323. The quotient of dividing a certain three-digit number by 11 is equal to the sum of the squares of the digits of the dividend. Find this number.
323. Given the condition $11\left(x^{2}+y^{2}+z^{2}\right)=100 x+10 y+z$, and either $x+z=y$ (1), or $x+z-y=11$ (2), or $y-x-z=11$ (3).

In case (1), we get $2 x^{2}+2 z^{2}+2 x z=10 x+z$, from which it is clear that $z$ is an even digit. Representing the last equation as $2 x^{2}+2(z-5) x+2 z^{2}-z=0$, we obtain $x=\frac{5-z \pm \sqrt{-3 z^{2}-8 z+25}}{2}$, hence $3 z^{2}+8 z-25 \leqslant 0, 0 \leqslant z \leqslant \frac{-4+\sqrt{91}}{3}, 0 \leqslant z \leqslant 1$; but since $z$ is an even digit, then $z=0$, so $x=5, y=5$; the desired number is 550.

In case (2), we arrive at the equation $2 x^{2}+2 x z+2 z^{2}-32 x-23 z+131=0$, from which it is clear that $z$ is an odd digit. Representing the last equation as $2 x^{2}+2(z-16) x+2 z^{2}-23 z+131=0$, we get

$$
x=\frac{-2(z-16) \pm \sqrt{4(z-16)^{2}-8\left(2 z^{2}-23 z+131\right)}}{4}
$$

hence, $3 z^{2}-14 z+6 \leqslant 0, 0<z \leqslant 4$; but since $z$ is an odd digit, then $z_{1}=1, z_{2}=3$. If $z=1$, we get the equation $x^{2}-15 x+55=0$, which does not provide solutions. If $z=3$, we get the equation $x^{2}-13 x+40=0$, from which $x_{1}=5, x_{2}=8$. For $x=5$, we have $y=-3$, which is discarded; for $x=8$, we find $y=0$; the desired number is 803.

In case (3), we arrive at the equation $2 x^{2}+2 x(z+6)+2 z^{2}+21 z+111=0$, from which, considering the discriminant, we get $z^{2}+10 z+62 \leqslant 0$, which is discarded.

Answer. 550 and 803.