2017 AMC 12B Problem 25

Let $T$ be the number of teams. Each team is counted $\binom{n-5}{4}$ times in the size-$9$ sum and $\binom{n-5}{3}$ times in the size-$8$ sum

The condition becomes $T =$ $\dfrac{n(n-1)(n-2)(n-3)(n-4)}{2^5 \cdot 3^2 \cdot 5 \cdot 7};$ count $n$ making this an integer

Let $T$ be the number of teams. Summing over size-$9$ subsets counts each team $\binom{n-5}{4}$ times and over size-$8$ subsets $\binom{n-5}{3}$ times. The averages are $\dfrac{\binom{n-5}{4}T}{\binom n9}$ and $\dfrac{\binom{n-5}{3}T}{\binom n8};$ setting the first equal to the reciprocal of the second and simplifying gives $$\begin{aligned} &T \\ &\quad {}= \scriptsize \frac{n(n-1)(n-2)(n-3)(n-4)}{2^5 \cdot 3^2 \cdot 5 \cdot 7}. \end{aligned}$$ We need this to be a positive integer with $n \ge 9.$ Let $N =$ $n(n-1)(n-2)(n-3)(n-4);$ as a product of five consecutive integers, $N$ is always divisible by $5.$ The condition that $N$ is divisible by $7$ holds for $5$ residues modulo $7;$ divisibility by $9$ holds for $7$ residues modulo $9;$ and divisibility by $32$ holds for $8$ residues modulo $16.$ The Chinese Remainder Theorem therefore gives $5\cdot7\cdot8=280$ solutions modulo $\operatorname{lcm}(7,9,16)=1008.$ So there are $560$ values in $1 \le n \le 2016;$ removing $n = 1, 2, 3, 4$ (which are below $9$) and adding $n = 2017$ (since $2017 \equiv 1 \pmod{1008}$) gives $560 - 4 + 1 = 557$ valid values.

Thus, the correct answer is D.