$33 \cdot 13$ If $N=\frac{1987^{4(\sqrt{3}+2)^{1987}+1988}}{1987^{(\sqrt{3}+2)^{1988}}+(\sqrt{3}+2)^{1988}}$, then the last digit of $N$ is
(A) 1 .
(B) 3 .
(C) 7 .
(D) 9 .
(China Sichuan Province Junior High School Mathematics Competition, 1988)
［Solution］By the properties of exponential operations, we also have:
$$
\begin{array}{l}
=1987^{(\sqrt{3}+2)^{1986}}\left[4(\sqrt{3}+2)-(\sqrt{3}+2)^{2}-1\right]+1988 \\
=1987^{1988}=1987^{497 \cdot 4} \text {, } \\
\end{array}
$$

Thus, the last digit of $N$ is 1 (the last digit of $7^{k}$ cycles through $7,9,3,1$). Therefore, the answer is $(A)$.