Question

How many pairs of natural numbers exist such that the HCF is $6$ and the LCM is $216?$

• A. $8$
• B. $16$
• C. $2$
• D. $1$

Answer 1

The correct option is C $2$

6 = 2 x 3

216 = 2 x 2 x 2 x 3 x 3 x 3

For HCF both numbers should have 2 x 3 in their prime factorization form.

For LCM after the common part is removed that is 2 x 3 the remaining should multiply to be 2 x 2 x 3 x 3.

So there can be 2 ways in which 2 x 2 x 3 x 3 can b distributed among both numbers such that they have nothing in common.

A = 2 x 3 = 6

B = 2 x 3 x (2 x 2 x 3 x 3) = 216

Or

A = 2 x 3 x (2 x 2) = 24

B = 2 x 3 x (3 x 3) = 54

Thus two such pair exists