern. Let's start with $1$ . If we include 1, that means we cannot include $2$ (which is $1$ $1$ ), so we write the next valid number, $3$ . Similarly, we cannot include $4$ or $6$ , so we write $5$ . Proceeding, our list looks like this:
$S$ = { $1$ $3$ $5$ $7$ ....
We've found a pattern! It's easy understand why this pattern of odd numbers will continue, seeing as all the even numbers can be written as sums between odds. Calculating how many odd numbers fit into the range $1$ $25$ , we conclude that the answer is $\boxed{13}$

Answer: B 13
Problem: Let $S_n$ be the sum of the first $n$ terms of an arithmetic sequence that has a common difference of $2$ . The quotient $\frac{S_{3n}}{S_n}$ does not depend on $n$ . What is $S_{20}$
$\textbf{(A) } 340 \qquad \textbf{(B) } 360 \qquad \textbf{(C) } 380 \qquad \textbf{(D) } 400 \qquad \textbf{(E) } 420$

Solution: Suppose that the first number of the arithmetic sequence is $a$ . We will try to compute the value of $S_{n}$ . First, note that the sum of an arithmetic sequence is equal to the number of terms multiplied by the median of the sequence. The median of this sequence is equal to $a + n - 1$ . Thus, the value of $S_{n}$ is $n(a + n - 1) = n^2 + n(a - 1)$ . Then, \[\frac{S_{3n}}{S_{n}} = \frac{9n^2 + 3n(a - 1)}{n^2 + n(a - 1)} = 9 - \frac{6n(a-1)}{n^2 + n(a-1)}.\] Of course, for this value to be constant, $6n(a-1)$ must be $0$ for all values of $n$ , and thus $a = 1$ . Finally, we have $S_{20} = 20^2 = \boxed{400}$

Answer: D 400
Problem: Let $S_n$ be the sum of the first $n$ terms of an arithmetic sequence that has a common difference of $2$ . The quotient $\frac{S_{3n}}{S_n}$ does not depend on $n$ . What is $S_{20}$
$\textbf{(A) } 340 \qquad \textbf{(B) } 360 \qquad \textbf{(C) } 380 \qquad \textbf{(D) } 400 \qquad \textbf{(E) } 420$

Solution: Recall that the sum of the first $n$ odd numbers is $n^2$
Since $\frac{S_{3n}}{S_{n}} = \frac{9n^2}{n^2} = 9$ , we have $S_n = 20^2 = \boxed{400}$

Answer: D 400
Problem: Consider systems of three linear equations with unknowns $x$ $y$ , and $z$ \begin{align*} a_1 x + b_1 y + c_1 z & = 0 \\ a_2 x + b_2 y + c_2 z & = 0 \\ a_3 x + b_3 y + c_3 z & = 0 \end{align*} where each of the coefficients is either $0$ or $1$ and the system has a solution other than $x=y=z=0$ .
For example, one such system is \[\{ 1x + 1y + 0z = 0, 0x + 1y + 1z = 0, 0x + 0y + 0z = 0 \}\] with a nonzero solution of $\{x,y,z\} = \{1, -1, 1\}$ . How many such systems of equations are there?
(The equations in a system need not be distinct, and two systems containing the same equations in a
different order are considered different.)
$\textbf{(A)}\ 302 \qquad\textbf{(B)}\ 338 \qquad\textbf{(C)}\ 340 \qquad\textbf{(D)}\ 343 \qquad\textbf{(E)}\ 344$

Solution: Let $M_1=\begin{bmatrix}a_1 & b_1 & c_1\end{bmatrix}, M_2=\begin{bmatrix}a_2 & b_2 & c_2\end{bmatrix},$ and $M_3=\begin{bmatrix}a_3 & b_3 & c_3\end{bmatrix}.$
We wish to count the ordered triples $(M_1,M_2,M_3)$ of row matrices. We perform casework:
There are $9+3=12$ ordered triples $(M_1,M_2,M_3).$
Similarly, for each of $M_1+M_3=M_2$ and $M_2+M_3=M_1,$ there are $12$ ordered triples $(M_1,M_2,M_3).$
In this subcase, we have $\boldsymbol{12\cdot3=36}$ ordered triples $\boldsymbol{(M_1,M_2,M_3).}$
Together, the answer is $8+168+126+36=\boxed{338}.$

Answer: B 338
Problem: Consider systems of three linear equations with unknowns $x$ $y$ , and $z$ \begin{align*} a_1 x + b_1 y + c_1 z & = 0 \\ a_2 x + b_2 y + c_2 z & = 0 \\ a_3 x + b_3 y + c_3 z & = 0 \end{align*} where each of the coefficients is either $0$ or $1$ and the system has a solution other than $x=y=z=0$ .
For example, one such system is \[\{ 1x + 1y + 0z = 0, 0x + 1y + 1z = 0, 0x + 0y + 0z = 0 \}\] with a nonzero solution of $\{x,y,z\} = \{1, -1, 1\}$ . How many such systems of equations are there?
(The equations in a system need not be distinct, and two systems containing the same equations in a
different order are considered different.)
$\textbf{(A)}\ 302 \qquad\textbf{(B)}\ 338 \qquad\textbf{(C)}\ 340 \qquad\textbf{(D)}\ 343 \qquad\textbf{(E)}\ 344$

Solution: The total number of possible systems is $2^9 = 512$ , with $8$ possible sets of coefficients per equation. We will use complementary counting to find the number of systems which only have the solution $(0, 0, 0)$ and subtract that from the total. Similar to what is observed in Solution 2, if any equation is repeated or $0x + 0y + 0z = 0$ , there will only be two or fewer equations for three variables, making one unique solution impossible. Therefore, we must choose $3$ different equations from $7$ possible ones, giving $7 \cdot 6 \cdot 5 = 210$ systems. However, there are two exceptions to consider, which will have more than one solution. The first is of the form $x + y + z = 0$ $x + y = 0$ $z = 0$ ; the second is of the form $x = 0$ $y = 0$ $x + y = 0$ . In both cases, there are $3$ ways to choose the variables in the equations, and then $6$ ways to arrange them, giving $2 \cdot 3 \cdot 6 = 36$ exceptions. Subtracting this gives $210 - 36 = 174$ systems with only one solution, and the answer is then $512 - 174 = \boxed{338}$

Answer: B 338