$4-$ Arithmetic progression $\quad]$ [Pairings and groupings; bijections]

Authors: Dynkin E.B., Moomanov S.A., Rozental A.L., Tomingo A.K.

For which $n$ can weights of 1 g, 2 g, 3 g, ..., $n$ g be divided into three equal-mass piles?
Let's show that this can be done if and only if $n>3$ and one of the numbers $n$ or $n+1$ is divisible by 3. The necessity of these conditions is obvious, since the total weight of the weights $1+2+3+\ldots+n=1 / 2 n(n+1)$ must be divisible by 3.

To prove sufficiency, note first that the partition is possible for $n$ equal to $5, 6, 8$, and 9 (this is shown in the figure). All other values of $n$ of interest can be obtained from these four by adding some number of sixes, and any group of six consecutive integers $a+1, \ldots, a+6$ can easily be divided into three pairs of numbers with equal sums: $\{a+1, a+6\},\{a+2, a+5\}$,
![](https://cdn.mathpix.com/cropped/2024_05_06_ed1a9e9dd82795d4db2eg-11.jpg?height=750&width=754&top_left_y=-2&top_left_x=652)

## Answer

For $n>3, n \equiv 0,2(\bmod 3)$.