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Matchstick Frenzy II (Posted on 2012-02-26) Difficulty: 3 of 5
A heap of a positive integer number of matches (that is, no broken matches) are divided into five groups.

If we take as many matches from the first group as there are in the second group and add them to the second, and then take as many from the second group as there are in the third group and add them to the third, and, so on ...... until finally, we take as many from the fifth group as there are in the first group and add them to the first group - the number of matches in each of the groups would be equal to the same positive integer.

What is the minimum number of matches in each group at the beginning?

No Solution Yet Submitted by K Sengupta    
Rating: 4.0000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Some Thoughts re: General n pile solution with proof | Comment 5 of 6 |
(In reply to General n pile solution with proof by Jer)

Jer:


Nice work!  I agree with your belief that this is minimal, because:

a) I believe that for any given number of piles, all solutions have the same ratio of matches among piles.  (In other words, all solutions are an integer multiple of the minimal solution)

and

b) x(2) and x(3) are relatively prime, so the solution you have given must be the minimal one.



  Posted by Steve Herman on 2012-02-27 16:30:33
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