4. Vasya remembers that his friend Petya lives on Kurchatovskaya Street, house number 8, but he forgot the apartment number. In response to a request to clarify the address, Petya replied: “My apartment number is a three-digit number. If you rearrange the digits, you can get five other three-digit numbers. So, the sum of these five numbers is exactly 2017.” Help Vasya remember Petya's apartment number.
Answer: 425.

Let the first, second, and third digits of Petya's apartment number be denoted by $a, b, c$ respectively. The condition of the problem can be algebraically written as:

$$
\overline{a c b}+\overline{b a c}+\overline{b c a}+\overline{c a b}+\overline{c b a}=2017
$$

Adding $\overline{a b c}$ to both sides and expanding all three-digit numbers using the formula $\overline{a b c}=100 a+10 b+c$, after combining like terms, we get

$$
222 \cdot(a+b+c)=\overline{a b c}+2017
$$

Therefore,

$$
3016=2017+999 \geqslant 222(a+b+c) \geqslant 2017+100=2117
$$

from which we conclude that $a+b+c$ can only take the values $10, 11, 12$ or 13. Let's consider all 4 cases:

(1) $a+b+c=10$, then

$$
\overline{a b c}=2220-2017=203
$$

does not fit, as the sum of the digits of 203 is 5, not 10;

(2) $a+b+c=11$, then

$$
\overline{a b c}=2442-2017=425
$$

- fits;

(3) $a+b+c=12$, then

$$
\overline{a b c}=2664-2017=647
$$

does not fit, as the sum of the digits of 647 is 17, not 12;

(4) $a+b+c=13$, then

$$
\overline{a b c}=2886-2017=869
$$

does not fit, as the sum of the digits of 869 is 23, not 13.

## Criteria

+ Full correct solution - 7 points.

$\pm$ The problem is reduced to a short enumeration, but the solution is lost due to arithmetic errors - 5 points.

$\mp$ The condition is rewritten in the form $222 \cdot(a+b+c)=\overline{a b c}+2017$ or equivalent, with no further progress - 3 points.

Only the correct answer without explanation - 2 points.