All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Logic
Rocking moves (Posted on 2016-05-17) Difficulty: 3 of 5
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.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
re(2): SolutionJer2016-05-19 07:19:50
Hints/Tipsre: SolutionAdy TZIDON2016-05-19 04:48:46
SolutionPaul2016-05-17 15:26:44
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information