Find all three-digit numbers $A$ such that the average of the numbers obtained by permuting the digits is equal to $A$.

## For verification

Solutions to be used only after you have really tried to solve the proposed exercises.
We write $A=100 a+10 b+c$. The numbers corresponding to a permutation of the digits of $A$ are therefore $100 a+10 b+c, 100 a+10 c+b, 100 b+10 a+c, 100 b+10 c+a$, $100 c+10 a+b$ and $100 c+10 b+a$. The average of these numbers is $\frac{1}{6}(222 a+222 b+222 c)=37(a+b+c)$.

For the number to be equal to the average of the numbers obtained by permuting the digits, it is necessary (and sufficient) that

$$
37(a+b+c)=100 a+10 b+c
$$

hence

$$
63 a=27 b+36 c
$$

and by dividing by 9:

$$
7 a=3 b+4 c
$$

Moreover, we have $a \in\{1,2, \ldots, 9\}$ and $b, c \in\{0,1,2, \ldots, 9\}$.

- If $b=0$ we have $7 a=4 c$ and the only solution is $a=4, c=7$.
- If $b=1$ we have $7 a=3+4 c$ and the two only solutions are then $a=c=1$ and $a=5, c=8$.
- Etc (for $b$ up to 9).

Therefore, the solutions are 111, 222, 333, 370, 407, 444, 481, 518, 555, 592, 629, 666, 777, 888, 999.

## 3 Arithmetic, second part

## Statements