1978 AHSME Problems/Problem 30

Solution

Since there are $n$ women, the number of matches between only women is $\frac{n(n-1)}{2},$ and similarly, there are $(2n - 1)n$ matches between only men. Since every woman plays every man exactly once, there are $2n\cdot n = 2n^2$ matches which are between a man and a woman. Call these $2n^2$ matches co-ed matches, and let $w$ be the number of co-ed matches won by women.

Then it follows that $\frac 75 = \frac{\frac{n(n-1)}{2} + w}{(2n - 1)n + 2n^2 - w},$ which can be simplified to $w = \frac{17}{8}n^2 - \frac38 n.$

The number of matches won by women must be less than the total number of matches, so we obtain the inequality $\frac{17}{8}n^2 - \frac38 n\leq 2n^2.$ Rearranging and factoring gives $0\leq -n(n-3),$ and the only integers which satisfy this inequality are $n = 0,1,2,$ and $3.$

Clearly, there could not have been $0$ people in the tournament, so $n\neq 0.$ If $n = 1,$ then there would have been only one woman and two men in the tournament, in which case the woman could not have won the majority of the matches.

We can now plug $n = 2$ back into the equation $\frac 75 = \frac{\frac{n(n-1)}{2} + w}{(2n - 1)n + 2n^2 - w},$ and solving for $w$ gives $w = \frac{31}{4}.$ Since $w$ must be an integer, $n$ cannot be $2.$ It follows that $n = 3,$ so the answer is (E). (When $n = 3,$ solving gives $w = 18.$ )

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1978 AHSME Problems/Problem 30

Solution