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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.)
Solution answers and general approach- spoiler | Comment 2 of 6 |

ans: 47,31, 30, 28, 24

My way of solving was to start with 3 groups and then generaliize for n. I have the answers for  3,4,5 and 6 groups , the total number is always  n*2^n    ;  for n=5  it is 160.


  Posted by Ady TZIDON on 2012-02-26 15:25:28
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