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Number theory Difficulty 6.0 National olympiad Prove it Ukraine

We chose several numbers among 1,2,,20221, 2, \ldots, 2022. It turned out that sum of any two of the chosen numbers isn't divisible by the difference between any two of the chosen numbers. What largest possible number of numbers could be selected?

(Oleksii Masalitin)

Solution

Suppose that more than 674674 numbers were chosen. Then there exists a triple 3n+1,3n+2,3n+33n + 1, 3n + 2, 3n + 3, among which at least two numbers were chosen, so the absolute difference between some two of the chosen numbers doesn't exceed 22. Clearly, there exist some two chosen numbers with the same parity, so their sum will be divisible by that difference not exceeding 22. So, not more than 674674 numbers were chosen.

Let's show that we can choose a given number of numbers. Consider the set {1,4,7,,2020}\{1, 4, 7, \ldots, 2020\}, consisting of 674674 numbers. As the difference between any two of these numbers is divisible by 33, and the sum of any two of these numbers isn't divisible by 33, this set satisfies the conditions from the statement.

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