Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $a=(b+c+d)/3$ ?

share|cite|improve this question
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
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 $n$.

share|cite|improve this answer
Thanks a lot @Julian Aguirre. – HarshCurious Mar 15 '11 at 3:55

What have you tried so far?


  • 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?
share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.