12. (5 points) In the multiplication problem shown below, $A, B, C, D, E, F, G, H, I$ each represent different single-digit numbers. Then the five-digit number “FIGAA” is $\qquad$
$$
\begin{array}{r}
A B C \\
\times \quad D C \\
\hline B E A \\
\hline F I G A A
\end{array}
$$
【Analysis】First, find the special cases in the problem. Based on the first product being a three-digit number and the last digits being the same, we can enumerate and exclude possibilities. Then, determine $B$ based on $A$ and $C$, and proceed to solve the problem.

【Solution】Solution: According to the problem:
$A, B, C, D, E, F, G, H, I$ are 9 digits, and there is no digit 0 in the problem.
Based on $\overline{\mathrm{ABC}} \times \mathrm{C}$ resulting in a three-digit number, the leading digit can be $C=2, A=4$ or $C=3, A=9$, which does not satisfy the condition of being a three-digit number. Therefore, $A=4, C=2$.

Based on the carry-over, $B=9, E=8$.
According to $E+H=A=4$, then $H=6, A$ plus the carry-over equals $I=5$. Therefore, $D=3, F=1$.
Thus: $49 \times 32=15744$.
The answer is: 15744.