$33 \cdot 5$ and $3^{1991}+1991^{3}$, when expressed in decimal form, have their last digit as
(A) 8.
(B) 4.
(C) 2.
(D) 0.
(3rd "Five Sheep Cup" Junior High School Mathematics Competition, 1991)
[Solution] $3^{1091}=3^{4 \cdot 477+3}=81^{497} \cdot 3^{3}$, its unit digit is 7. Also, the unit digit of $1991^{3}$ is 1, so the last digit of $3^{1991}+1991^{3}$ is 8. Therefore, the answer is $(A)$.