Question

The HCF of two numbers is 6 and the product of the two numbers is 4320. How many pairs of numbers exist, which satisfy the above conditions?

Answer 1

HCF=6
let the numbers be 6a and 6b
LCM=6ab
ab=4320/6
(1,720) and (9,80) are Co primes
pairs of numbers (6×1,6×720);(6×9,6×80)
(6,4320);(54,480)

Answer 2

Given:

• HCF = 6
• Product of two numbers = 4320

To Find:

• The total number of pairs existing.

Solution:

• Let the first number be = x and the second number = y.
• HCF of x and y = 6 and xy = 4320
• We can say that both the numbers are multiples of 6.
• let x = 6a and y = 6b
• So, their (6a,6b) HCF is 1, this implies they are coprime factors.
• x*y=4320
• 6a*6b = 4320
• a*b = 720
• 720 = (1*720) and (9*80) [They are the coprime factors of 720).
• The pairs are (6,4320) and (54,480)

Two numbers of pairing is possible.