Find the H.C.F. of the three numbers 4 × 27 × 3125, 8 × 9 × 25 × 7 and 16 × 81 × 5 × 11 × 49.

Introduction / Context:
This question assesses your skill in working with composite numbers given in factored form and finding their highest common factor (H.C.F.). It is a typical number theory problem where prime factorization plays a major role, and it helps reinforce the method of extracting common prime powers from several complex products.

Given Data / Assumptions:

• First number: 4 × 27 × 3125.
• Second number: 8 × 9 × 25 × 7.
• Third number: 16 × 81 × 5 × 11 × 49.
• We must find the H.C.F. of these three numbers.

Concept / Approach:
The H.C.F. of a set of numbers is obtained by taking each prime factor that is common to all the numbers with the smallest exponent appearing in their prime factorizations. Since the numbers are given as products, our first step is to fully factor each component into primes. After that, we identify the primes common to all three numbers and multiply them with the lowest powers to obtain the H.C.F.

Step-by-Step Solution:
Step 1: Prime factorize each component.4 = 2^2, 27 = 3^3, 3125 = 5^5.Thus first number = 2^2 * 3^3 * 5^5.8 = 2^3, 9 = 3^2, 25 = 5^2, 7 = 7^1.Thus second number = 2^3 * 3^2 * 5^2 * 7.16 = 2^4, 81 = 3^4, 5 = 5^1, 11 = 11^1, 49 = 7^2.Thus third number = 2^4 * 3^4 * 5 * 7^2 * 11.Step 2: Identify primes common to all three numbers: 2, 3, and 5.Step 3: For prime 2, the powers are 2^2, 2^3, and 2^4, so smallest power is 2^2.Step 4: For prime 3, the powers are 3^3, 3^2, and 3^4, so smallest power is 3^2.Step 5: For prime 5, the powers are 5^5, 5^2, and 5^1, so smallest power is 5^1.Step 6: The H.C.F. is therefore 2^2 * 3^2 * 5.Step 7: Compute: 2^2 = 4, 3^2 = 9, so H.C.F. = 4 * 9 * 5 = 36 * 5 = 180.

Verification / Alternative check:
Check that 180 divides each number: 180 = 2^2 * 3^2 * 5.First number has 2^2, 3^3, 5^5 so it contains at least these prime powers.Second number has 2^3, 3^2, 5^2 so it also contains them.Third number has 2^4, 3^4, 5^1 so it contains them as well. No higher power of any prime is common to all three, so 180 is indeed the highest common factor.

Why Other Options Are Wrong:
Options b (360) and c (540) include higher powers of some primes, making them larger than the true H.C.F. and no longer dividing at least one of the numbers exactly. Option d (1260) includes extra prime factors (such as 7) with unnecessary powers and will not divide all three numbers without remainder. Only 180 satisfies the divisibility condition for all three and is maximal.

Common Pitfalls:
Students may overlook that only common primes with their smallest exponents contribute to the H.C.F. It is also easy to make arithmetic errors when expanding the products or to mix up H.C.F. with L.C.M. Staying organised with prime factorization and clearly tracking exponents prevents these mistakes.

Final Answer:
The H.C.F. of the three given numbers is 180.