2023 AMC 10A Problem 16

Say there are $L$ left-handers and $2L$ right-handers, so $3L$ players and $\binom{3L}{2}$ games. Every game has one winner, and left wins are $1.4$ times right wins, so the wins split $7 : 5$ and the total must be a multiple of $12$. Left-handers can win at most all games involving at least one left-hander, namely $\binom{L}{2} + 2L^2$. Hence $$\frac{7}{12}\binom{3L}{2} \leq \binom{L}{2} + 2L^2,$$ which gives $L \leq 3$. For $L = 1$ and $L = 2$, the total is not divisible by $12$. For $L = 3$, there are $\binom{9}{2} = 36$ games. This is attainable if the left-handers win all $18$ cross-group games and all $3$ games among themselves, while the right-handers win their $15$ internal games. The win totals are $21$ and $15$, so the answer is $36$. Therefore, the answer is B.