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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?

See The Solution Submitted by K Sengupta    
Rating: 4.3333 (3 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Some Thoughts general solution Comment 6 of 6 |
If there are N groups, designated 1, 2, ..., i, ..., N;

then the formula for the minimum number of matches in each group is:

1: 1.5 * (2^N) - 1
2: (2^N) - 1
3: (2^N) - 2
4: (2^N) - 4
...
i: (2^N) - 2^(i-2)
...
N: (2^N) - 2^(N-2)


  Posted by Larry on 2012-03-09 20:41:55
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