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

When $\frac{200-n}{3}, \frac{400-5 n}{6}$ are both integers, it is found that $n=6 k-4$.
Combining $n \leqslant 95 \Rightarrow k=1,2, \cdots, 16$, since $k=15,16 \Rightarrow n=86,92$.
$a_{86}=C_{200}^{86} \cdot 3^{38} \cdot 2^{-5}=\frac{200!}{86!\cdot 114!} \cdot 3^{38} \cdot 2^{-5}$, where the power of 2 in 200! is $\left[\frac{200}{2}\right]+\left[\frac{200}{2^{2}}\right]+\cdots+\left[\frac{200}{2^{7}}\right]=197,86$! The power of 2 in 86! is $\left[\frac{86}{2}\right]+\left[\frac{86}{2^{2}}\right]+\cdots+\left[\frac{86}{2^{6}}\right]$ $=82,114$! The power of 2 in 114! is $\left[\frac{114}{2}\right]+\left[\frac{114}{2^{2}}\right]+\cdots+\left[\frac{114}{2^{6}}\right]=110$.
Thus, the power of 2 in $C_{200}^{86}$ is $197-82-110=5$, so $a_{86}$ is an integer. $a_{92}=C_{200}^{92} \cdot 3^{36} \cdot 2^{-10}=\frac{200!}{92!\cdot 108!} \cdot 3^{36} \cdot 2^{-10}$, where the power of 2 in 92! is $\left[\frac{92}{2}\right]+\left[\frac{92}{2^{2}}\right]+\cdots+\left[\frac{92}{2^{6}}\right]=88,108$! The power of 2 in 108! is $\left[\frac{108}{2}\right]+\left[\frac{108}{2^{2}}\right]+\cdots+\left[\frac{108}{2^{6}}\right]$ $=104$. Thus, the power of 2 in $C_{200}^{92}$ is $197-88-104=5$, so $a_{92}$ is not an integer. In summary, the number of integer terms in the sequence $\left\{a_{n}\right\}$ is 15.