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 ?

(a) 6
(b) 7
(c) 8
(d) More than 8 ✓ UPSC's answer

Why the answer is (d)

• The number 1001 factors into distinct primes as $7 \times 11 \times 13$.

• Since the HCF of the three integers is 1, no prime factor can be common to all three numbers.

• Since the LCM is 1001, each of the three prime factors (7, 11, 13) must appear in at least one of the three integers.

• For each prime factor, it can be assigned to any of the three integers, giving $3^3 = 27$ total assignments.

• We must exclude assignments where a prime is assigned to all three integers (which would make the HCF greater than 1). There are 3 such invalid cases (one for each prime).

• Thus, the number of valid sets is $27 - 3 = 24$, which is more than 8.

Why the other options are wrong

(a) 6: The calculation $3^3 - 3 = 24$ yields a value significantly larger than 6.
(b) 7: The total number of valid combinations is 24, not 7.
(c) 8: The total number of valid combinations is 24, which exceeds 8.

Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2025, held on 25 May 2025. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.