2016/17 AHSMC Part 1 Problem 13

Let $m$, $n$ denote the number of girls and respectively boys in the club. The condition of the problem is $$n\binom{m}{2} - m\binom{n}{2} = 2016 \iff mn(m-n) = 2 \times 2016 = 63 \times 64 = 2^6 \times 3^2 \times 7.$$ Let $k = (m,n)$ be the greatest common divisor of $m$, $n$. Since $k$ is a divisor of $m$, $n$, $m-n$, then $k^3$ will be a divisor of $2^6 \times 3^2 \times 7 = mn(m-n)$ hence $k \in \{1,2,4\}$. If $k = 1$ the only convenient solutions are $m = 64$, $n = 1$ and $m = 64$, $n = 63$, otherwise $m > 64$, which is not possible. If $k = 2$ then $m = 2m_1$, $n = 2n_1$ with $(m_1,n_1) = 1$, and $m_1 n_1(m_1-n_1) = 8 \times 9 \times 7$. No convenient integer values for $m_1$, $n_1$ can be found in this case. Similarly, if $k = 4$ then $m = 4m_2$, $n = 4n_2$ with $(m_2,n_2) = 1$, and the equation can be simplified to $m_2 n_2(m_2-n_2) = 9 \times 7$, which does not have integer solutions.

Alternative Solution: The solutions of the equation $mn(m-n) = 2^6 \times 3^2 \times 7$ can be found using another approach. First one can remark by AGM inequality that $$4032 = mn(m-n) \le m \Bigl(\frac{n+m-n}{2}\Bigr)^2 = \frac{m^3}{4}$$ hence $m^3 \ge 16128$ and thus $m \ge 26$. On the other hand if $m > 64$ then $$64 \cdot 63 = mn(m-n) > 64n(64-n)$$ hence $$63 > n(64-n) \iff (n-1)(n-63) > 0$$ and thus $n > 63$, which is not possible since $mn(m-n) = 63 \cdot 64$.

We conclude that $m \in \{28,32,36,42,48,56,63,64\}$. The only convenient value is $m = 64$ which gives the equation $n(64-n) = 63$ and hence $n = 1$, $n = 63$.

The answer is (C).