Special case of a rather classical result by Erdos. The main trick here is to use Pigeonhole principle. To prove the problem, examine any set of $11$ integers. Note that there are $10$ odd positive integers between $0$ and $21$ exclusive, so by pigeonhole there must be two integers in this $11$-set with the same odd component, therefore the smaller one divides the larger one.

write each number in the form n=q.2^r where q is odd. as all the numbers are under 20 , there are only 10 possibilities for q. and the rest is trivial using PHP. Click to reveal hidden text