Find the HCF (GCD) of the following three numbers: 1) 4 × 27 × 3125 2) 8 × 9 × 25 × 7 3) 16 × 81 × 5 × 11 × 49 What is the HCF of these three values?

Introduction / Context:
This question involves computing the HCF of three composite numbers that are each given as products. The most systematic approach is to factor each component into prime factors, combine them, and then identify the common primes with the minimum exponents. This is exactly how the HCF is defined in terms of prime powers.

Given Data / Assumptions:

• A = 4 × 27 × 3125
• B = 8 × 9 × 25 × 7
• C = 16 × 81 × 5 × 11 × 49
• We need HCF(A, B, C)

Concept / Approach:
Write each of A, B, and C in prime factor form. Then the HCF is the product of all primes that appear in every number, each taken to the smallest exponent with which it appears in any of the three factorizations.

Step-by-Step Solution:
Step 1: Factorize A.4 = 2^2, 27 = 3^3, 3125 = 5^5.So A = 2^2 * 3^3 * 5^5.Step 2: Factorize B.8 = 2^3, 9 = 3^2, 25 = 5^2, 7 = 7^1.So B = 2^3 * 3^2 * 5^2 * 7.Step 3: Factorize C.16 = 2^4, 81 = 3^4, 5 = 5^1, 11 = 11^1, 49 = 7^2.So C = 2^4 * 3^4 * 5^1 * 11 * 7^2.Step 4: Identify primes common to A, B, and C: these are 2, 3, and 5.Prime 2: minimum exponent is min(2, 3, 4) = 2.Prime 3: minimum exponent is min(3, 2, 4) = 2.Prime 5: minimum exponent is min(5, 2, 1) = 1.Step 5: HCF = 2^2 * 3^2 * 5^1 = 4 * 9 * 5 = 180.

Verification / Alternative check:
Since 180 = 2^2 * 3^2 * 5, and each of A, B, and C contains at least 2^2, 3^2, and 5^1, 180 divides all three numbers. Any attempt to increase the exponent of 2, 3, or 5 in the HCF would break divisibility for at least one of the numbers, so 180 is indeed the greatest common divisor.

Why Other Options Are Wrong:
90: Contains only 2^1, but A has 2^2 as the minimum common exponent, so 90 is not the HCF.120: Has 2^3, which exceeds the exponent of 2 in A, so it does not divide A.360: Has higher exponents and does not divide B or A exactly.60: Too small and does not reflect the full common prime power structure.

Common Pitfalls:
Including primes like 7 or 11 in the HCF even though they are not common to all three numbers.Using maximum exponents instead of minimum ones, which gives the LCM instead of the HCF.Making mistakes when factoring large components such as 3125 or 81.

Final Answer:
180