1. Let $a_{n}=7^{n}+9^{n}\left(n \in \mathbf{N}_{+}\right)$. Then the remainder when $a_{2008}$ is divided by 64 is ( ).
(A) 0
(B) 2
(C) 16
(D) 22
$\begin{array}{l}\text { i.1.B. } \\ a_{2008}=7^{2008}+9^{2008} \\ =(8-1)^{2008}+(8+1)^{2008} \\ =64 k+2 .\end{array}$