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For which integral values of $n$ can a set $(1,2,3,...,4n)$ be split into $n$ disjoint $4$-element subsets $(a,b,c,d)$ such that in each of these sets $a=(b+c+d)/3$ ?

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I'm not sure what it is, but most certainly this question does not fall under elementary set theory, and even less under set theory tag. –  Asaf Karagila Mar 14 '11 at 18:13

2 Answers 2

up vote 2 down vote accepted

If $a=\frac{b+c+d}{3}$, then $a+b+c+d=4a$ is a multiple of $4$. If $\{1,\dots,4n\}$ can be split in quadruples $\{a,b,c,d\}$ with $a=\frac{b+c+d}{3}$, then $1+2+\dots+4n$ must also be a multiple of $4$. I leave you to check that this is so if and only if $n$ is even.

For $n=2$ we have the solution $\{1,\dots,8\}=\{4,1,3,8\}\cup\{5,2,6,7\}$. From here it is easy to construct a solution for any even $n$.

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Thanks a lot @Julian Aguirre. –  HarshCurious Mar 15 '11 at 3:55

What have you tried so far?

Hints:

  • since they all have to be integers, how many ways can $b+c+d$ be divisible by 3 and less than 40/3?
  • since there are no other restrictions on $b,c,d$, how many ways can these three be permuted and still give the same sum?
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