$29 \cdot 171997^{2000}$ when divided by 7 leaves a remainder of
(A) 1 .
(B) 2 .
(C) 4 .
(D) 6 .
(Shanxi Province Taiyuan City Junior High School Mathematics Competition, 1997)
［Solution］Since 1997 can be written in the form $7 n+2$, then
$$
1997^{2}=7 h+4 \text { and } 1997^{3}=(7 n+2)^{3}=7 m+1 \text {, }
$$

thus $1997^{1998}$ can be written in the form $7 k+1$.
Therefore, $1997^{2000}=1997^{2} \cdot(7 k+1)=(7 h+4)(7 k+1)$, where $m$, $n, k, h \in J$.

It follows that the remainder when $1997^{2000}$ is divided by 7 is 4. Hence, the answer is $(C)$.