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 Rocking moves (Posted on 2016-05-17)
From the 2001 Moscow Mathematical Olympiad:

Before you are three piles of stones: one containing 51 stones, second with 49 stones, and the third 5 stones.
On each move you can either combine two piles into one or divide any pile with an even number of stones into two equal piles.

Is it possible to end up with 105 piles, each containing a single stone?

 Submitted by Ady TZIDON Rating: 5.0000 (1 votes) Solution: (Hide) Answer: No. If at any point the number of stones in each pile is a multiple of the same odd number k, then this will remain true after any number of operations. No matter what first move we make we create two piles with an odd common divisor. All subsequent operations will produce piles divisible by that odd factor, preventing us from producing piles of single stones. See detailed solution by Paul.

 Subject Author Date re(2): Solution Jer 2016-05-19 07:19:50 re: Solution Ady TZIDON 2016-05-19 04:48:46 Solution Paul 2016-05-17 15:26:44

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