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Let $n$ be greater or equal to $1$, and let $S$ be an $(n+1)$-subset of $[2n]$. Prove that there exist two numbers in $S$ such that one divides the other.

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1 Answer 1

HINT: Create a pigeonhole for each odd positive integer $2k+1<2n$, and put into it all numbers in $[2n]$ of the form $(2k+1)2^r$ for some $r\ge 0$.

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Ok, so for any set S=(1, 2, ..., 2n), we choose all odd numbers from that S, so So=(1, 3,...,2k+1) such 2k+1 < 2n. For each element in S choose those that satisfy (2k+1)2^r, which are multiples of each element in So. Let this set be S2 = (2, 6,...,(2k+1)2^r), as long as (2k+1)2^r < 2n. S2 is necessarily bigger than So, and thus, each element in So can be 'mapped' to multiple multiples of itself. Consequently, any set bigger than S, of size n+1, must also have this property. Is this reasoning correct? –  user64093 Feb 26 '13 at 18:38

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