201. Find all three-digit numbers $A$ that have the following property: the arithmetic mean of all numbers obtained from the number $A$ by different permutations of its digits is equal to $A$.
201. Answer: 15 numbers ( $111,222, \ldots, 999,407,518,629,370$, $481,592$).

Let $\overline{a b c}$ be the desired number. The condition of the problem means that

$$
\overline{a b c}+\overline{a c b}+\overline{b a c}+\overline{b c a}+\overline{c a b}+\overline{c b a}=6 \overline{a b c}
$$

from which $222(a+b+c)=6(100 a+10 b+c)$, or $7 a=3 b+4 c$. This equation can be solved by trying all options (for example, sequentially setting $a=1,2, \ldots, 9$ and then solving the resulting equation with the consideration that $b$ and $c$ are digits). The search can be reduced by divisibility considerations. For example, the equation can be rewritten as: $7(a-b)=4(c-b)$, from which either $a=b=c$, or $a-b=4$ and $c-b=7$ (then $0 \leqslant b \leqslant 2$ ), or $b-a=4, b-c=7$ (then $7 \leqslant b \leqslant 9$ ).