There are n sets of numbers each containing only three positive integers whose LCM is equal to 1001 and HCF is equal to 1. What is the value of n ?

1001 = 7 × 11 × 13. We need ordered/unordered triples of positive integers with LCM = 1001 and HCF = 1. The divisors of 1001 are {1, 7, 11, 13, 77, 91, 143, 1001} — 8 divisors. We need to choose three divisors such that their LCM is 1001 and HCF is 1. Considering unordered triples: many combinations like (1, 7, 1001), (1, 11, 1001), (1, 13, 1001), (7, 11, 13), (7, 11, 143), (7, 13, 77), (11, 13, 91), (1, 77, 13), (1, 91, 11), (1, 143, 7), etc. Counting carefully yields more than 8 such triples. Hence the answer is 'More than 8'.