The HCF of two numbers is 13 and their product is 2,028. How many different pairs of such numbers are possible?

Explanation

1. 01

   Write the numbers as 13a and 13b, where a and b are co-prime.

2. 02

   $13a \times 13b = 2028$, so ab = $2028 \div 169 = 12$.

3. 03

   Co-prime pairs with product 12: (1, 12) and (3, 4). The pair (2, 6) fails because $\operatorname{HCF}(2, 6) = 2$.

4. 04

   So 2 pairs: (13, 156) and (39, 52).