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?

If A is the initial number of matches in the 1st group, and B is the initial number of matches in the 2nd group, and N is the initial number of matches in the nth group, then the final number of matches in the 1st group can be given by the equation:
(A - B²) + (A - B²)²
and the final number of matches in the nth group can be given by the equation:
(N + N²) - (A - B²)²

Let x = (A - B²), then, given that the 1st group and the nth group equal each other:
N^2 + N - x^2 = x^2 + x

A quadratic equation can be formed and solved for N:
N^2 + N + (-2x^2 + x) = 0
As N must be a positive integer, only one of the two possible quadratic values for N is possible:
N = [SQRT(8x^2 + 4x + 1) - 1]/2.

Using a spreadsheet to iterate through values of x, one finds at least two possible solutions. The first is where x = 10 giving N = 14, and the second is where x = 492 giving N = 348.
By substitution of the latter I find that
(348² + 348) - (492)² = -120612. As -120612 is a negative value it can not be a solution to the problem. All values where x > 348 would also result in negative (though not necessarily integer) values for the number of matches in the nth group, thus only the first integer solution,  (14² + 14) - (10)² = 110, results in the positive integer. Thus it is the only possible solution.
 
Of course this does not prove how many groups of matches are formed, only that the final number in each group can be no other quantity.

Edited on March 11, 2012, 4:03 am

Posted by Dej Mar on 2012-03-11 03:58:45