$33 \cdot 4$ and the unit digit of $1988^{1989}+1989^{1988}$ is
(A) 9.
(B) 7.
(C) 5.
(D) 3.
(Chinese Jiangsu Province Junior High School Mathematics Competition, 1989)
[Solution] Let $\langle a\rangle$ denote the unit digit of $a$.
Then
$\left\langle 1988^{1989}\right\rangle=\left\langle 8^{1989}\right\rangle=\left\langle 8^{4 \cdot 497+1}\right\rangle=\left\langle 8^{1}\right\rangle=8$.
Also, $\left\langle 1989^{1988}\right\rangle=\left\langle 9^{1988}\right\rangle=\left\langle 9^{4 \cdot 497}\right\rangle=\left\langle 9^{4}\right\rangle=1$,

Thus, $\left\langle 1988^{1989}+1989^{1988}\right\rangle=\left\langle 1988^{1989}\right\rangle+\left\langle 1989^{1988}\right\rangle=8+1=9$.
Therefore, the answer is $(A)$.