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Leave no stone unturned (Posted on 2017-03-19) Difficulty: 2 of 5
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.)
Some Thoughts re: Other cases | Comment 3 of 5 |
(In reply to Other cases by Steve Herman)

I probably picked a bad example.  It is obvious that (1110,110, 10) cannot be reduced to zero.  Just divide by 10 to reach (111,11,1) and then apply Charlie's parity argument.

A better example is (1110, 108, 10), which can be reduced to (48, 50, 10) but no further.  

I do not see an obvious test for whether any set can reduced so that one of its terms is zero, other than:
a) divide each pile by their greatest common divisor
b) if all piles are odd after the division, it is impossible
c) Otherwise, repeatedly subtracted the two smallest piles from whatever is the largest at each step, until zero is reached or until the two smaller piles have a sum greater than the largest pile. 

  Posted by Steve Herman on 2017-03-19 14:29:54
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