13、 $\overline{A A A A A A A A A} \times \overline{B B B B B B B B B}$ The sum of the digits of the product equals

The sum of the digits of the product of $\overline{A A A A A A A A A} \times \overline{B B B B B B B B B}$ equals
[Answer] 81
[Solution]
$$
\begin{array}{l}
\overline{A A A A A A A A A} \times \overline{B B B B B B B B B} \\
=A \times 111111111 \times B \times 111111111 \\
=A \times B \times 999999999 \times 12345679^{\circ} \\
=A \times B \times 12345679 \times\left(10^{9}-1\right)
\end{array}
$$

Since $A \times B \times 12345679$ is at most $9 \times 9 \times 12345679=999999999$ (a nine-digit number), let's assume $A \times B \times 12345679=\overline{a_{1} a_{2} a_{3} a_{4} a_{5} a_{6} a_{7} a_{8} a_{9}}$ (where $a_{1}$ can be 0). Therefore,
$$
\begin{array}{l}
A \times B \times 12345679 \times\left(10^{9}-1\right) \\
=\overline{a_{1} a_{2} a_{3} a_{4} a_{5} a_{6} a_{7} a_{8} a_{9}} \times 10^{9}-\overline{a_{1} a_{2} a_{3} a_{4} a_{5} a_{6} a_{7} a_{8} a_{9}} \\
=\overline{a_{1} a_{2} a_{3} a_{4} a_{5} a_{6} a_{7} a_{8} a_{9} 000000000}-\overline{a_{1} a_{2} a_{3} a_{4} a_{5} a_{6} a_{7} a_{8} a_{9}} \\
=\overline{a_{1} a_{2} a_{3} a_{4} a_{5} a_{6} a_{7} a_{8}\left(a_{9}-1\right)\left(9-a_{1}\right)\left(9-a_{2}\right) \cdots\left(9-a_{8}\right)\left(10-a_{9}\right)} \\
\end{array}
$$
(If you don't understand this step, you can try $123456789000000000-123456789$ to get a feel for it) Therefore, the final sum of the digits is $9 \times 9=81$