On the board, a three-digit natural number was written. We added to it all other three-digit numbers that can be obtained by changing the order of its digits. Then, in addition to the original number, there were three new ones on the board. The sum of the two smallest of all four numbers is 1088. What digits does the original number contain?

(L. Hozová)

Hint. Determine whether the original number contains zero and whether any digits are repeated.
Let's denote the used digits as $a, b, c$. From the problem statement, it is clear that the digits in the three-digit number do not repeat, i.e., $a, b$, and $c$ are all different. If any two digits were the same, then by changing their order, we could get at most three different numbers: for example, if $a=b \neq c$ and all digits were non-zero, the numbers would be

$$
a a c, a c a, c a a \text {. }
$$

Similarly, we can conclude that one of the used digits must be 0. If $a, b$, and $c$ were all different and non-zero digits, then by changing their order, we would get exactly six different numbers:

$$
a b c, a c b, b a c, b c a, c a b, c b a
$$

The used digits are therefore all different and include 0; without loss of generality, we can assume that $a=0$ and $b<c$. By changing the order of such digits, we get exactly the following four numbers (written in ascending order):

$$
b 0 c, b c 0, c 0 b, c b 0 \text {. }
$$

The rest of the problem is solved as an algebrogram:

![](https://cdn.mathpix.com/cropped/2024_04_17_db2558a90e80c9854485g-1.jpg?height=175&width=221&top_left_y=1877&top_left_x=920)

From the sum of the units $c+0=8$, it follows that $c=8$ (the sum in the tens column agrees: $0+c=8$). From the sum of the hundreds $b+b=10$, it follows that $b=5$. The original three-digit number must contain the digits 0, 5, and 8.