The product of two numbers is 2028 and their HCF is 13. How many such pairs of numbers are possible?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2
With HCF 13, the numbers are \(13a\) and \(13b\), where \(a\) and \(b\) have no common factor (otherwise the HCF would be larger than 13).
\(13a\times13b=2028\), so \(ab=\frac{2028}{169}=12\)
Factor pairs of 12: (1, 12), (2, 6), (3, 4)
(2, 6) has the common factor 2, so it is rejected
(1, 12) gives 13 and 156; (3, 4) gives 39 and 52
There are 2 such pairs.