10.48 Find all natural numbers that are 11 times the sum of their digits.
(Kiev Mathematical Olympiad, 1977)
[Solution] Let the required natural number be $N$, and $N$ is an $n$-digit number, then
$$
N \geqslant 10^{n-1} \text {. }
$$

Let the sum of the digits of $N$ be $S$, then
$$
S \leqslant 9 n \text {. }
$$

By the problem statement,
$$
N=11 S \text {, }
$$

which means
$$
11 \cdot 9 n \geqslant 10^{n-1} \text {. }
$$

It can be proven that for $n \geqslant 4$,
$$
10^{n-1}>99 n \text {. }
$$

Therefore, the number $N$ can have at most three digits, i.e., $n \leqslant 3$.
Let $N=\overline{a b c}$, where $a, b, c \in\{0,1,2, \cdots, 9\}$.
Thus, we have the equation
$$
\begin{array}{c}
100 a+10 b+c=11(a+b+c), \\
89 a=b+10 c .
\end{array}
$$

Since $b \leqslant 9, c \leqslant 9$, then
$$
89 a \leqslant 99, \quad a \leqslant 1 .
$$

If $a=0$, then $b=0, c=0$, hence $N=0$, which is not a natural number, and does not meet the problem's requirements.
If $a=1$, then

From this, we get
$$
\begin{array}{c}
b+10 c=89 \\
c=8, \quad b=9
\end{array}
$$

Thus, the unique solution to this problem is:
$$
N=198 \text {. }
$$