Two numbers have a product of 2028 and their highest common factor (HCF) is 13. How many distinct pairs of such numbers exist?

Explanation

Let the two numbers be 13a and 13b, since their HCF is 13. Their product is given as 2028, so (13a) × (13b) = 2028, which simplifies to 169ab = 2028, or ab = 12. Since a and b must be coprime pairs whose product is 12, the possible pairs are (1, 12) and (3, 4). Multiplying these by 13 gives the pairs (13, 156) and (39, 52). Thus, there are exactly two such pairs.