0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.