5. Let the sequence $\left\{a_{n}\right\}: a_{0}=2, a_{1}=16, a_{n+2}=$ $16 a_{n+1}-63 a_{n}(n \in \mathbf{N})$. Then the remainder when $a_{2008}$ is divided by 64 is $(\quad)$.
(A) 0
(B) 2
(C) 16
(D) 48
5.C.

The sequence $\left\{a_{n}\right\}$ has a remainder of $2,16,-2,-16,2,16$, $-2,-16, \cdots$ when divided by 64, with a period of 4. Since 2005 leaves a remainder of 1 when divided by 4, the solution is obtained.