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?
Answer & explanation
Answer: (d) More than 8
Every number in such a set is a divisor of 1001 = 7 × 11 × 13. The sets that include 1 alone number 12, already more than 8; a full count gives 32 sets of three different numbers.
- 1001 = 7 × 11 × 13, so every number must be one of its divisors: 1, 7, 11, 13, 77, 91, 143, 1001.
- The LCM must contain 7, 11 and 13, and the HCF must be 1, so no prime may divide all three numbers.
- Sets containing 1 automatically have HCF 1. Those with LCM 1001: {1, 7, 143}, {1, 11, 91}, {1, 13, 77}, {1, 77, 91}, {1, 77, 143}, {1, 91, 143}, and {1, x, 1001} for x = 7, 11, 13, 77, 91, 143 — 12 sets.
- 12 is already more than 8; sets such as {7, 11, 13} and {7, 11, 143} add still more (32 in all with three different numbers).
- So n is more than 8.
Remember · With a 'more than' option, you only need to cross the threshold: count the easiest family of cases and stop.
Question and answer: UPSC's official GS Paper II (2025, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). ·