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Question

# How many pairs of integers (x, y) exist such that the product of x, y and HCF (x, y) = 1080

CAT 2021

A
B
C
D
##### 12
Solution

Correct option is

(C)

We have 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.$h^3(ab)&space;=&space;1080&space;=&space;(2^3)(3^3)(5).$

We need to write 1080 as a product of a perfect cube and another number.

Four cases which are possible :

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 may be done in two ways - $1&space;;and&space;;(3^3&space;)*&space;(5)&space;,&space;(3^3)&space;;and&space;;&space;(5)$

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).

so , C is the correct option