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)
• 1001 = 7 × 11 × 13. For the LCM to be 1001, each of 7, 11 and 13 must divide at least one of the three numbers; for the HCF to be 1, none of them may divide all three.
• So every number in such a set is a divisor of 1001: 1, 7, 11, 13, 77, 91, 143 or 1001.
• Any set containing 1 has HCF 1, so it is enough to pick two other divisors whose LCM is 1001: {1, 7, 143}, {1, 11, 91}, {1, 13, 77}, {1, 77, 91}, {1, 77, 143}, {1, 91, 143}, {1, 7, 1001}, {1, 11, 1001}, {1, 13, 1001}.
• That is already 9 sets, and more exist without 1, such as {7, 11, 13} and {77, 91, 143}.
• So n is more than 8, option (d).
Why the other options are wrong
- (a) 6
- At least nine valid sets can be listed (for example {1, 7, 143}, {1, 11, 91}, {1, 13, 77} …), so n cannot be 6.
- (b) 7
- At least nine valid sets can be listed (for example {1, 7, 143}, {1, 11, 91}, {1, 13, 77} …), so n cannot be 7.
- (c) 8
- At least nine valid sets can be listed, and sets such as {7, 11, 13} add more, so n cannot be 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.