Is it true that however we pick 1011 numbers alongside of $\{1,2,...,2020\}$ then for at least two of them we'll have that one of those two numbers is a divisor of the other?
Meaning that however we pick $A\subseteq ${$1,2,...,2020$} with $|A|=1011$, then $\exists x,y\in A$ s.t. $x|y$ or $y|x$.
What I thought of is somehow using the Pidgeonhole principle, but I've no constructive thoughts from that on. I'll appreciate any help. Thanks.
Every integer is a power of two times an odd number. There are only 1010 candidates for the odd part. Apply pigeon hole