# 2023 AMC 12A Problems/Problem 13

The following problem is from both the 2023 AMC 10A #16 and 2023 AMC 12A #13, so both problems redirect to this page.

## Problem

In a table tennis tournament every participant played every other participant exactly once. Although there were twice as many right-handed players as left-handed players, the number of games won by left-handed players was $40\%$ more than the number of games won by right-handed players. (There were no ties and no ambidextrous players.) What is the total number of games played?

$\textbf{(A) }15\qquad\textbf{(B) }36\qquad\textbf{(C) }45\qquad\textbf{(D) }48\qquad\textbf{(E) }66$

## Solution 1

We know that the total amount of games must be the sum of games won by left and right handed players. Then, we can write $g = l + r$, and since $l = 1.4r$, $g = 2.4r$. Given that $r$ and $g$ are both integers, $g/2.4$ also must be an integer. From here we can see that $g$ must be divisible by 12, leaving only answers B and D. Now we know the formula for how many games are played in this tournament is $n(n-1)/2$, the sum of the first $n-1$ triangular numbers. Now, setting 36 and 48 equal to the equation will show that two consecutive numbers must have a product of 72 or 96. Clearly, $72=8*9$, so the answer is $\boxed{\textbf{(B) }36}$.

~~ Antifreeze5420

## Solution 2

First, we know that every player played every other player, so there's a total of $\dbinom{n}{2}$ games since each pair of players forms a bijection to a game. Therefore, that rules out D. Also, if we assume the right-handed players won a total of $x$ games, the left-handed players must have won a total of $\dfrac{7}{5}x$ games, meaning that the total number of games played was $\dfrac{12}{5}x$. Thus, the total number of games must be divisible by $12$. Therefore leaving only answer choices B and D. Since answer choice D doesn't satisfy the first condition, the only answer that satisfies both conditions is $\boxed{\textbf{(B) }36}$

## Solution 3

Let $r$ be the amount of games the right-handed won. Since the left-handed won $1.4r$ games, the total number of games played can be expressed as $(2.4)r$, or $12/5r$, meaning that the answer is divisible by 12. This brings us down to two answer choices, $B$ and $D$. We note that the answer is some number $x$ choose $2$. This means the answer is in the form $x(x-1)/2$. Since answer choice D gives $48 = x(x-1)/2$, and $96 = x(x-1)$ has no integer solutions, we know that $\boxed{\textbf{(B) }36}$ is the only possible choice.

## Solution 4

Here is the rigid way to prove that $36$ is the only answer. Let the number of left-handed players be $n$, so the number of right-handed players is $2n$. The number of games won by the left-handed players comes in two ways:

• The games played by two left-left pairs, which is $\frac{n(n-1)}{2}$, and
• The games played by left-right pairs, which we'll call $x$.

Note that $x\leq 2n^2,$ which is the total number of games played by left-right pairs. Using the same logic for right-right pairs and right-left pairs, we have that $$\frac{\frac{n(n-1)}{2}+x}{\frac{2n(2n-1)}{2}+2n^2-x}=1.4,$$ which gives $$x=\frac{17n^2}{8}-\frac{3n}{8}.$$ We know that $x\leq 2n^2$, applying that becomes \begin{align*} \frac{17n^2}{8}-\frac{3n}{8} &\le 2n^2 \\ 17n^2 - 3n &\le 16n^2 \\ n^2 - 3n &\le 0 \\ n &\le 3. \end{align*} (We can safely divide by $n$ because it must be positive). So the total number of players $p$ can only be $3$, $6$, and $9$.

Since the total number of games $p(p-1)/2$ is $(1.4+1 = 12/5)$ times a non-negative integer number of games won by righties, $p(p-1)$ must be a multiple of $12(2) = 24$. Among $\{3,6,9\}$, only $p = 9$ satisfies this condition, so the total number of games is $(9)(8)/2 = \boxed{\textbf{(B)}\ 36}.$

~ggao5uiuc, oinava, yingkai_0_ (minor edits)

## Solution 5 (🧀Cheese🧀)

If there are $n$ players, the total number of games played must be $\binom{n}{2}$, so it has to be a triangular number. The ratio of games won by left-handed to right-handed players is $7:5$, so the number of games played must also be divisible by $12$. Finally, we notice that only $\boxed{\textbf{(B) }36}$ satisfies both of these conditions.

~MathFun1000

## Video Solution ⚡️ Under 2 min ⚡️

~Education, the Study of Everything

## Video Solution

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)