3. Determine all two-digit numbers that are equal to twice the product of their digits.
Solution. Let $\overline{x y}$ be the desired number. Then $10 x+y=2 x y$, from which we get $y=2 x(y-5)$. Clearly, $x, y \neq 0$ and since $y$ is a digit, $y>0$,
so it must be $y>5$. From the last equality, it follows that $y$ is an even number, so $y=6$ or $y=8$. For $y=6$ we get $x=3$, and for $y=8$ we get $8=6 x$ and this equation has no solution in the set of natural numbers. Finally, the desired number is 36.