Question

The product of two numbers is $2028$ and their H.C.F. is $13$. The number of such pairs is:

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (B)

$2$

Let the numbers $13a$ and $13b$.
Then, $13a\times 13b=2028$
$\Rightarrow ab=12$.
Now, the co-primes with product $12$ are $(1,12)$ and $(3,4)$.
[Note: Two integers $a$ and $b$ are said to be coprime or relatively prime of they have no common positive factor other than $1$ or, equivalently, if their greatest common divisor is $1$]
So, the required numbers are $(13\times 1,13\times 12)$ and $(13\times 3,13\times 4)$.
Clearly, there are $2$ such pairs.