10.78 Find all two-digit numbers $M$ such that the last two digits of any power of the number $M$ are the same as the last two digits of the number $M$.
(Kyiv Mathematical Olympiad, 1961)
[Solution] Since the last two digits of the product of two numbers are only related to the last two digits of each factor, if the last two digits of $M^{2}$ are the same as those of $M$, then the last two digits of any power of $M$ are the same as those of $M$.

Therefore, we only need to find the number $M$ such that the last two digits of $M^{2}$ are the same as those of $M$. This is equivalent to
$$
\begin{array}{l}
100 \mid M^{2}-M, \\
100 \mid M(M-1) .
\end{array}
$$

Thus, only two numbers meet the requirement: $M=25$ or $M=76$.