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 Leave no stone unturned (Posted on 2017-03-19)
There are three virtual piles of stones. In one operation one may add to, or remove from, one of the piles the number of stones equivalent to the quantity in the other two piles combined, leaving the numbers in those two piles unchanged.
Thus, e.g., (12,3,5) can become (12,20,5) by adding 12+5=17 stones to the second pile, or (12,3,5) can become (4,3,5) by removing 3+5=8 stones from the first pile.

Assume a starting state (1111,111,11).
Is it possible, by a sequence of such operations, reach a state where one of the piles is empty?

 See The Solution Submitted by Ady TZIDON Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 re: Other cases | Comment 4 of 5 |
(In reply to Other cases by Steve Herman)

Clearly if the numbers of stones, each divided by the GCD of the three, are all odd, my argument holds as well. Such is the case with  (1110,110,10), where the GCD is 10.

I wonder if, when the numbers are divided by their GCD, that the results are not all odd, whether in all those cases, a zero pile is possible.

 Posted by Charlie on 2017-03-19 14:33:33
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