How many pairs of integers x y exist such that the product of x y and HCF x y 1080 ........

Explanation:

We need to find ordered pairs (x, y) such that xy * HCF(x, y) = 1080.
Let x = ha and y = hb where h = HCF(x, y) => HCF(a, b) = 1.
So h³(ab) = 1080 = (2³)(3³)(5).
We need to write 1080 as a product of a perfect cube and another number.
Four cases:
1. h = 1, ab = 1080 and b are co-prime. We gave 4 pairs of 8 ordered pairs (1, 1080), (8, 135), (27, 40) and (5, 216). (Essentially we are finding co-prime a,b such that a*b = 1080).
2. h = 2, We need to find number of ways of writing (3³) * (5) as a product of two co-prime numbers. This can be done in two ways - 1 and (3³) * (5) , (3³) and (5)
number of pairs = 2, number of ordered pairs = 4
3. h = 3, number of pairs = 2, number of ordered pairs = 4
4. h = 6, number of pairs = 1, number of ordered pairs = 2
Hence total pairs of (x, y) = 9, total number of ordered pairs = 18.
The pairs are (1, 1080), (8, 135), (27, 40), (5, 216), (2, 270), (10, 54), (3, 120), (24, 15) and (6, 30).