If the base number of a certain number system is an arbitrary positive integer $a$, then the product of the number $12345 \cdots(a-3)(a-1)$ and $k(a-1)$ (where $k<a$) is a $(a-1)$-digit number, all of whose digits are $k$.
Let's write the number in its detailed form. This is the following:

$$
\begin{gathered}
a^{a-3}+2 a^{a-4}+3 a^{a-5}+\ldots+(a-4) a^{2}+ \\
+(a-3) a+(a-1)
\end{gathered}
$$

If I multiply this number by $(a-1)$, I get the following result:

$$
\begin{gathered}
a^{a-2}+2 a^{a-3}+3 a^{a-4}+\ldots+(a-4) a^{3}+ \\
+(a-3) a^{2}+(a-1) a- \\
-a^{a-3}-2 a^{a-4}-\ldots-(a-5) a^{3}- \\
-(a-4) a^{2}-(a-3) a-(a-1)
\end{gathered}
$$

or

$$
a^{a-2}+a^{a-3}+a^{a-4}+\ldots+a^{3}+a^{2}+a+1
$$

The shorter form of this number is:

$$
111 \ldots 1111
$$

from which it is immediately clear that its product with $k(<a)$ is

$$
k k k \ldots k k k k
$$

(Márkus Kohn, real school. VI. grade. Pécsett).

The problem was also solved by: Bernát Friedmann, S.-A.-Ujhely; Béla Grünhut and Aladár Visnya, Pécsett.