- A. 6
- B. 7
- C. 8
- D. More than 8
Answer: D
Explanation
First, find the prime factorization of 1001: 1001 = 7 × 11 × 13.
Let the three positive integers be a, b, and c. We are given that LCM(a, b, c) = 1001 and HCF(a, b, c) = 1.
For LCM(a, b, c) = 7 × 11 × 13, each of the prime factors (7, 11, 13) must appear in at least one of the numbers a, b, or c, and their highest power in any of the numbers must be 1.
For HCF(a, b, c) = 1, no prime factor (7, 11, 13) can be common to all three numbers.
Let’s list some possible sets {a, b, c} that satisfy these conditions:
1. {7, 11, 13}: HCF(7, 11, 13) = 1, LCM(7, 11, 13) = 1001. (1 set)
2. {1, 7, 143}: (143 = 11 × 13). HCF(1, 7, 143) = 1, LCM(1, 7, 143) = 1001. (1 set)
3. {1, 11, 91}: (91 = 7 × 13). HCF(1, 11, 91) = 1, LCM(1, 11, 91) = 1001. (1 set)
4. {1, 13, 77}: (77 = 7 × 11). HCF(1, 13, 77) = 1, LCM(1, 13, 77) = 1001. (1 set)
5. {1, 1, 1001}: HCF(1, 1, 1001) = 1, LCM(1, 1, 1001) = 1001. (1 set)
6. {7, 11, 1001}: HCF(7, 11, 1001) = 1, LCM(7, 11, 1001) = 1001. (1 set)
7. {7, 13, 1001}: HCF(7, 13, 1001) = 1, LCM(7, 13, 1001) = 1001. (1 set)
8. {11, 13, 1001}: HCF(11, 13, 1001) = 1, LCM(11, 13, 1001) = 1001. (1 set)
9. {1, 77, 1001}: HCF(1, 77, 1001) = 1, LCM(1, 77, 1001) = 1001. (1 set)
10. {1, 91, 1001}: HCF(1, 91, 1001) = 1, LCM(1, 91, 1001) = 1001. (1 set)
11. {1, 143, 1001}: HCF(1, 143, 1001) = 1, LCM(1, 143, 1001) = 1001. (1 set)
We have already found 11 distinct sets. Since the options are 6, 7, 8, or More than 8, the answer is clearly ‘More than 8’. This problem is quite challenging for UPSC CSAT, requiring careful enumeration or a more advanced combinatorial approach to ensure all valid sets are considered. It tests advanced concepts of LCM and HCF.