9. Find the number of positive integers $k<100$ such that $2\left(3^{6 n}\right)+k\left(2^{3 n+1}\right)-1$ is divisible by 7 for any positive integer $n$.
(A) 10
(B) 12
(C) 13
(D) 14
(E) 16
9. Answer: (D)
We have
$$
\begin{aligned}
2\left(3^{6 n}\right)+k\left(2^{3 n+1}\right)-1=2\left(27^{2 n}\right)+2 k\left(8^{n}\right)-1 & =2(-1)^{2 \pi}+2 k\left(1^{n}\right)-1(\bmod 7) \\
& =2 k+1(\bmod 7) .
\end{aligned}
$$

Thus, for any positive integer $n, 2\left(3^{6 n}\right)+k\left(2^{3 n+1}\right)-1$ is divisible by 7 if and only if $2 k+1=0(\bmod 7)$. As $k<100$, it is clear that the congruence holds for $k=3$, $10,17, \ldots, 94$. Thus the required number of positive integers $k$ is 14 .