10.35 In the decimal representation, the natural number
$$
n=\frac{99 \cdots 9}{100 \uparrow} \text {. }
$$

How many 9's are there in the representation of $n^{3}$?
(Dutch Mathematical Olympiad, 1983)
[Solution]
$$
\begin{array}{c}
n=\frac{99 \cdots 9}{100 \uparrow}=10^{100}-1 \\
n^{3}=\left(10^{100}-1\right)^{3} \\
=\left(10^{300}-1\right)-3 \cdot 10^{200}+3 \cdot 10^{100}
\end{array}
$$

Since $10^{300}-1$ has 300 nines, when subtracting $3 \cdot 10^{200}$, the first 99 nines remain unchanged, and the last 200 nines remain unchanged.

When adding $3 \cdot 10^{100}$, the last 100 nines remain unchanged, the first 99 nines remain unchanged, and the numbers in the middle are not nines.
Thus, the representation of $n^{3}$ contains a total of 199 nines.