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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)

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 solution | Comment 1 of 5
In the starting state, each pile contains an odd number of stones. Adding two of these numbers together will produce and even number, and when that even number is added to or subtracted from the remaining odd number, that number will remain odd.

The situation then continues, always with an odd number in each pile.

Zero is an even number; no pile will ever get to that count.

 Posted by Charlie on 2017-03-19 13:32:00
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