$3.2^{2011}+3^{2011}$ 's last digit is ( ).
(A) 1
(B) 3
(C) 5
(D) 7
3. C.
The last digit of $2^{n}$ cycles in the order of $2, 4, 8, 6$, while the last digit of $3^{n}$ cycles in the order of $3, 9, 7, 1$.

Since 2011 is a number of the form $4k+3$, the last digit of $2^{2011}$ is 8. The last digit of $3^{2011}$ is 7, so the last digit of $2^{2011}+3^{2011}$ is 5: