The following square is to be understood in the number system with base $A+1$, where $A, B, C$ represent different digits. Determine the digits.

$$
A A A A^{2}=A A A B C C C B
$$
Let's write $G$ instead of $A+1$, so $A=G-1$. The left side is based on the largest four-digit number in the $G$-based number system, which is 1 less than the smallest five-digit number in the system: $10000=G^{4}$. Thus

$$
\text { b) } \begin{aligned}
\overline{A A A A}^{2}={ }^{G)} \overline{A A A A} \cdot\left(G^{4}-1\right)={ }^{G)} \overline{A A A \quad A \quad 0000} \\
\text { }{ }^{G)} \overline{-\overline{A A A A}} \\
\end{aligned}
$$

The subtraction was performed using the usual written procedure, mentally adding $G$ to the 0 digits of the minuend with place values $1, G, G^{2}, G^{3}$, so that in these columns the digit in the minuend is larger than the digit in the subtrahend, while simultaneously increasing the digit of the subtrahend with place values $G, G^{2}, G^{3}, G^{4}$ by 1 (carrying the "remainder").

From the last four digits of the difference (1), we see that $B=1$ and $C=0$, so from the 5th digit from the right, $A-1=B=1, A=2$, and finally, the base of the number system used is 3.

Béla Fialovszky (Esztergom, Temesvári Pelbárt g. IV. o. t.)

Remark. Similarly, $B=1, C=0, A=2$ can be derived from the equality

$$
A+1) \underbrace{\overline{A A \cdots A}^{2}}_{k+1 \text { digit }}=\underbrace{A A \cdots A}_{k \text { digit }} B \underbrace{C C \cdots C}_{k \text { digit }} B
$$

József Balogh (Hatvan, Bajza J. g. III. o. t.)