The product of two numbers is 2028 and their HCF is 13. The number of such of pair is = ?

Answer: Option B

HCF = 13 (given)
Let number are 13x & 13y respectively
$$\begin{array}{rl}& \text{Also given}\\ & 13x\times 13y=2028\\ & 13\times 13\times xy=2028\\ & xy=\frac{2028}{13\times 13}=12\\ & \therefore \text{Possible pairs are}\\ & \text{ = }\left(1,12\right)\left(3,4\right)\end{array}$$
Only two pair are possible