Question

If the HCF of three numbers $144,x$ and $192$ is $12$, then the number $x$ cannot be

• A. $180$
• B. $84$
• C. $60$
• D. $48$

Answer 1

Answer: (D)

$48$

If we would take x=48, then the HCF of the given numbers would become 48 as all the numbers would be divided by 48 to give a positive integer.

Hence x cannot be 48.