10.61 Find a two-digit number that is equal to twice the product of its digits.
(Kyiv Mathematical Olympiad, 1952)
[Solution] Let the required two-digit number be $10x + y$, where $x, y \in \{0,1,2, \cdots, 9\}$, $x \neq 0$.
$$
\begin{array}{c}
10x + y = 2xy, \\
y = 2x(y - 5).
\end{array}
$$

From (1), we know that $y$ is even, and $5 \leqslant y \leqslant 9$, so we have
$$
y = 6 \text{ or } 8 \text{. }
$$
When $y = 8$, equation (1) becomes $8 = 6x$. Here, $x$ is not an integer.
When $y = 6$, from (1) we get $x = 3$.
Therefore, the required two-digit number is 36.