Question

The H.C.F of two numbers is 8 and their L.C.M is 3600. How many such pairs are possible?

Answer 1

HCF of two numbers is 8 and LCM is 3600

Let two numbers is 8a and 8b , where a and b are not divisible by each other.

Now, LCM of {8a , 8b} × HCF of {8a , 8b } = product of 8a and 8b [ from theorem, ]

Given, HCF of {8a, 8b} = 8

LCM of {8a , 8b} = 3600

so, 3600 × 8 = 8a × 8b

⇒ 3600 × 8 = 64ab

⇒450 = ab

Now, prime factors of 450

450 = 2 × 3 × 3 × 5 × 5 = 2 × 3² × 5²

= 2 × 9 × 25

if we assume a = 2 , then b must be equal to 9 × 25

If we assume a = 9, then b = 50

if we assume a = 25 , then b = 2 × 9

If we assume a = 2 × 9 × 25 then, b = 1

So there are four pair of numbers

e.g., (3600, 8), (200, 144), (72, 400) and (16, 1800)