There are 314 coins in 21 open boxes. In each move you can take 1 coin from each of any two boxes and put them into a third box and in the final move you take all the coins from one box. What is the maximum number of coins you can get?

The answer is 314 and I am struggling to prove it is possible to get 314 coins at last for every possible distribution of coins among the 21 boxes.

  • $\begingroup$ Is the number of moves limited? $\endgroup$ – orlp Apr 26 at 11:21
  • $\begingroup$ No. There is no limit for the number of moves. $\endgroup$ – Kumudini Wickramasuriya Apr 26 at 11:25
  • $\begingroup$ This is not clear. Must I take two coins or can I simply take $1$ if I prefer? Say we had three coins in three boxes, distributed as $(2,1,0)$. Then taking one coin from each of the non-empty ones gives you $(1,0,2)$ which is essentially the same. So is this configuration hopeless? $\endgroup$ – lulu Apr 26 at 11:25
  • $\begingroup$ It is necessary to take two coins(one coin from each box) at a time. $\endgroup$ – Kumudini Wickramasuriya Apr 26 at 11:27

$$(10,1,0,0)\to(9,0,2,0)\to(8,2,1,0)\to(7,1,1,2)\to (9,0,1,1)\to(11,0,0,0)$$

  • $\begingroup$ How does this sequence help to solve the problem?? $\endgroup$ – Kumudini Wickramasuriya Apr 26 at 14:41
  • $\begingroup$ As soon as you have at least four boxes, and at least three coins in box 1, you can fetch any other coin into box 1. $\endgroup$ – Empy2 Apr 26 at 14:43
  • $\begingroup$ +1 nice! And I see you have made an editorial choice between being clear and being cute... and I kinda like your choice, actually. :) $\endgroup$ – antkam Apr 26 at 16:34

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