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Find the HCF of three composite products: Compute the HCF of (2 × 3 × 7 × 9), (2 × 3 × 9 × 11), and (2 × 3 × 4 × 5).

Difficulty: Easy

Correct Answer: 2 × 3

Explanation:


Introduction / Context:
The Highest Common Factor of several numbers is obtained by taking the minimum exponent of each prime that appears in all numbers. Breaking each product into primes makes the comparison straightforward.


Given Data / Assumptions:

  • A = 2 × 3 × 7 × 9
  • B = 2 × 3 × 9 × 11
  • C = 2 × 3 × 4 × 5


Concept / Approach:
Prime factorize each: 9 = 3^2 and 4 = 2^2. Then take the intersection of primes with their smallest exponents across A, B, and C. Any prime not present in all three drops out of the HCF.


Step-by-Step Solution:

A = 2^1 * 3^3 * 7^1.B = 2^1 * 3^3 * 11^1.C = 2^3 * 3^1 * 5^1.Common primes across all: 2 and 3.Minimum exponents: 2^min(1,1,3) = 2^1; 3^min(3,3,1) = 3^1.Therefore HCF = 2 * 3 = 6 (shown as 2 × 3).


Verification / Alternative check:

Divide each number by 6 to confirm no larger common factor exists: quotients are integers, and any attempt to include another 2 or 3 or include 5/7/11 fails for one of the numbers.


Why Other Options Are Wrong:

  • 2 × 3 × 7 or 2 × 3 × 9: Include primes or powers not common to all three numbers.
  • 2 × 7 × 9 × 11: Contains primes not present across all three simultaneously.


Common Pitfalls:

  • Using maximum exponents (LCM logic) instead of minimum (HCF logic).


Final Answer:

2 × 3
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