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?

Unless I'm mistaken, you've shown all you need.

Let all piles end with y^2+y matches.

y is Pellian: y=((2^(1/2)+1)^(4*n+1)-(2^(1/2)-1)^(4*n+1)-2)/8, for all possible y: {10,348,11830,401880...}. This being so, the corresponding number in the last-but-one column has to approximate y*2^(1/2): {14,492,16730,568344...} resulting in negative groups of matches for all higher values of y.

So let the final number of matches in each pile be y^2+y=110, with y positive, i.e. y=10. Then on the move before that, z^2+z= 2y^2+y=210, for some z, giving:
z= (1/2)*((8y^2+4y+1)^(1/2)-1) in in the last pile, F, before the final transfer, giving z = 14. Now for pile (F-1), 14^2 or 196 matches must have been transferred, giving x^2+x=306, or x=17.  Last but not least, for pile (F-2), 17^2 or 289 matches were transferred, so w^2+w=399; but then w is not an integer (it is about 19.48).

So there cannot be more than 3 groups.

Edited on March 11, 2012, 6:18 am

Posted by broll on 2012-03-11 05:39:24