3. Find the digits $A, B, C$, so that the multiplication process of two three-digit numbers given by

$$
\begin{gathered}
\underline{\overline{A B C} \cdot \overline{B A C}} \\
---- \\
--A \\
---B \\
\hline-----
\end{gathered}
$$

where $\overline{A B C} \cdot C=---, \overline{A B C} \cdot A=--A, \overline{A B C} \cdot B=---B$. Different letters represent different digits and each dash represents one digit.
Solution. From the fact that the numbers $\overline{A B C}$ and $\overline{B A C}$ are three-digit, it follows that $A \neq 0$ and $B \neq 0$. It is also obtained that $C \neq 1$, because for $C=1$ the equality $\overline{A B C} \cdot C=---$ is not possible, since $\overline{A B 1} \cdot 1=\overline{A B 1}$, which is not a four-digit number.

From the equality $\overline{A B C} \cdot A=--A$, if $A>3$ it follows that

$$
\overline{A B C} \cdot A>400 \cdot 4=1600>999
$$

Thus, $A \leq 3$. If $A=1$, then $C=1$ as well, which is not possible. Therefore, $A=2$ or $A=3$. If $A=3$, then from $\overline{A B C} \cdot A=--A$ it must be that $C=1$, which is not possible. Therefore, $A=2$. Now, from $\overline{2 B C} \cdot 2=--2$, it follows that $C=1$ or $C=6$. Thus, $C=6$ must be true.

Finally, from the equality $\overline{2 B 6} \cdot B=---B$, it must be that $B \in\{1,2,4,6,8\}$. For $B=1$, the resulting number is not four-digit. From the remaining cases, by checking, it is obtained that $B=8$.