$10.47 A=\underbrace{999 \cdots 9}_{81 \uparrow}$. Find the sum of the digits of $A^{2}$.
(Japan Mathematical Olympiad, 1991)
[Solution]
$$
\begin{aligned}
A^{2} & =\left(10^{81}-1\right)^{2} \\
& =10^{162}-2 \cdot 10^{81}+1 \\
& =\underset{80 \uparrow}{99 \cdots} 9800 \cdots 01 .
\end{aligned}
$$

So the sum of the digits of $A^{2}$ is
$$
9 \cdot 80+8+1=729 \text {. }
$$