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i.e. Let $A=\{1,2,3,...,n\}$, what is the greatest cardinality of $B\subset A$ s.t. no element in $B$ divides another element in $B$. Preferably using pigeonhole principle.

nasdas asdas
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  • What have you tried? An obvious first thought would be to write out the answers for small $n$. – lulu Jan 19 '20 at 23:37
  • This is specifically for a few n's around 2000, so I was wondering if there was a general principle I could use to develop them given an arbitrary n. – nasdas asdas Jan 19 '20 at 23:39
  • And the best way to get an understanding is to work the same problem for small $n$. – lulu Jan 19 '20 at 23:40
  • @lulu had initially given away the answer, but edited the comment to remove it. – Rushabh Mehta Jan 19 '20 at 23:42
  • @DonThousand In truth, my first guess was wrong. You can do much better than the set of primes $≤n$. For $n=2m$ you can find a set with $m$ elements that works. – lulu Jan 19 '20 at 23:45
  • This well known Putnam Problem is similar and relevant. – lulu Jan 19 '20 at 23:47
  • @lulu I found the correct approach (I think). Iterate from $n\to1$, and check whether the given element is a divisor of any of the elements in the subcollection you are generating (at the beginning, it is empty). If it isn't, add it to the subcollection. – Rushabh Mehta Jan 19 '20 at 23:52
  • @DonThousand That's how I was thinking at first, but it's simpler than that. At least for even $n=2m$. Pigeonhole tells us that $m+1$ fails and its easy to do $m$. I haven't thought about odd $n$ but it has to be more or less the same. – lulu Jan 19 '20 at 23:54
  • @lulu I think it's exactly the same for $2n+1$, since the elements $n+1\to2n+1$ work, while any more clearly doesn't by pigeonhole. – Rushabh Mehta Jan 19 '20 at 23:59
  • @DonThousand Agreed, it's not materially different. – lulu Jan 20 '20 at 00:01

1 Answers1

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As pointed out in the comments, all we need is a minor addition to an earlier question

Suppose first that $n=2m$. Clearly the subset $m+1,\dots,2m$ works. Now make pigeonholes for $k=1,\dots,m$, so that the pigeonhole for $k$ contains all numbers $(2k-1)2^r$ (with $r\ge0)$) from $A$. Between them the pigeonholes contain the whole of $A$ and we can pick at most one number from each pigeonhole.

For $n=2m+1$, again the subset $m+1,\dots,2m+1$ works. Now we know we cannot pick more than $m$ from $1,2,\dots,2m$, so we cannot pick more than $m+1$ from $1,2,\dots,2m+1$.

almagest
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