11.1. Three natural numbers are such that the last digit of the sum of any two of them is the last digit of the third number. The product of these three numbers was written on the board, and then everything except the last three digits of this product was erased. What three digits could have remained on the board? Find all possible answers.

(N. Agakhanov)
Answer: 000, 250, 500, or 750.

Solution: Let \(a, b, c\) be the given numbers. According to the problem, the numbers \(a+b-c\), \(b+c-a\), and \(c+a-b\) are divisible by 10. Therefore, their sum, which is \(a+b+c\), is also divisible by 10. On the other hand, from the equality \(a+b+c = (a+b-c) + 2c\) and the problem's condition, it follows that the last digit of the sum of all three numbers is the same as the last digit of the number \(2c\). Thus, the number \(c\) ends in 5 or 0. Similarly, the numbers \(a\) and \(b\) end in 0 or 5.

Finally, since the sum \(a+b+c\) is even, one of the numbers \(a, b, c\) is also even. Therefore, one of these numbers is divisible by 10, and the other two are divisible by 5. Then the product is divisible by 250, and thus can end only in 250, 500, 750, or 000. It remains to provide examples of triples of numbers that satisfy the conditions and give the last digits: \(10 \cdot 10 \cdot 10 = 1000\); \(5 \cdot 5 \cdot 10 = 250\); \(5 \cdot 5 \cdot 20 = 500\); \(5 \cdot 5 \cdot 30 = 750\).

Comment: Only the answer - 0 points.

Four examples are provided, showing that the product can end in each of the numbers 000, 250, 500, and 750, with no further progress - 2 points.

Only three out of the four such examples are provided, with no further progress - 1 point.

It is only proven that all three numbers are divisible by 5 - 2 points. If, in addition, it is proven that one of them is divisible by 10 - another 1 point.

It is proven that the product is divisible by 250 (or an equivalent statement) and only examples are provided showing that three out of the possible options 000, 250, 500, and 750 are realized - 5 points.