The product of two numbers is 2028 and their HCF is 13. How many such pairs of numbers are possible?

Concept / Approach

• Let the numbers be 13a and 13b with gcd(a, b) = 1.
• Then (13a)(13b) = 2028 ⇒ 169ab = 2028 ⇒ ab = 12.
• Each prime power in 12 must be assigned wholly to either a or b to keep gcd(a, b) = 1.

Step-by-step reasoning
12 = 2^2 × 3Possible coprime unordered factor pairs (a, b): (1, 12) and (3, 4)Corresponding number pairs: (13×1, 13×12) and (13×3, 13×4)

Note
If ordered pairs were asked, the count would be 4. For unordered pairs (usual convention), the count is 2.

Final Answer
2