8.31 In decimal, the tens digit of the square of a number with more than one digit is 7, what is the units digit of this square number?
(Hungarian Mathematical Olympiad, 1917)
[Proof] Let a number be $A=10 a+b$, where $a$ is an integer, and $b \in\{0,1,2, \cdots, 9\}$, then
$$
\begin{aligned}
A^{2} & =(10 a+b)^{2} \\
& =100 a^{2}+20 a b+b^{2} .
\end{aligned}
$$
The tens digit of $A^{2}$ is determined by the sum of the tens digit of $b^{2}$ and the units digit of $2 a b$.
Given that the tens digit of $A^{2}$ is 7, which is odd, and the units digit of $2 a b$ is even, the tens digit of $b^{2}$ must be odd.

Examining the tens digit of $b^{2}$ for $b=0,1,2, \cdots, 9$, we find that only when $4^{2}=16$ and $6^{2}=36$, the tens digit of $b^{2}$ is odd. Thus, the units digit of $A^{2}$ is 6.

From the above, when the tens digit of $A^{2}$ is odd, the units digit of $A^{2}$ is 6. For this problem, the units digit of $A^{2}$ is 6. In fact, $24^{2}=576$ and $26^{2}=676$ are such numbers.