The Matchstick Man Strikes Back (Posted on 2012-03-07)

	Submitted by broll
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The first pile must have started with 299 matches. Further, there were at most 3 piles to start with, in which case the other 2 piles began with 17 and 14 matches respectively. 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 values of y greater than 10. 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, and if there are 3, their contents are fixed.

	Subject	Author	Date
	re(3): a solution for n=3	broll	2012-03-11 05:39:24
	re(2): a solution for n=3	Dej Mar	2012-03-11 03:58:45
	re: a solution for n=3	broll	2012-03-10 15:26:23
	a solution for n=3	Dej Mar	2012-03-10 14:58:20
	Solution	Larry	2012-03-09 20:29:57