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

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

Solution:
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 × 13b = 2028
⇒ 169ab = 2028
⇒ ab = 2028
∴ ab = 12

Therefore, the 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.