# find the largest positive integer $(k>2)$, such that sum of any two distinct numbers is ot divisible by their difference.

For $$A=\{1,2,\cdots ,2012\}$$ , find the largest positive integer $$(k>2)$$ such that it is possible to choose $$k$$ elements in $$A$$ whose sum of any two distinct numbers among the k chosen numbers is not divisible by their difference.

I think it has something to do with divisibility by $$3$$. $$2$$ different numbers that divide by $$3$$ leave a remainder of $$2$$ whose sum is not divisible by the difference.

Define $$B$$ as a subset of $$A$$ such that the sum of any two numbers in $$B$$ is not divisible by their difference.
Observe, we can choose only one of the three consecutive numbers, $$x$$, $$x+1$$ and $$x+2$$ to be elements of $$B$$. Therefore, $$k=|B|\le \left\lceil \dfrac{2012}{3}\right\rceil=671$$. It remains to show that such a set $$B$$ with $$671$$ elements exists. The following construction proves that $$k_{\max}=671$$. $$B=\{1,4,7,\cdots,2011\}\implies |B|=671$$
• Why we can chose only one of the three consecutive numbers $x,x+1, x+2$? Can you explain fo me more details? Commented Jun 15, 2021 at 5:28