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Prove that from any set of $11$ natural numbers, there exists 6 numbers such that their sum is divisible by $6$.

Satvik Mashkaria
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    First note that this is quite a hard question (at least I think so, if you haven't seen the solution). Second, it would be good to know what you have tried and the context in which the pigeonhole principle is mentioned. – Mark Bennet Apr 22 '14 at 16:41

1 Answers1

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Lemma. From any set of five natural numbers we can pick three so that their sum is a multiple of $3$.

Proof: If three of the numbers have the same remainder $\pmod 3$, their sum is a multiple of $3$ and we are done. Thus assume each remainder occurs at most twice, hence - by the pigeon-hole principle - each remainder occurs at least once. But $0+1+2\equiv 0\pmod 3$. $_\square$

By the lemma, pick three numbers $a_1,a_2,a_3$ with $3\mid a_1+a_2+a_3$. Form the remaining $8$ numbers pick $b_1,b_2,b_3$ with $3\mid b_1+b_2+b_3$. Froim the remaining five numbers pick $c_1,c_2,c_3$ with $3\mid c_1+c_2+c_3$. Among the three numbers $a_1+a_2+a_3$, $b_1+b_2+b_3$, $c_1+c_2+c_3$, two must have the same parity (again by the pigeon-hole principle). Together we obtain six numbers whose sum is divisible by both $3$ and $2$, hence by $6$.

Hagen von Eitzen
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    And note that $10$ natural numbers won't necessarily do it e.g. $1,1,1,1,1,6,6,6,6,6$ – Mark Bennet Apr 22 '14 at 16:57
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    Given the question title, maybe "hence each remainder occurs at least once" should have a lampshade hung on it. If any remainder occurs 0 times then we're trying to put 5 pigeons into two holes and hence one remainder occurs at least thrice, contrary to what was assumed. – Steve Jessop Apr 22 '14 at 18:17
  • @SteveJessop I don't understand the lampshade analogy. Does that mean you want to make that point less visible? I get the feeling you meant the opposite – Cruncher Apr 22 '14 at 18:20
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    @Cruncher: oh sorry, by "hang a lampshade" I mean make it more obvious as you say. http://tvtropes.org/pmwiki/pmwiki.php/Main/LampshadeHanging defines it as "dealing with any element of the story that threatens the audience's Willing Suspension of Disbelief ... by calling attention to it and simply moving on". Here I mean call attention to the fact that Pigeonhole principle is used here in the proof, although for a small finite case you normally wouldn't (and Hagen didn't) actually cite it. – Steve Jessop Apr 22 '14 at 18:21
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    @SteveJessop Well, the pigeon-hole principle is in fact used twice (the other instance being: Three integer-pigeons are put into two parity-holes). Added lampshades accordingly – Hagen von Eitzen Apr 23 '14 at 15:12