13.139. Find a two-digit number, the quotient of which when divided by the product of its digits is $8 / 3$, and the difference between the number and the number written with the same digits but in reverse order is 18.
## Solution.

Let's represent the desired number as $10 x + y$. Then, according to the condition,

$$
\left\{\begin{array}{l}
\frac{10 x + y}{x y} = \frac{8}{3}, \\
10 x + y = 10 y + x + 18,
\end{array} \text { from which } x = 6, y = 4. \text { The desired number is } 64. \right.
$$

Answer: 64.