10 The last two digits of $9^{2008}$ is
(A) 01
(B) 21
(C) 41
(D) 61
(E) 81
10. Answer: (B)
Here we try to find $9^{2008}(\bmod 100)$. We see that the $9^{2}(\bmod 100)=81$, $9^{4}(\bmod 100)=61,9^{6}(\bmod 100)=41,9^{8}(\bmod 100)=21$ and $9^{10}(\bmod 100)=01$. Hence $9^{2008}=9^{8}(\bmod 100)=21$.