UPSC Prelims 2025 CSAT, question 28

Topic: Number system

Q28

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 $n$ ?

Solution.

Answer: (d) More than 8

Note that $1001 = 7 \times 11 \times 13$ , which is the product of three distinct primes and has $2^3 = 8$ divisors.
Any unordered triple of distinct divisors of $1001$ has $\operatorname{LCM} = 1001$ and $\operatorname{HCF} = 1$ whenever each prime $7, 11, 13$ appears in at least one and at most two of the three numbers.
There are $32$ such sets of three distinct positive integers (for example, $\{1, 7, 143\}, \{1, 11, 91\}, \{1, 13, 77\}, \{7, 11, 13\}, \{7, 11, 143\}, \dots$ ), which is far more than $8$ .