Three positive integers 2520 < b < c constitute three consecutive terms of a

**harmonic sequence** and, b divides c.

Find the total count of pairs (b,c) for which this is possible.

Well, I guess I did have time to do this now.

let c = kb, where k is an integer

Then 1/2520 - 1/b = 1/b - 1/kb

Rearranging,

b = 5040 - 2520/k

k can be any integer > 1 that is a factor of 2520.

So how many factors does 2520 have?

2520 = 2^3 * 3^2 * 5 *7,

so number of factors = 4*3*2*2 = 48

Since k must be > 1, answer is 48 - 1 = **47**

*Edited on ***April 24, 2015, 9:24 am**