Take the same basic idea as in Matchstick Frenzy II - A heap of a positive integer number of matches (that is, no broken matches) is divided into n groups.

This time, we take from the first group the square of the number of matches in the second group and add to the second, then take from the second group the square of the number of matches in the third group and add to the third, and so on up to the nth group.

Finally we take from the nth group the square of the number of matches left in the first group and add them to the first group to make the number of matches in each group the same positive integer.

Now for the question: how many matches were there in the first group to start with?

I apologize, I didn't read the question carefully enough.
I thought the rules were the same as in Matchstick Frenzy II.  So I originally had here a general solution to that problem

Edited on March 9, 2012, 8:43 pm

Posted by Larry on 2012-03-09 20:29:57