The LCM of two numbers is 1710 and their HCF is 57. How many such pairs of numbers are possible?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 4 pairs
Two numbers with HCF \(h\) can be written as \(ha\) and \(hb\), where \(a\) and \(b\) have no common factor. Their LCM is then \(hab\), so the pairs are counted from the co-prime pairs \((a, b)\) with the right product.
Numbers \(=57a\) and \(57b\), LCM \(=57ab=1710\)
\(ab=1710\div57=30\)
Pairs with product 30: (1, 30), (2, 15), (3, 10), (5, 6)
Each of these pairs has no common factor, so all 4 are allowed
The number pairs are (57, 1710), (114, 855), (171, 570), (285, 342)
So 4 pairs of numbers are possible.