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For which values of the positive integer $n$ is it possible to divide the first $3n$ positive integers into three groups each of which has the same sum?


This puzzle is from a UK Junior Mathematical Olympiad.

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1 Answer 1

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It is possible for

all $n\ge2$.

Clearly it is not possible for $n=1$, and $n=2$ has the easy solution $1+6$; $2+5$; $3+4$.

$n=3$ has solutions, such as $3+5+7$; $1+6+8$; $2+4+9$.

If you have a solution for any particular $n$, you can extend it to $n+2$ by distributing the $6$ extra numbers over the three groups in a similar way as the $n=2$ solution: $(n)+(n+5)$; $(n+1)+(n+4)$; $(n+2)+(n+3)$.
Therefore by induction it is possible for all $n\ge2$.

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