China Mathematical Competition

We have $a_n=C_{200}^n\cdot 3^{\frac{200-n}{3}}\cdot 2^{\frac{400-5n}{6}}$. When $a_n$ $(1\le n\le 95)$ is an integer, $\frac{200-n}{3}$ and $\frac{400-5n}{6}$ must be integers. Then $6\mid n+4$.

When $n=2,8,14,20,26,32,38,44,50,56,62,68,74,80$, $\frac{200-n}{3}$ and $\frac{400-5n}{6}$ are all non-negative integers. So the corresponding $a_n$, totally 14, are integers.

When $n=86$, we have $a_{86}=C_{200}^{86}\cdot 3^{38}\cdot 2^{-5}$. The number of the factors of 2 in $200!$ is $$\left[\frac{200}{2}\right]+\left[\frac{200}{2^2}\right]+\left[\frac{200}{2^3}\right]+\left[\frac{200}{2^4}\right]+\left[\frac{200}{2^5}\right]+\left[\frac{200}{2^6}\right]+\left[\frac{200}{2^7}\right]=197.$$ By the same reason, the numbers of the factors of 2 in $86!$ and $114!$ are 82 and 110, respectively. Therefore, the number of the factors of 2 in $C_{200}^{86}=\frac{200!}{86!\cdot 114!}$ is $197-82-110=5$. So $a_{86}$ is an integer.

When $n=92$, we have $a_{92}=C_{200}^{92}\cdot 3^{36}\cdot 2^{-10}$. In the same way, we find the numbers of the factors of 2 in $92!$ and $108!$ are 88 and 105, respectively, which means that in $C_{200}^{92}$ is $197-88-105=4$. Therefore, $a_{92}$ is not an integer.

Overall, the required number is $14+1=15$. ☐