$\mathrm{A, B, C}$ letters can be replaced by which digits to make the following multiplication correct: $\overline{A B B C} \cdot \overline{C C A}=\overline{C C C C A C}$. - In which other base number systems can the problem still be solved?
I. solution. I. According to the requirement,

$$
\overline{C C C C A C}: \overline{C C A}=\overline{A B B C}
$$

the division ends without a remainder. Here, the quotient is a four-digit number, so the first partial dividend was $\overline{C C C}$, thus $\overline{C C C} \geq \overline{C C A}$, i.e., $C \geq A$. Furthermore, the first digit of the quotient $A=1$, because the difference between $\overline{C C C}$ and $\overline{C C A}$ is a single digit, and the divisor cannot fit into it again. Let the first remainder be $C-A=C-1=D$, so the second partial dividend is $\overline{D C}$. This is at most a two-digit number, so it is smaller than the divisor, hence the second digit of the quotient $B=0$.

The above conclusions are consistent with the fact that the third digit of the quotient is also $B$, i.e., 0, since from the third partial division, $\overline{D C A}: \overline{C C A}$, where $D<C$, nothing else can follow. The fourth partial division is

$$
\overline{D C A C}: \overline{C C A}=C
$$

and according to the check,

$$
\overline{C C A} \cdot C=\overline{D C A C}
$$

From this, using the place values, and considering $A=1$ and $D=C-1$, we get a quadratic equation for $C$:

$$
\begin{gathered}
(110 C+1) \cdot C=1000(C-1)+100 C+10+C \\
110 C^{2}-1100 C+990=0, \quad C^{2}-10 C+9=0
\end{gathered}
$$

thus $C=9$ or $C=1$.

Both solutions are valid, as the problem did not specify that different digits should be written in place of the letters. Indeed, $1009 \cdot 991=999919$, and $1001 \cdot 111=111111$.

II. In the above, we only used the fact that the requirement was understood in the decimal system to determine $C$. In a base $X$ number system - where $X \geq 2$, an integer - (1) and the subsequent calculations are replaced by the following:

$$
\begin{gathered}
\left(C \cdot X^{2}+C \cdot X+1\right) \cdot C=(C-1) \cdot X^{3}+C \cdot X^{2}+X+C \\
\left(X^{2}+X\right) \cdot C^{2}-\left(X^{3}+X^{2}\right) \cdot C+\left(X^{3}-X\right)=0 \\
X(X+1) \cdot C^{2}-X^{2}(X+1) \cdot C+X(X+1)(X-1)=0 \\
C^{2}-X C+(X-1)=0, \text { since } X(X+1) \neq 0
\end{gathered}
$$

and from this $C_{1}=X-1, C_{2}=1$. According to this, our result is valid in every number system; the first value of $C$ is the largest digit of the number system; it is only necessary to note that for $X=2$ the two solutions coincide: $C_{1}=C_{2}$.

II. solution. The sixth digit (the coefficient of the fifth power of the base) of the product of a three-digit and a four-digit number is determined from the product of the third and fourth digits (the coefficients of the second and third powers) of the factors and the remainder $M$ from the previous digit. Since there is no seventh digit in the product, there is no remainder to carry over when determining the sixth digit, so $A \cdot C+M=C$, i.e., $A=1, M=0$, since $C$, as the first digit of a number, cannot be 0.

The fifth digit is similarly determined from the product of the fourth and second, and the third and third digits, and the remainder $N$ from the previous digit. Since $M=0$, there is no remainder here either, so $A \cdot C+B \cdot C+N=(1+B) \cdot C+N=C$. From this, $B=N=0$.

Now let's look at the formation of the second digit of the product. The product of the first digits $1 \cdot C$ does not produce a remainder. Thus, the second digit of the product, which is 1, is the last digit of $B \cdot A+C \cdot C=0 \cdot 1+C^{2}=C^{2}$. In the decimal system, $C$ can only be 1 or 9, and both are valid.

For an arbitrary base $x$, the third $C$ digit is the remainder $P$ from $C^{2}$, and the last digit of $B \cdot A+B \cdot C+C \cdot C=C^{2}$, i.e.,

$$
\begin{gathered}
P+1=C, \quad P=C-1 . \quad \text { Thus } \quad C^{2}=P \cdot x+1=(C-1) x+1 . \text { Rearranging } \\
C^{2}-1-(C-1) \cdot x=(C-1)(C+1-x)=0
\end{gathered}
$$

thus $C=1$ or $C=x-1$, and again both values are valid.

Tibor Faragó (Budapest, Bláthy O. erősár. ip. techn. II. o. t.)