9.20 Find all three-digit numbers that satisfy the following condition: the quotient when divided by 11 equals the sum of the squares of its digits.
(2nd International Mathematical Olympiad, 1960)
[Solution] Let the required three-digit number be
$$
n=100 a+10 b+c,
$$

where $b, c \in\{0,1,2, \cdots, 9\}, a \in\{1,2, \cdots, 9\}$.
The necessary and sufficient condition for $n$ to be divisible by 11 is that $a-b+c$ is divisible by 11.
Since $\quad-8 \leqslant a-b+c \leqslant 18$, we have only
$$
\begin{array}{l}
a-b+c=0 . \\
a-b+c=11 .
\end{array}
$$

Also, by the problem statement,
$$
100 a+10 b+c=11\left(a^{2}+b^{2}+c^{2}\right) .
$$

We will discuss the following two cases:
First case: Solve (1) and (3), i.e., solve
$$
\left\{\begin{array}{l}
a-b+c=0, \\
100 a+10 b+c=11\left(a^{2}+b^{2}+c^{2}\right) .
\end{array}\right.
$$

From (1), we get
$$
b=a+c .
$$

Substituting into (3), we obtain
$$
\begin{array}{c}
100 a+10(a+c)+c=11\left[a^{2}+(a+c)^{2}+c^{2}\right], \\
11(10 a+c)=11\left(2 a^{2}+2 c^{2}+2 a c\right), \\
10 a+c=2 a^{2}+2 c^{2}+2 a c, \\
2 a^{2}+2(c-5) a+2 c^{2}-c=0 .
\end{array}
$$

Considering the discriminant of equation (4)
$$
\Delta=4(c-5)^{2}-8\left(2 c^{2}-c\right) \geqslant 0,
$$

we solve for the range of $c$ as
$$
\frac{-\sqrt{91}-4}{3} \leqslant c \leqslant \frac{\sqrt{91}-4}{3} .
$$

Under the condition of (5), from equation (4) we get
$$
a=\frac{5-c \pm \sqrt{25-8 c-3 c^{2}}}{2} .
$$

Since $c$ is a non-negative integer, by (5) we have
$$
c=0 \quad \text { or } c=1 \text {. }
$$

Noting that $\quad 10 a+c=2 a^{2}+2 c^{2}+2 a c$,
$c$ must be even, so we have only
$$
c=0 .
$$

Thus, we solve to get $\quad a=5$ or $a=0$.
Since $a \in\{1,2, \cdots, 9\}$, we discard $a=0$.
Also, from $b=a+c$ and $a=5, c=0$ we get
$$
b=5 \text {. }
$$

Thus, the required three-digit number is 550.
Second case: Solve (2) and (3), i.e., solve
$$
\left\{\begin{array}{l}
a-b+c=11, \\
100 a+10 b+c=11\left(a^{2}+b^{2}+c^{2}\right) .
\end{array}\right.
$$

From (2), we get
$$
b=a+c-11,
$$

Substituting into (3), we obtain
$$
100 a+10(a+c-11)+c=11\left[a^{2}+(a+c-11)^{2}+c^{2}\right],
$$

which simplifies to
$$
2 a^{2}+2(c-16) a+2 c^{2}-23 c+131=0 .
$$

Considering the discriminant of equation (6)
$$
\Delta=4(c-16)^{2}-8\left(2 c^{2}-23 c+131\right) \geqslant 0,
$$

we solve for the range of $c$ as
$$
\frac{7-\sqrt{31}}{3} \leqslant c \leqslant \frac{7+\sqrt{31}}{3} .
$$

Under the condition of (7), from equation (6) we get
$$
a=\frac{16-c \pm \sqrt{-3 c^{2}+14 c-6}}{2} .
$$

Since $c$ is a non-negative integer, by (7) we have
$$
c=1, \quad c=2, \quad c=3, \quad c=4 .
$$

Also, from (6) we get
$$
23 c=2 a^{2}+2(c-16) a+2 c^{2}+131,
$$

so $c$ must be odd, i.e., we have only
$$
c=1 \text { or } c=3 \text {. }
$$

When $c=1$, $a=\frac{15 \pm \sqrt{5}}{2}$ is not an integer.

When $c=3$, we can solve to get
$$
a_{1}=8 \text {, or } a_{2}=5 \text {. }
$$

From $b=a+c-11$ and $a=8, c=3$ we get
$$
b=0 \text {. }
$$

In this case, the three-digit number is 803.
From $b=a+c-11$ and $a=5, c=3$ we get
$$
b=-3 \text {. }
$$

which is not a non-negative integer, so we discard it.
Thus, the required three-digit numbers are:
550 and 803.