Given $n>0$, let $S$ be a set whose elements are positive integers $\leq 2n$ such that: S is a set with the property that for all a,b∈S with $a<b$, a doesn't divide b. What is the maximum number of integers that $S$ can contain ?
I thought it was the number of prime numbers smaller than $2n$ because the primes satisfy the property. 
But thinking better this could be one of the least number of elements possible, cause every other number greater than 1 and smaller than 2n can be divided by one of the primes, this would mean there would be a lot of excluded numbers.
I believe this number is smaller than $n+1$

$S$ can certainly contain at least $n$ integers: take $S=\{n+1,n+2,\ldots,2n\}$. The question is whether it can contain $n+1$. Each integer in $\{1,\ldots,2n\}$ can be written uniquely in the form $2^km$, where $m$ is odd and $k\ge 0$. There are $n$ possible choices for $m$, so if $S$ contains more than $n$ integers, two of them must share the same odd part, and the larger of the two will be a multiple of the smaller. Thus, $n$ is the largest possible size of $S$, and it’s always achievable.
