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

Post a Comment

Popular posts from this blog

Fraction Brainteaser

Source: Sent to me by Gaurav Sinha Problem: Siddhant writes a Maths test and correctly answers 5 out of 6 Arithmetic questions and 20 out of 28 Geometry questions.  In total, Siddhant scores 25 out of 34.  Vaibhav writes another Maths test and correctly answers 20 out of 25 Arithmetic questions and 6 out of 9 Geometry questions. in total, Vaibhav scores 26 out of 34. Note that a) Vaibhav scores more than Siddhant b) Siddhant score better than Vaibhav in both individual topics -  5/6 > 20/25 and 20/28 > 6/9 How is it possible? 
Read more

Buying Dimsums

Source: Alok Goyal (Stellaris VP, Ex-Helion VC) puzzle blog Problem: A fast food restaurant sells dimsums in boxes of 7 and 3. What’s the greatest number of dimsums a person cannot buy. Generalize it for p and q where p and q are relatively prime. I loved the puzzle. Hope you enjoy it too.
Read more

Consecutive Heads

Let's say A keep tossing a fair coin, until he get 2 consecutive heads, define X to be the number of tosses for this process; B keep tossing another fair coin, until he get 3 consecutive heads, define Y to be the number of the tosses for this process. 1) Calculate P{X>Y} 2) What's the expected value of X 3) What's the expected value of Y This is probably the hardest puzzle I have ever put on my blog. Hence, I will post its solution in the post directly rather than on comment. Solution: 1) (Solved by me finally after 13 months :)) Make a state diagram. Let the state be (a,b) where a is the number of consecutive heads streak "A" is on currently and b is the number of consecutive heads streak "B" is on currently. So, (0,3) (1,3) are final accepted states and (2,0) (2,1) (2,2) (2,3) are failure states. If you get tails, your contribution to the state reaches to "0" f(State) = P(X>Y | "State" configuration initially)
Read more