Question

The product of two numbers is $2028$ and their $HCF$ is $13$. The number of such pairs is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

Let the two numbers be $x$ and $y$ respectively.

It is given that the product of the two numbers is $2028$, therefore,

$xy=2028$

Also $13$ is their HCF, thus both numbers must be divisible by $13$.

So, let $x=13a$ and $y=13b$, then

$13a\times 13b=2028\Rightarrow 169ab=2028\Rightarrow ab=\frac{2028}{69}\Rightarrow ab=12$

Therefore, required possible pair of values of $x$ and $y$ which are prime to each other are $(1,12)$ and $(3,4)$.

Thus, the required numbers are $(12,156)$ and $(39,52)$.

Hence, the number of possible pairs is $2$.