21.3. Find all two-digit numbers that are equal to three times the product of their digits.

$$
(7-8 \text {th grades })
$$
21.3. Let $\overline{a b}$ be the required two-digit number. According to the problem, we have the equation: $10 a+b=3 a b$, from which

$$
b=\frac{10 a}{3 a-1}
$$

Since the numbers $a$ and $3 a-1$ are coprime, 10 must be divisible by $3 a-1$. The following 4 cases are possible: $3 a-1=1, 3 a-1=2$, $3 a-1=5, 3 a-1=10$. Only in two of these cases will we get a natural value for $a$ not greater than $9: a=1$ and $a=2$. Then for $b$ we will get the respective values: $b=5$ and $b=4$. Answer: 15 and 24.