# 2000 JBMO Problems/Problem 4

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## Problem

At a tennis tournament there were $2n$ boys and $n$ girls participating. Every player played every other player. The boys won $\frac 75$ times as many matches as the girls. It is knowns that there were no draws. Find $n$.

## Solution

The total number of games played in the tournament is $\tfrac{3n(3n-1)}{2}.$ Since the boys won $\tfrac75$ as many matches as the girls, the boys won $\tfrac{7}{12}$ of all the games played, so the total number of games that a boy won is $\tfrac{7}{12} \cdot \tfrac{3n(3n-1)}{2} = \tfrac{7n(3n-1)}{8}.$

Since the number of games that a boy won is a whole number, $n(3n-1)$ must be a multiple of 8. Testing each residue, we find that $n \equiv 0,3 \pmod{8}.$

For $n$ to be valid, the number of games the boys won must be less than or equal to the number of games where a boy has played. The number of games with only girls is $\tfrac{n(n-1)}{2},$ so the number of games where there is at least one boy playing is $\tfrac{3n(3n-1)}{2} - \tfrac{n(n-1)}{2}.$ This means we can write and solve the following inequality. \begin{align*} \frac{3n(3n-1)}{2} - \frac{n(n-1)}{2} &\ge \frac{7n(3n-1)}{8} \\ \frac{5n(3n-1)}{8} &\ge \frac{n(n-1)}{2} \end{align*} Since $n > 0,$ we do not have to change the inequality sign when we divided by $n.$ \begin{align*} \frac{5(3n-1)}{4} &\ge n-1 \\ \frac{15n}{4} - \frac54 &\ge n-1 \\ \frac{11n}{4} &\ge \frac14 \\ n &\ge \frac{1}{11} \end{align*} Thus, we can confirm that all positive integers congruent to 0 or 3 modulo 8 satisfy the conditions. In summary, $\boxed{n \in \{\mathbf{N} \equiv 0,3 \pmod{8}\}}.$