Pebble Piles

Problem: You are given three piles with 5, 49 and 51 pebbles respectively. Two operations are allowed:
(a) merge two piles together or
(b) divide a pile with an even number of pebbles into two equal piles.

Is there a sequence of operations that would result in 105 piles with one pebble each?

Update(12/02/10):
Solution: Posted by Shantanu Gangal (CSE IITB Alumnus and BCG Consultant) in comments!!

Comments

  1. Shantanu11:42 PM, February 11, 2010

    Not possible.

    Lemma: If the gcd of all the pile sizes is a odd number (>1), then the objective isn't possible.
    Proof: Assume that the common gcd of piles is n, an odd no > 1. Hence, both the operation will keep this invariance. As a result, you can never create a pile of size < n.

    Now, the only operation you can do at {5, 49, 51} is (a).
    But, 3| {51, 54}; 5|{5, 100} and 7|{49, 56}. Hence the desired piling isn't possible.

    ReplyDelete
  2. Pratik Poddar11:38 AM, February 12, 2010

    Thanx Shantanu.. Correct Solution :)

    ReplyDelete
  3. stp11:43 AM, February 24, 2010

    im not sure of my solution..
    this is mine..
    if after all such operations we get 105 single pebbled piles...
    the last operation would be dividing a pile of 2 into 2 piles of 1 pebble.
    in that way the proceeding from back
    we get 52 piles of 2 pebbles and 1 single pebbled pile.There is no pile with equal number if pebbles as this 1 pebbles pile.
    So this is not possible...

    ReplyDelete
  4. Pratik Poddar12:23 PM, February 26, 2010

    I am sorry.. I did not get this. Why is this not possible?

    ReplyDelete
  5. stp1:03 PM, February 26, 2010

    i was not sure of my soln....
    later after posting it, i came 2 know dat i was wrong...

    ReplyDelete
  6. Pratik Poddar1:46 PM, February 26, 2010

    :) chalega :P

    ReplyDelete

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