Let's determine those one-digit natural number pairs, different from 1, where one of the digits in their product matches one of the factors.
Let the sought pair of numbers be $a, b$. Since $1<a<10$, and $1<b<10$, it follows that $a b<10 a$, and thus $a b=10 a+c$ is impossible. Therefore, only

$$
a b=10 c+a
$$

can be the equation, which means

$$
a(b-1)=10 c
$$

The right-hand side is divisible by 5, so either $a$ or $(b-1)$ must be equal to 5.

1. If $a=5$, then $b-1$ is even, i.e., $b=3,5,7,9$.
2. If $b-1=5$, i.e., $b=6$, then $a$ must be even, i.e., $a=2,4,6,8$.

Thus, the sought pairs of numbers are:

$$
\begin{array}{llll}
5 \cdot 3, & 5 \cdot 5, & 5 \cdot 7, & 5 \cdot 9 \\
6 \cdot 2, & 6 \cdot 4, & 6 \cdot 6, & 6 \cdot 8
\end{array}
$$

László Pálmai (Bp. VIII., Széchenyi g. I. o. t.)