Without experimentation, determine those one-digit natural numbers, different from 1 and 9, whose product has a sum of digits equal to one of the factors.
Let the sought single-digit numbers be $x$ and $y$. Then their product is

$$
x y=10 a+b
$$

the sum of the digits of this product is equal to one of the factors.

$$
x=a+b
$$

where $a$ and $b$ are single-digit non-negative integers.

Subtract (2) from (1), we get

$$
x y-x=9 a
$$

from which

$$
a=\frac{x(y-1)}{9}
$$

Thus, $x(y-1)$ must be divisible by $9$. The case $x=9$ is excluded by the problem, and if $y-1=9$, then $y$ is already a two-digit number, so $x$ and $(y-1)$ must each be divisible by 3.

Therefore, the solutions are:

$$
\begin{array}{llllll}
x=3, & y-1=3, & y=4 ; & x=6, & y-1=3, & y=4 \\
x=3, & y-1=6, & y=7 ; & x=6, & y-1=6, & y=7
\end{array}
$$

Thus, the sought pairs of numbers are:

$$
\begin{array}{ll}
3 \cdot 4=12, & 6 \cdot 4=24 \\
3 \cdot 7=21, & 6 \cdot 7=42
\end{array}
$$

András Sárközy (Gyöngyös, Vak Bottyán g. II. o. t.)