I am attempting to prove the statement: "If $n$ is a positive integer, and you choose any subset of size $n+1$ from the set $\{1,2, \cdots, 2n\}$, it will always be possible to find two distinct elements in the subset such that one is a factor of the other".

I have found a method which works provided that the following proposition is true: if $a < b \leq n$ and $a$ does not divide $b$, then there exists some multiple of $a$ within the set $\{n+1,n+2,...,2n\}$ which is not also a multiple of $b$. How can I prove this proposition?

  • $\begingroup$ $2n$ and $n$ of course $\endgroup$
    – user
    Nov 13, 2018 at 13:21
  • 1
    $\begingroup$ You don't mean "size $\geq 2$", but "size $>n$" perhaps? $\endgroup$
    – B. Goddard
    Nov 13, 2018 at 13:24

1 Answer 1


This is false. Take $n=10$ and pick $\{3, 7\}$. None of them divide each other.

  • $\begingroup$ Yes, sorry, I made a typo in the question. It should say "any subset of size $n+1$". Could you have a look again? $\endgroup$
    – Prasiortle
    Nov 13, 2018 at 15:27
  • $\begingroup$ @Prasiortle Sure. I've already tried to think of a solution to that problem using the pigeonhole principle, but I've concluded that you can't solve it using just a simple pigeonhole. The idea is to make $n$ pairs of two, and in each pair one element divides the other. But you cannot make such pairs; for example, if $n = 10$, you must put $17$ with $1$, but then you don't have anything to put $13$ with. $\endgroup$
    – Ovi
    Nov 13, 2018 at 17:09

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