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A game is played with a collection of $n \geq 1$ beans. The beans are first divided into piles. A move consists of taking a bean from each pile, and forming a new pile with these beans. This process is repeated indefinitely. So, for example, if 12 beans were divided into piles of 3, 4, and 5 beans, then after one move there would be piles with 2, 3, 3, and 4 beans, and after another move there would be piles with 1, 2, 2, 3, and 4 beans. A collection of piles of beans is stable if making a move does not change the number of piles of any given size. So, for example, three piles of sizes 1, 2, and 3 would form a stable collection. Prove that if n beans can be divided into a stable collection of piles then n = 1 + 2 + · · · + k, for some positive integer k.

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    What reasoning have you done so far? – 2017-01-18
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    You should be able to make some progress on this. – 2017-01-18
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    I understand why a set of piles that has 1, 2, 3, 4, 5, ..., k beans is stable, but I don't understand why this also works for other sets of piles that also add up to $\frac{(k)(k+1)}{2}$. – 2017-01-18
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    Suppose you have a stable set of m piles. The smallest pile must be 1. Why? There can't be two piles of 1. Why? Hence there must be a 2. Why? But there can't be two piles of 2. Why? Continue ... – 2017-01-19
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    Only piles $1...n$ are stable. Other piles that add to $\frac{n^2+n}2$ are not stable. For example, consider a single pile with 10 beans: not stable, and if you apply the process it turns into $1,2,3,4$, which is stable. – 2017-01-19
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    The so-called duplicate is about showing $\binom{k+1}{2} = \frac{k(k+1)}{2}$ will ultiimately give a stable pile, while this Question asks if such a number of "beans" is the only way a stable pile will be achieved. – 2017-01-19

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