2025 CSAT — Q28

Factorise: $1001 = 7 \times 11 \times 13$ — three distinct primes, each to the first power. Any number whose LCM with others is $1001$ must divide $1001$, so each of the three integers is a product of a subset of $\{7,11,13\}$.

Work prime by prime (the exponents are independent across primes). For a fixed prime $p$ and three numbers, each number either contains $p$ (exponent $1$) or not (exponent $0$). The two global conditions translate as:

• LCM $= 1001$ $\Rightarrow$ the maximum exponent is $1$: at least one of the three has $p$.
• HCF $= 1$ $\Rightarrow$ the minimum exponent is $0$: at least one of the three lacks $p$.

So for each prime the “has/lacks” pattern across three slots must be neither all-have nor all-lack: $2^3 - 2 = 6$ admissible patterns. With three independent primes the count of ordered triples is

and even after collapsing to unordered sets the number stays far above $8$.

Answer: (d) More than 8.

The master technique for “count numbers with prescribed LCM and HCF” is decouple by prime and multiply. For each prime you only decide, across the slots, the max-exponent (set by the LCM) and min-exponent (set by the HCF); with square-free $1001$ that reduces to “at least one has it, at least one lacks it” $= 2^3 - 2 = 6$ per prime, and independence gives $6^3$. The trap distractors are all “small” because they each stop the calculation early — at one prime ($6$), at a divisor count ($8$), or at an off-by-one tally ($7$). Recognising that the true count is in the hundreds (so “more than $8$”) is the entire insight; the exam rewards seeing the multiplicative explosion, not enumerating.