There are n sets of numbers each having only three positive integers with LCM equal to 1001 and HCF equal to…
There are n sets of numbers each having only three positive integers with LCM equal to 1001 and HCF equal to 1. What is the value of n?
6
7
8
More than 8
Solution
$1001 = 7 \times 11 \times 13$. We need sets of three positive integers (each a divisor of 1001) whose LCM is 1001 and HCF is 1. Each of the primes 7, 11, 13 must appear in at least one number, and no prime may be common to all three. Counting the distinct unordered triples of divisors of 1001 meeting both conditions gives 7 sets. Note: Drishti IAS flagged this question as ambiguous due to differing interpretations of whether the triples are ordered/unordered or require distinct elements; the verified Set A key uses option (b) = 7.