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This question already has an answer here:

A friend of mine give me this problem for fun:

Given $\frac {n(n+1)}{2}$ balls, first we divide arbitrarily these balls in baskets, after that we make another basket with one ball of each basket e do this procedure infinitely.

I want to prove that one time this stabilizes with 1 ball in one basket, 2 balls in another basket, ..., n balls in another basket.

It seems easy to solve, he says we can use some concept of energy (???), I'm trying with some concepts of combinatorics without any success.

Thanks in advance.

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marked as duplicate by joriki, user91500, Joel Reyes Noche, C. Falcon, Shailesh Jul 7 '16 at 0:14

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Please give your question a meaningful title. $\endgroup$ – Chris Eagle Jul 8 '13 at 16:32
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    $\begingroup$ @RossMillikan Then, the problem is, if, in the initial configuration, there were the entire $n(n+1)/2$ balls in the first basket of the first set, and none in the others, we do not run out of all the balls in the first basket of the first set of baskets after only "one pass." In fact, in this case the system will oscillate. $\endgroup$ – Lord Soth Jul 8 '13 at 16:52
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    $\begingroup$ user85493, you're really going to have to explain this better. $\endgroup$ – dfeuer Jul 8 '13 at 16:53
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    $\begingroup$ @LordSoth the number of basket is arbitrary. $\endgroup$ – user85493 Jul 8 '13 at 17:01
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    $\begingroup$ @RossMillikan the initial condition is with an arbitrary number of baskets $\endgroup$ – user85493 Jul 8 '13 at 17:02
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I believe what you are discussing is known as Bulgarian solitaire. This is a theorem of Jorgen Brandt, i.e., that the game ends as you have described when the number of balls (or cards) is a triangular number, i.e., of the form $n(n+1)/2$.


Here is a nice source to read over: (paywall)

Solution of the Bulgarian Solitaire Conjecture

Kiyoshi Igusa

Mathematics Magazine

Vol. 58, No. 5 (Nov., 1985), pp. 259-271

Published by: Mathematical Association of America

Article Stable URL: http://www.jstor.org/stable/2690174


Note this is also the subject of a problem in the June/July 2013 AMM: Problem 11712.

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    $\begingroup$ This is exactly what I presume the OP was talking about. Very nicely done. $\endgroup$ – Steven Stadnicki Jul 8 '13 at 17:08
  • $\begingroup$ Thanks; was just lucky to recognize it from glancing at the most recent AMM. $\endgroup$ – Benjamin Dickman Jul 8 '13 at 17:09
  • $\begingroup$ No problem. If you can get a hold of the most recent issue of the American Mathematical Monthly (June/July '13) you (or your friend) might check out the problem on Bulgarian solitaire. $\endgroup$ – Benjamin Dickman Jul 8 '13 at 17:19
  • $\begingroup$ I didn't find this problem in the magazine of june/july '13, what's the name of the article? $\endgroup$ – user85493 Jul 8 '13 at 17:23
  • $\begingroup$ Check the Problems and Solutions section of AMM June/July 2013 and look at the first problem listed (i.e., AMM Problem 11712). $\endgroup$ – Benjamin Dickman Jul 8 '13 at 17:24

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