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Choose eight different numbers from one to twenty, prove that there must exist six different numbers such that $a+b+c=d+e+f$.

Since $\binom{20}{8}$ is only $125970$, I can do it with brute force, but that's not a proof.

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Do the six numbers all have to be distinct? –  Nishant Jun 25 at 4:45
    
@Nishant Yes, question edited. –  Ziqian Xie Jun 25 at 4:47
    
Sorry, I meant the six numbers, not the eight numbers, but I assume that's a yes as well? –  Nishant Jun 25 at 4:48
    
@Nishant Yes... –  Ziqian Xie Jun 25 at 4:52

2 Answers 2

up vote 5 down vote accepted

A brute force search may not be particularly satisfying as a proof, but it yields the counter-example $$[ 1, 2, 4, 5, 13, 16, 17, 20 ].$$

This can be verified using the GAP code:

counterexample:=[ 1, 2, 4, 5, 13, 16, 17, 20 ];
triples:=Combinations(counterexample,3);
pairs_of_disjoint_triples:=Filtered(Combinations(triples,2),p->Intersection(p[1],p[2])=[]);
0 in List(pairs_of_disjoint_triples,p->Sum(p[1])-Sum(p[2]));

which returns false.

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If we choose 6 different values a,b,c,d,e,f among 1-20, there are 51 possible values for a+b+c. I think we have to use "The pigeon hall" principal to solve this problem

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from 6 to 57, shouldn't it be 52? –  Ziqian Xie Jun 25 at 5:24

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