1B. Find all three-digit numbers $A$, for which the arithmetic mean of the numbers obtained by permuting the digits of the number $A$, is equal to the number $A$.
Solution. Let $A=\overline{a b c}$ be the given number that satisfies the conditions of the problem. Numbers obtained by permutation of the digits of the number $A$ are $\overline{a c b}, \overline{b a c}$, $\overline{b c a}, \overline{c a b}, \overline{c b a}$. Then

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

from which we get

$$
222(a+b+c)=6(100 a+10 b+c)
$$

or $7 a=3 b+4 c$. The last equation can be solved in the set $\{0,1,2$, $3,4,5,6,7,8,9\}$, by substituting $a=1, a=2, \ldots, a=9$ and solving the resulting equations.

The process can be shortened if the equation $7 a=3 b+4 c$ is written in the form $7(a-b)=4(c-b)$. Then, according to divisibility criteria, we have

$$
a=b=c \text { or }\left\{\begin{array} { l } 
{ a - b = 4 } \\
{ c - b = 7 }
\end{array} \text { or } \left\{\begin{array}{l}
b-a=4 \\
b-c=7
\end{array} .\right.\right.
$$

In the first system

$$
\left\{\begin{array}{l}
a-b=4 \\
c-b=7
\end{array}\right.
$$

the inequality $0 \leq b \leq 2$ is satisfied, and in the second system the inequality $7 \leq b \leq 9$ is satisfied.

All solutions to the problem are

111,222 , 333 , 444 , $555,666,777$, 888 , 999 , $407,518,629$, 370 , 481, 592 .