More Harmonic Integers (Posted on 2007-03-12)

	Submitted by K Sengupta
Rating: 3.5000 (2 votes)
Solution:	(Hide)
Since x, y and z are in harmonic sequence with x < y< z, it follows that y = 2xz/(x+z). Since, xy=z and z must be positive, we obtain: 2*x^2 = x+ z,so that: z = 2*x^2 - x and y = 2x-1. Now, x+ z-72 = 2y Or, 2*x^2 - 4x - 70 = 0. The only positive root of the above equation is given by x = 7, so that y = 2*7-1 = 13 and z = 7*13 = 91 Consequently, (x,y,z) =(7,13,91) is the only possible solution to the given problem.

	Subject	Author	Date
	Solution (and a half)	Brian Smith	2007-03-14 20:55:50
	Solution	Dej Mar	2007-03-12 11:58:56
	spoiler	Ady TZIDON	2007-03-12 08:26:48