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 More Harmonic Integers (2) (Posted on 2007-03-28)
Consider three positive integers x< y< z in Harmonic Sequence.

Determine all possible values of the positive integer constant S for which the equation 15x + Sy = 15z admits of valid solutions.

 See The Solution Submitted by K Sengupta Rating: 2.5000 (2 votes)

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 Partly solved | Comment 3 of 6 |

Let the three harmonic integers be

x/(a+b) < x/a < x/(a-b) with x>b [a and b must both be factors of x] ***sorry I used x differently than the original problem***

So we want

15x/(a+b) + Sx/a = 15x(a-b)

[15xa(a-b) + Sx(a+b)(a-b) - 15xa(a+b)] / [a(a+b)(a-b)] = 0

-30xab + Sx(a^2 - b^2) = 0

S = 30ab/(a^2 - b^2)

So any S that fits this pattern should work, although not every choice for a and b gives an integer for S.  I'm not sure what S fits this.

The simplest a=2, b=1 gives S=20
Any x (my x) that gives integers works.  In this case x must be a multiple of 6.

 Posted by Jer on 2007-03-29 11:24:46

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