If product of two numbers is $1014$ and their $\text{H.C.F.}$ is $13,$ then number of such pairs is

The correct option is C $2$
Let the numbers be $13a$ and $13b,$ where $a$ and $b$ are co-prime.
Then, $13a\times 13b=1014$
$\Rightarrow ab=6$
Now, the co-primes with product $6$ are $(1,6)$ and $(2,3).$

So, the required pairs are $(13\times 1,13\times 6)$ and $(13\times 2,13\times 3)$
or, $(13,78)$ and $(26,39)$
Clearly, there are two such pairs.