Understanding the Problem: Sets of Numbers, LCM, and HCF

The question asks for the number of sets, n, each containing three positive integers, let's call them {a, b, c}, such that their Least Common Multiple (LCM) is 1001 and their Highest Common Factor (HCF) is 1. The phrase "sets of numbers each having only three positive integers" implies that the three integers in each set must be distinct.

Prime Factorization of 1001

First, let's find the prime factorization of 1001.

\(1001 = 7 \times 143 = 7 \times 11 \times 13\)

The prime factors of 1001 are 7, 11, and 13, all raised to the power of 1.

HCF and LCM Conditions in Terms of Prime Exponents

Let the three positive integers be a, b, and c. Since their LCM is 1001, the prime factors of a, b, and c can only be 7, 11, and 13. The maximum power of each prime factor in any of the numbers cannot exceed the power of that prime in the LCM (which is 1 for each prime). So, each number a, b, c must be of the form \(7^{e_7} \times 11^{e_{11}} \times 13^{e_{13}}\), where the exponents \(e_7, e_{11}, e_{13}\) are either 0 or 1.

Let \(v_p(x)\) denote the exponent of prime \(p\) in the prime factorization of number \(x\).

• The condition LCM\((a, b, c) = 1001\) means that for each prime factor \(p \in \{7, 11, 13\}\), the maximum exponent among a, b, and c must be 1. Mathematically, \(\max(v_p(a), v_p(b), v_p(c)) = 1\). Since the exponents can only be 0 or 1, this means not all exponents for a given prime can be 0.
• The condition HCF\((a, b, c) = 1\) means that there is no prime factor common to all three numbers. Mathematically, for each prime factor \(p \in \{7, 11, 13\}\), the minimum exponent among a, b, and c must be 0. Mathematically, \(\min(v_p(a), v_p(b), v_p(c)) = 0\). Since the exponents can only be 0 or 1, this means not all exponents for a given prime can be 1.

Combining these conditions for each prime \(p\), the exponents \((v_p(a), v_p(b), v_p(c))\) must be a combination of 0s and 1s such that the minimum is 0 and the maximum is 1. The possible combinations of exponents \((e_1, e_2, e_3)\) from \(\{0, 1\}\) for a prime across the three numbers (a, b, c) that satisfy \(\min(e_1, e_2, e_3) = 0\) and \(\max(e_1, e_2, e_3) = 1\) are the permutations of (1,0,0) or (1,1,0):

• Permutations of (1,0,0): (1,0,0), (0,1,0), (0,0,1) - 3 possibilities.
• Permutations of (1,1,0): (1,1,0), (1,0,1), (0,1,1) - 3 possibilities.

Thus, for each prime (7, 11, and 13), there are 6 possible ways to distribute the exponents (0 or 1) among the three numbers (a, b, c) to satisfy the HCF and LCM conditions for that specific prime.

Counting Ordered Triplets (a, b, c)

Since the choice of exponents for each prime is independent, the total number of ordered triplets \((a, b, c)\) satisfying the HCF and LCM conditions is the product of the number of choices for each prime:

Total ordered triplets = (Choices for prime 7) × (Choices for prime 11) × (Choices for prime 13) = \(6 \times 6 \times 6 = 216\).

Each of these 216 ordered triplets \((a, b, c)\) consists of positive integers whose HCF is 1 and LCM is 1001.

Identifying Sets with Distinct Integers

The question asks for the number of sets of numbers, each having only three positive integers. This phrasing implies that the three integers in the set {a, b, c} must be distinct. We need to find how many of the 216 ordered triplets consist of distinct integers and then group these triplets into sets.

An ordered triplet \((a, b, c)\) has repeated elements if \(a=b\), or \(b=c\), or \(c=a\). There is no case where \(a=b=c\), because if \(a=b=c\), then \(v_p(a)=v_p(b)=v_p(c)\) for all primes \(p\). This would mean the exponent triplet \((v_p(a), v_p(b), v_p(c))\) is either (0,0,0) (violating max=1) or (1,1,1) (violating min=0).

Let's count the number of ordered triplets where there are exactly two equal numbers.

• Case 1: \(a = b\) but \(a \neq c\). For \(a=b\), the exponents must be equal for all primes: \(v_p(a) = v_p(b)\). The valid exponent triplet \((v_p(a), v_p(b), v_p(c))\) must satisfy \(v_p(a) = v_p(b)\), \(\min=0\), \(\max=1\).
  • If \(v_p(a) = v_p(b) = 0\), then \(v_p(c)\) must be 1 for \(\max=1\). This gives the triplet (0,0,1). \(\min(0,0,1)=0\), \(\max(0,0,1)=1\). Valid.
  • If \(v_p(a) = v_p(b) = 1\), then \(v_p(c)\) must be 0 for \(\min=0\). This gives the triplet (1,1,0). \(\min(1,1,0)=0\), \(\max(1,1,0)=1\). Valid.
  For each prime, the exponent triplet \((v_p(a), v_p(b), v_p(c))\) must be either (0,0,1) or (1,1,0). There are 2 choices for each of the 3 primes. So there are \(2^3 = 8\) ordered triplets where \(a=b\). We confirmed \(a \neq c\) for these 8 cases because if a=c, then a=b=c, which is impossible.
• Case 2: \(b = c\) but \(b \neq a\). Similar to Case 1, there are 8 ordered triplets where \(b=c\).
• Case 3: \(c = a\) but \(c \neq b\). Similar to Case 1, there are 8 ordered triplets where \(c=a\).

