$10 \cdot 39$ Find all three-digit numbers $A$ such that the arithmetic mean of all numbers obtained by rearranging the digits of $A$ is still equal to $A$.
(8th All-Soviet Union Mathematical Olympiad, 1974)
［Solution］Let the required three-digit number be $A=\overline{a b c}$.
By the problem, we have
$$
\overline{a b c}+\overline{a c b}+\overline{b a c}+\overline{b c a}+\overline{c a b}+\overline{c b a}=6 \overline{a b c} \text {. }
$$

From $\overline{a b c}=100 a+10 b+c$ and so on, we get
$$
222(a+b+c)=6(100 a+10 b+c) \text {. }
$$

That is
$$
7 a=3 b+4 c \text {. }
$$

Thus
$$
7(a-b)=4(c-b) .
$$

Since $|a-b| \in\{0,1,2, \cdots, 9\}, \quad|c-b| \in\{0,1, \cdots, 9\}$ and 7 and 4 are coprime, we get the following system of equations
$$
\left\{\begin{array} { l } 
{ a - b = 0 , } \\
{ b - c = 0 . }
\end{array} \quad \left\{\begin{array}{l}
|a-b|=4, \\
|c-b|=7 .
\end{array}\right.\right.
$$

For the first system of equations, we have $a=b=c$, thus we get 9 numbers
$$
111,222,333, \cdots, 999 \text {. }
$$

For the second system of equations, it can be transformed into
$$
\left\{\begin{array} { l } 
{ a - b = 4 , } \\
{ c - b = 7 , }
\end{array} \quad \left\{\begin{array} { l } 
{ b - a = 4 , } \\
{ c - b = 7 , }
\end{array} \quad \left\{\begin{array} { l } 
{ a - b = 4 , } \\
{ b - c = 7 , }
\end{array} \quad \left\{\begin{array}{l}
b-a=4, \\
b-c=7 .
\end{array}\right.\right.\right.\right.
$$

Since $a-b$ and $c-b$ have the same sign, the second and third systems of equations above do not hold.
For
$$
\left\{\begin{array}{l}
a-b=4, \\
c-b=7,
\end{array}\right.
$$

we get
$$
b=0,1,2 \text {. }
$$

At this point, we get three three-digit numbers $407,518,629$.
For
$$
\left\{\begin{array}{l}
b-a=4, \\
b-c=7,
\end{array}\right.
$$

we get
$$
b=7,8,9 \text {. }
$$

At this point, we get three three-digit numbers $370,481,592$.
From the above, we get a total of 15 three-digit numbers that meet the conditions:
$$
111,222,333, \cdots, 999,407,518,629,370,481,592
$$