How many pairs of integers (x, y) exist such that the product of x, y and HCF…

Given two numbers, finding HCF, LCM is very easy. Given some property about HCF and the numbers, finding the numbers is much tougher.

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).

The question is "How many pairs of integers (x, y) exist such that the product of x, y and HCF (x, y) = 1080?"

Hence the answer is "9"

Choice C is the correct answer.