The total number of ordered triplets with repeated elements is \(8 + 8 + 8 = 24\). These 24 triplets correspond to sets with exactly two distinct elements (e.g., {1, 1, 1001}).

The number of ordered triplets with distinct elements is the total number of triplets minus those with repetitions: \(216 - 24 = 192\).

Each set of three distinct integers {a, b, c} corresponds to \(3! = 6\) different ordered triplets (permutations of a, b, and c).

Therefore, the number of distinct sets {a, b, c} is the number of distinct ordered triplets divided by 6: \(192 / 6 = 32\).

Conclusion: Determining the Value of n

There are 32 sets of three distinct positive integers satisfying the given conditions.

The value of n is 32.

Comparing this value to the options:

• Option 1: 6
• Option 2: 7
• Option 3: 8
• Option 4: More than 8

Since 32 is greater than 8, the correct option is "More than 8".

Revision Table: LCM, HCF, and Sets

Reviewing the key concepts used in this problem:

• Prime Factorization: Breaking down a number into its prime components is fundamental for problems involving HCF and LCM.
• HCF and LCM properties: For any positive integers x, y, z, and a prime p, \(v_p(\text{HCF}(x,y,z)) = \min(v_p(x), v_p(y), v_p(z))\) and \(v_p(\text{LCM}(x,y,z)) = \max(v_p(x), v_p(y), v_p(z))\).
• Counting principle: If independent choices are made, the total number of outcomes is the product of the number of choices for each step.
• Sets vs. Ordered Triplets: An ordered triplet \((a, b, c)\) is distinct from \((b, a, c)\) unless \(a=b\). A set {a, b, c} considers the elements regardless of order; {a, b, c} = {b, a, c}. The phrase "sets of numbers each having only three positive integers" implies distinct elements in the set.

Additional Information: Generalizing the Problem

This problem can be generalized. Consider finding the number of sets of k distinct positive integers \(\{x_1, x_2, \dots, x_k\}\) such that HCF\((x_1, \dots, x_k) = h\) and LCM\((x_1, \dots, x_k) = m\). For such sets to exist, h must divide m. Let \(m/h = M\). Let the prime factorization of M be \(p_1^{e_1} p_2^{e_2} \dots p_r^{e_r}\).

The numbers \(x_j/h\) must be divisors of M, such that their HCF is 1 and their LCM is M. Let \(y_j = x_j/h\). We need sets \(\{y_1, \dots, y_k\}\) of distinct positive integers such that HCF\((y_1, \dots, y_k) = 1\) and LCM\((y_1, \dots, y_k) = M\).

For each prime factor \(p_i\) of M, the exponents \((v_{p_i}(y_1), \dots, v_{p_i}(y_k))\) must satisfy \(\min(v_{p_i}(y_j)) = 0\) and \(\max(v_{p_i}(y_j)) = e_i\). If \(e_i=1\) (as in our problem), then for each prime \(p_i\), the exponents \((v_{p_i}(y_1), \dots, v_{p_i}(y_k))\) must be a k-tuple of 0s and 1s containing at least one 0 and at least one 1.

For a single prime with exponent 1 in M, there are \(2^k\) total combinations of exponents (0 or 1) for k numbers. The combinations (0,0,...,0) and (1,1,...,1) are excluded. So there are \(2^k - 2\) valid ordered k-tuples of exponents for each such prime. If M has r distinct prime factors (all with exponent 1), there are \((2^k - 2)^r\) ordered k-tuples \((y_1, \dots, y_k)\).

In our problem, k=3, h=1, m=1001=7¹ 11¹ 13¹. So M=1001, r=3, \(e_i=1\).

Number of ordered triplets \((y_1, y_2, y_3)\) (which are the same as (a,b,c) since h=1) is \((2^3 - 2)^3 = (8 - 2)^3 = 6^3 = 216\). This matches our calculation.

The general problem of counting distinct sets \(\{y_1, \dots, y_k\}\) from these ordered k-tuples is more complex and involves combinatorial arguments related to partitions and inclusion-exclusion, especially when \(e_i > 1\) or when \(y_j\) don't have to be distinct (multisets). For distinct integers and \(e_i=1\), the approach of counting distinct ordered k-tuples and dividing by k! (if all k are distinct) or considering cases with repetitions is applicable.