Sure, here is the translated text:

```
Let's write digits in place of the letters in the following schema (different letters should be replaced by different digits), so that we get a correct division. Then, continue the division until a previous remainder repeats.

\[
\begin{aligned}
& A B: C A B=D, C E B K \\
& A B D \\
& B E F D \\
& \quad G H D \\
& \quad C E K D
\end{aligned}
\]
```
Obviously, $D$ represents 0. According to the first partial division, subtracting $100 C^{2}$ from the three-digit number $A B D$ leaves $B E F$, so more than $100$ remains, which means $C^{2}<9, C<3$. On the other hand, the distinctness of the digits implies $0<B<C$, so $B=1, C=2$. By inserting the known digits, the first partial division is as follows:

$$
\overline{A 10}=2 \cdot \overline{2 A 1}+\overline{1 E F}
$$

from which

$$
80 A=492+\overline{E F}, \quad A=6+\frac{12+\overline{E F}}{80}
$$

This can only be an integer if $\overline{E F}=68, A=7$, so the next division can only be:

```
71:271=0,26199, 71/271=0,26199.
710
1680
540
2690
2510
71
```

This meets the requirements, $G=5, H=4, K=9$, and the sixth remainder matches the first.

Ujvári Edit (Budapest, Kossuth L. g. II. o. t.)