Which are those three-digit natural numbers that are equal to eleven times the sum of the squares of their digits?
Let the digits of the desired number $N$ from left to right be $a, b, c$, then

$$
N=100 a+10 b+c=11\left(a^{2}+b^{2}+c^{2}\right)
$$

Writing the left side as $99 a+11 b+(a+c-b)$ and dividing the equation by 11:

$$
9 a+b+\frac{a+c-b}{11}=a^{2}+b^{2}+c^{2}
$$

According to this, the expression $k=a+c-b$ is divisible by 11 because the other expressions are integers.

Since the digits satisfy $1 \leq a \leq 9$ and $0 \leq b, c \leq 9$, the largest value of $k$ is with the largest $a, c$ and the smallest $b$, which is $9+9-0=18$, and the smallest value of $k$ is with the smallest $a, c$ and the largest $b$, which is $1+0-9=-8$, so for $k / 11$

$$
-1<-\frac{8}{11} \leq \frac{k}{11} \leq \frac{18}{11}<2
$$

thus the value of $k / 11$ is 0 or 1, and the value of $k$ is 0 or 11.

If $k=0$, that is, $b=a+c$, then from (2) by eliminating $b$ we get the two-variable Diophantine equation for $a$ and $c$:

$$
2 a^{2}+2 a c+2 c^{2}-10 a-c=0
$$

Rearranging this for $a$:

$$
2 a^{2}+(2 c-10) a+\left(2 c^{2}-c\right)=0
$$

the discriminant is:

$$
D=(2 c-10)^{2}-8\left(2 c^{2}-c\right)=4\left(-3 c^{2}-8 c+25\right)
$$

This takes the value 0 around $c_{1} \approx-4.51$ and $c_{2} \approx 1.84$, and is positive only between these values. Thus, the only possible values for $c$ are 0 and 1. With $c=1$, $D$ is not a perfect square, so it does not yield an integer $a$; with $c=0$, (3) becomes:

$$
2 a^{2}-10 a=0
$$

from which $a=5$ or 0. But only $a=5$ can be the leading digit, with which $b=5$, and $N=550$. This satisfies (1).

With $k=1$, $b=a+c-11$, and similarly to the previous:

$$
\begin{gathered}
2 a^{2}+(2 c-32) a+\left(2 c^{2}-23 c+131\right)=0 \\
D=4\left(-3 c^{2}+14 c-6\right)
\end{gathered}
$$

$D$ is positive between $c_{1} \approx 0.48$ and $c_{2} \approx 4.19$, but only $c=3$ among the integer values $c=1,2,3,4$ makes $D$ a perfect square; with $c=3$, (4) becomes:

$$
2 a^{2}-26 a+80=0
$$

from which the value of $a$ is 5 or 8. However, with $a=5$ and $c=3$, $b=-3$, which is not a digit. With $a=8$ and $c=3$, the resulting $b=0$ gives $N=803$, which satisfies (1).

Considering all possibilities, we found that there are two three-digit natural numbers whose digits' square sum multiplied by 11 equals the number itself, these are: 550 and 803.

Gábor Kóta (Tatabánya, Árpád g. IV. o. t.)