$6 \cdot 53$ Find all positive integers $M$ less than 10 such that 5 divides $1989^{M}+$ $M^{1989}$.
(China National Training Team Problem, 1990)
[Solution] Consider $\bmod 5$.
$$
1989^{M} \equiv(-1)^{M}(\bmod 5) .
$$

If $M=5$, then

then
$$
\begin{array}{c}
M^{1989}=5^{1989} \equiv 0 \quad(\bmod 5) \\
1989^{M}=1989^{5} \equiv-1 \quad(\bmod 5) \\
5 \nmid 1989^{M}+M^{1989}
\end{array}
$$

Thus,
$$
M \neq 5 \text {. }
$$

By Fermat's Little Theorem, 5 is a prime, and $M \neq 5$ with $1 \leqslant M \leqslant 9$, then $(5, M)=1$, thus
$$
M^{1989} \equiv M \quad(\bmod 5) .
$$

If $M$ is odd, then therefore

If $M$ is even, then therefore
$$
\begin{array}{c}
5 \mid M-1 \\
M=1 \\
5 \mid M+1 \\
M=4
\end{array}
$$

Thus, $M=1$ or 4.