IAS 2025 — CSAT Maths

Correct Option:

D) More than 8

Explanation:

The problem requires finding the number of distinct sets of three positive integers {a, b, c} such that their Least Common Multiple (LCM) is 1001 and their Highest Common Factor (HCF) is 1.

First, determine the prime factorization of 1001:

• 1001 = 7 × 11 × 13.

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. Each number can be expressed in the form:

• a = 7^x_a × 11^y_a × 13^z_a
• b = 7^x_b × 11^y_b × 13^z_b
• c = 7^x_c × 11^y_c × 13^z_c

Where the exponents x_i, y_i, z_i are either 0 or 1.

The conditions for LCM and HCF impose constraints on these exponents for each prime factor (p ∈ {7, 11, 13}):

• For LCM(a, b, c) = 1001: For each prime p, the maximum exponent among (x_a, x_b, x_c) must be 1. This implies that for each prime, at least one of the numbers a, b, or c must have that prime as a factor (exponent 1). Thus, (0,0,0) is not allowed for any prime's exponents.
• For HCF(a, b, c) = 1: For each prime p, the minimum exponent among (x_a, x_b, x_c) must be 0. This implies that for each prime, not all three numbers a, b, and c can have that prime as a factor (exponent 1). Thus, (1,1,1) is not allowed for any prime's exponents.

Considering the possible exponent triplets (x_a, x_b, x_c) for any single prime p, where each x_i ∈ {0, 1}:

• Triplets satisfying max(x_a, x_b, x_c) = 1 and min(x_a, x_b, x_c) = 0 are:
  • 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.
• Total valid exponent distributions for each prime = 3 + 3 = 6.

Since there are three distinct primes (7, 11, 13), and the exponent choices for each prime are independent, the total number of ordered triplets (a, b, c) satisfying the conditions is:

• Total ordered triplets = (choices for prime 7) × (choices for prime 11) × (choices for prime 13) = 6 × 6 × 6 = 216.

These 216 triplets include cases where elements are repeated. We need to find the number of distinct sets {a, b, c}.

Consider cases of repeated elements:

1. All three elements are equal (a = b = c):

   • If a = b = c, then HCF(a,a,a) = a and LCM(a,a,a) = a. This would imply a = 1 and a = 1001, which is a contradiction. Therefore, no triplets exist where a = b = c.

2. Exactly two elements are equal (e.g., a = b, but a ≠ c):

   • For a = b, the exponents for each prime must be equal for 'a' and 'b' (x_a = x_b).
   • The conditions become min(x_a, x_c) = 0 and max(x_a, x_c) = 1.
   • This means for each prime, the exponent triplet (x_a, x_b, x_c) must be either (0,0,1) or (1,1,0). There are 2 such choices for each prime.
   • Number of ordered triplets where a = b and a ≠ c = 2 × 2 × 2 = 8.
   • Similarly, for b = c and b ≠ a, there are 8 ordered triplets.
   • Similarly, for c = a and c ≠ b, there are 8 ordered triplets.
   • Total ordered triplets with exactly two equal elements = 8 + 8 + 8 = 24.

Now, calculate the number of ordered triplets (a, b, c) where all three elements are distinct:

• Number of distinct ordered triplets = Total ordered triplets - (Triplets with exactly two equal elements) - (Triplets with all three equal elements)
• Number of distinct ordered triplets = 216 - 24 - 0 = 192.

Since the question asks for the number of distinct sets {a, b, c}, and each set of three distinct integers {a, b, c} can be arranged in 3! = 6 different ordered triplets, we divide by 6:

• Number of distinct sets n = (Number of distinct ordered triplets) / 3!
• n = 192 / 6 = 32.

The value of n is 32, which is "More than 8".

Incorrect Options:

Options 1, 2, and 3 (6, 7, 8) are incorrect because the calculated number of such sets is 32. These options represent values significantly lower than the actual count, indicating a misunderstanding of the combinatorial principles involved in distributing prime factors while satisfying both HCF and LCM conditions for three distinct numbers.