Which are the two-digit even numbers $\overline{a b}$, whose fifth power ends in $\overline{a b}$?
Solution. The two-digit number, $\overline{a b}$ can be written in the form $10 a+b$. According to the binomial theorem:

$$
(10 a+b)^{5}=(10 a)^{5}+5(10 a)^{4} b+10(10 a)^{3} b^{2}+10(10 a)^{2} b^{3}+5(10 a) b^{4}+b^{5}
$$

It is obvious that the first four terms are divisible by 100. Since the number is even, $b$ is even, so the fifth term is also divisible by 100. Therefore, the last two digits of the power do not depend on $a$, but only on the value of $b$.

If the value of $b$ were 0, then the last two digits of the power would be 0, which would also result in 0 for $a$. Thus, the possible values of $b$ are: $2,4,6$, 8, and the values of $b^{5}$ in these cases are:

$32, \quad 1024, \quad 7776, \quad 32768$

The last digits of these powers are the same as the bases, respectively.

Therefore, the corresponding two-digit numbers are: $32,24,76,68$.