2023 AMC 10A Problem 16

1. Count the players. Say there are $L$ left-handers. Then there are $2L$ right-handers, so $3L$ players in all. The number of players is a multiple of $3$.
2. Count the games. Every pair of players meets once. With $n$ players, each plays $n - 1$ others, and each game gets counted by both players, so there are $\frac{n(n-1)}{2}$ games. Try $n = 3, 6, 9, 12$:

Of these, $15$, $36$ and $66$ are among the answer choices.
3. Use the wins. Each game has exactly one winner, so total wins = total games. Suppose right-handers won $5$ games. Then left-handers won $40\%$ more, which is $7$. The wins come in the ratio $5 : 7$, so the total number of wins is a multiple of $5 + 7 = 12$.
4. Match the two lists. Among $15$, $36$, $66$, only $36$ is a multiple of $12$.
5. Check it can really happen. With $9$ players ($3$ left, $6$ right) and $36$ games, right-handers win $15$ and left-handers win $21$. The left-handers play $3$ games among themselves and $3 \times 6 = 18$ games against right-handers. If they win all $18$ of those, they have $3 + 18 = 21$ wins, and the right-handers have the $15$ games among themselves. It works.

The answer is $\boxed{\textbf{(B)}\ 36}$.

Two separate facts each limit the total. The number of games must come from a whole number of players, and the win ratio $5 : 7$ forces a multiple of $12$. Only one answer choice passes both tests. In any tournament problem, remember that total wins equal total games.

Let there be $L$ left-handed and $2L$ right-handed players, $3L$ in all. Every pair plays once, so the number of games is
$$\binom{3L}{2} = \frac{3L(3L-1)}{2}.$$
Trying $L = 1, 2, 3, 4$ gives $3, 15, 36, 66$ games; only $15, 36, 66$ appear among the choices.

Each game produces exactly one win. If right-handers won $R$ games, left-handers won $1.4R$, so the total is $2.4R = \frac{12R}{5}$. For this to be an integer $R$ must be a multiple of $5$, and then the total is a multiple of $12$. Of $15, 36, 66$, only $36$ is divisible by $12$.

Sanity check for $L = 3$: nine players, $36$ games, left-handers win $21$ and right-handers win $15$. The three left-handers play $3$ games among themselves and $18$ against right-handers; winning all $18$ plus their $3$ internal games gives exactly $21$, so the scenario is achievable.

The answer is $\boxed{\textbf{(B)}\ 36}$.