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More Harmonic Integers (Posted on 2007-03-12) Difficulty: 2 of 5
Determine three positive integers x, y and z in harmonic sequence such that x<y<z, xy=z, and (x, y, z-72) are in arithmetic sequence.

  Submitted by K Sengupta    
Rating: 2.0000 (1 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.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
Solution (and a half)Brian Smith2007-03-14 20:55:50
SolutionSolutionDej Mar2007-03-12 11:58:56
SolutionspoilerAdy TZIDON2007-03-12 08:26:48
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