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?

To solve this, first find the prime factorization of 1001, which is 7 times 11 times 13. These are three distinct prime numbers. Let the three integers in a set be x, y, and z. For their Least Common Multiple (LCM) to be 1001, the prime factors 7, 11, and 13 must be distributed among x, y, and z. Specifically, for each prime factor, at least one of the three numbers must contain it, and none can have a higher power of that prime. For their Highest Common Factor (HCF) to be 1, it is required that no single prime factor is present in all three numbers simultaneously. The number of ways to distribute these three primes across three numbers under these conditions is quite large. For a single prime factor like 7, there are several valid distributions across x, y, and z. Since there are three independent prime factors, the total number of combinations is the product of the ways to distribute each prime. Mathematically, for each prime, there are 6 ways to distribute it such that it appears in at least one number but not in all three. With three primes, there are 6 times 6 times 6, which equals 216 ordered triplets. Even after accounting for the fact that the sets are unordered and the numbers must be positive integers, the total number of unique sets far exceeds 8. Therefore, the value of n is more than 8.