Let n be greater or to 1, and let S be an (n+1)-subset of [2n]. Prove that there exist two numbers in S whose sum is 2n+1.

I know I have to use the pigeonhole principle - no idea how to start...


Consider the $n$ pairs $\{1,2n\}$, $\{2,2n-1\}$, $\{3,2n-2\}$, and so on up to $\{n,n+1\}$. The two numbers in each pair add up to $2n+1$.

If we choose $n+1$ numbers, then they cannot all belong to different pairs.

You can reword this in terms of pigeons. The pair $\{1,2n\}$ are a couple, with their own pigeonhole, as are $\{2,2n-1\}$, and so on up to $\{n,n+1\}$. If there are $n+1$ pigeons currently in the pigeonholes that are their homes, then some couple must be at home.

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    $\begingroup$ If this is a hint, then what's the answer? :) $\endgroup$ – Amit Kumar Gupta Feb 26 '13 at 6:15
  • $\begingroup$ Now it is even longer, and I have removed the hint label. Actually, it was a hint before, because they are probably expected to identify the pigeonholes, and students often have trouble with that. $\endgroup$ – André Nicolas Feb 26 '13 at 6:20
  • $\begingroup$ Thanks a bunch! You're right - I'm having trouble identifying the "boxes", and consequently having trouble applying the principle. $\endgroup$ – user64025 Feb 26 '13 at 6:30

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