## Task $1 / 61$

Determine all three-digit numbers that, when divided by 11, yield a number equal to the sum of the squares of the digits of the original number!
A number $z=a_{n} 10^{n}+a_{n-1} 10^{n-1}+\ldots+a_{2} 10^{2}+a_{1} \cdot 10+a_{0}$ is divisible by 11 if the alternating sum of digits $a_{0}-a_{1}+a_{2}-+\ldots+(-1)^{n} a_{n}=0$ or is divisible by 11.

For three-digit numbers divisible by 11, represented as $z=100 a+10 b+c$, where $a, b$ and $c$ are single-digit numbers, it follows that $c-b+a=0$ or a number divisible by 11, i.e.,

$$
a-b+c=0
$$

or $\quad a-b+c=11$

(other cases, such as $a-b+c=22$, are not possible, as the maximum of $a-b+c<22$ must be). The digits of the desired number must satisfy the equation

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

Case 1: $b=a+c$ substituted into (3), yields

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

$10 a+c$ must therefore be even, so $c$ must also be even. And further

$$
a^{2}+(c-5) a+c^{2}-\frac{c}{2}=0 \rightarrow a_{1 ; 2}=\frac{5-c}{2} \pm \frac{1}{2} \sqrt{25-8 c-3 c^{2}}
$$

The following cases must be examined (even $c$): $c=0, c=2, c=4, c=6, c=8$: $c=0: a_{1 ; 2}=\frac{5}{2} \pm \frac{5}{2}$ with $a_{1}=5$ and $a_{2}=0$ (useless); $c \geq 2$: root is imaginary. Therefore, Case 1: $c=0, a=5, b=5$, one desired number 550

Case 2: $b=a+c-11$ substituted into (3), yields

$100 a+10 a+10 c-100+c=11\left(a^{2}+(a+c-11)^{2}+c^{2}\right) \rightarrow 10 a+c=131+2\left(a^{2}+c^{2}+a c-11 a-11 c\right)$ $c$ must therefore be odd. And further

$$
a^{2}+(c-16) a+c^{2} \frac{131}{2}-\frac{23 c}{2}=0 \rightarrow a_{1 ; 2}=\frac{16-c}{2} \pm \frac{1}{2} \sqrt{14 c-3 c^{2}-6}
$$

Cases to be examined (odd $c$): $c=1, c=3, c=5, c=7, c=9$:

$c=1: a$ is not a natural number

$c=3: a_{1}=8$ and $a_{2}=5$ (useless, as then $b<0$)

$c \geq 5$: root is imaginary

Therefore, Case 2: $c=3, a=8, b=0$, second desired number 803