(8 Given $a_{n}=\mathrm{C}_{200}^{n} \cdot(\sqrt[3]{6})^{200-n} \cdot\left(\frac{1}{\sqrt{2}}\right)^{n}(n=1,2, \cdots, 95)$, then the number of integer terms in the sequence $\left\{a_{n}\right\}$ is $\qquad$ .
(8) 15 Hint: $a_{n}=\mathrm{C}_{200}^{n} \cdot 3^{\frac{200 n}{3}} \cdot 2^{\frac{400-5 n}{6}}$.

To make $a_{n}(1 \leqslant n \leqslant 95)$ an integer, it must be that $\frac{200-n}{3}, \frac{400-5 n}{6}$ are both integers, thus $6 \mid n+4$.

When $n=2,8,14,20,26,32,38,44,50,56,62,68,74,80$, $\frac{200-n}{3}$ and $\frac{400-5 n}{6}$ are both non-negative integers, so $a_{n}$ is an integer, with a total of 14. When $n=86$, $a_{86}=\mathrm{C}_{200}^{86} \cdot 3^{38} \cdot 2^{-5}$, in $\mathrm{C}_{200}^{86}=\frac{200!}{86!\cdot 114!}$, the number of factor 2 in 200! is
$$
\left[\frac{200}{2}\right]+\left[\frac{200}{2^{2}}\right]+\left[\frac{200}{2^{3}}\right]+\left[\frac{200}{2^{4}}\right]+\left[\frac{200}{2^{5}}\right]+\left[\frac{200}{2^{6}}\right]+\left[\frac{200}{2^{7}}\right]=197 \text {, }
$$

Similarly, the number of factor 2 in 86! is 82, and the number of factor 2 in 114! is 110, so the number of factor 2 in $\mathrm{C}_{200}^{86}$ is $197-82-110=5$, hence $a_{86}$ is an integer.

When $n=92$, $a_{92}=\mathrm{C}_{200}^{92} \cdot 3^{36} \cdot 2^{-10}$, in $\mathrm{C}_{200}^{92}=\frac{200!}{92!\cdot 108!}$, similarly, the number of factor 2 in 92! is 88, and the number of factor 2 in 108! is 105, so the number of factor 2 in $\mathrm{C}_{200}^{86}$ is $197-88-105=4$, hence $a_{92}$ is not an integer.
Therefore, the number of integer terms is $14+1=15$.