833. Find all two-digit numbers that are equal to triple the product of their digits.
$\triangle$ Let the desired number be denoted by $\overline{a b}$. Then

$$
\overline{a b}=3 a b, \quad 10 a+b=3 a b
$$

We have obtained a quadratic equation with two unknowns of the type considered in § 18, point 18.1, with the difference that this equation needs to be solved in integers from 0 to 9.

Express $b$ in terms of $a$ from the equation:

$$
10 a=3 a b-b, \quad 10 a=b(3 a-1), \quad b=\frac{10 a}{3 a-1}
$$

Multiply the last equation by 3 and separate the integer part of the fraction formed in the right-hand side:

$$
3 b=\frac{30 a}{3 a-1}=\frac{(30 a-10)+10}{3 a-1}=10+\frac{10 a}{3 a-1}
$$

From this, $3 a-1$ can only take the values $1, 2, 5,$ and 10. Try all these four cases on your own.

Answer: 15, 24