Let's write in place of the letters all possible digits that make the following two multiplications correct:

$$
A \cdot \overline{B A}=\overline{C D} \quad F \cdot \overline{B E}=\overline{D C}
$$

(Equal letters should be replaced by equal digits, different letters by different digits.)
I. solution. We cannot write 0 in place of any letter, because all five letters occur as the first digit of a factor or product, which cannot be 0. Furthermore, we cannot write 1, 5, and 6 in place of $A$ and $E$, because the product of these digits with themselves ends in 1, 5, or 6, respectively, whereas the product $A \cdot A$ must end in $D$, which is different from $A$, and the product $E \cdot E$ must end in $C$. Finally, we cannot write a digit greater than 6 in place of $A$ and $E$, because even with the smallest value of $B$, $B=1$, the product of the factors on the left side would be a three-digit number. Therefore, only 2, 3, or 4 can be written in place of $A$ and $E$.

The quotients $\overline{C D} / A$ and $\overline{D C} / E$ are integers, and their first digit is $B$. The numerators are less than 100, and one of the denominators is at least 3, so the smaller quotient is less than $100 / 3$, or at most 33, hence $B \leq 3$. But $B=3$ is only possible if one of $A$ and $E$ is 2 and the other is 3; however, in this case, $B$ would not differ from both. Therefore, $B$ can only be 1 or 2, and the two multiplications in the problem must be chosen from $2 \cdot 12, 3 \cdot 13, 4 \cdot 14$ or $2 \cdot 22, 3 \cdot 23, 4 \cdot 24$. Of these, only the pair $3 \cdot 23=69, 4 \cdot 24=96$ meets the requirements.

Gábor Csanády (Budapest, Móricz Zs. Gymnasium, 2nd grade)

II. solution. Both $C$ and $D$ are the last digits of square numbers, but they are different from 5 (and 0), so their values are among 1, 4, 6, and 9 (if, for example, $C=6$, then only $E=4$ is possible). Thus, among $\overline{C D}$ and $\overline{D C}$, the smaller one (in which the first digit is the smaller one), the following values come into question: 14, 16, 19, 46, 49, 69. Of these, 14, 16, and 49 cannot be factored into the product of a one-digit and a two-digit integer, since $A \neq 1$ and $E \neq 1$, otherwise $C=B$, or $D=B$ would follow, and for the same reason, 19 does not fit; in the factorization $46=2 \cdot 23$, the one-digit factor differs from the unit digit of the two-digit number. Finally, trying $\overline{C D}=69$, we find a solution: $69=3 \cdot 23, A=3, B=2$, and since $B=2$, the only divisor of $\overline{D C}=96$ between 20 and 30 is 24, with which $96: 24=4, E=4$ also fits.