4. Answer: (A)
$7^{2008}=(8-1)^{2008}=64 k_{1}+1$ for some integers $k_{1}$. Similarly, we have $9^{2008}=(8+1)^{2008}=64 k_{2}+1$ for some integers $k_{2}$. Hence the remainder is 2 .
4 The remainder when $7^{2008}+9^{2008}$ is divided by 64 is
(A) 2
(B) 4
(C) 8
(D) 16
(E) 32