The H.C.F. of 3240, 3600, and a third number is 36.\nThe L.C.M. of these three numbers is 2^4 × 3^5 × 5^2 × 7^2.\nWhat is the value of the third number?

Introduction:
This is a higher-level H.C.F. and L.C.M. question involving three numbers. It requires careful use of prime factorisation and understanding how H.C.F. and L.C.M. relate to the prime exponents of all the numbers involved. Such questions are common in advanced aptitude and number theory practice.

Given Data / Assumptions:

• Numbers: 3240, 3600, and an unknown third number N.
• H.C.F.(3240, 3600, N) = 36.
• L.C.M.(3240, 3600, N) = 2^4 × 3^5 × 5^2 × 7^2.
• We need to determine the value of N.

Concept / Approach:
We use prime factorisation of the known numbers:

• For H.C.F. across all numbers, take the minimum exponent of each prime among the three numbers.
• For L.C.M., take the maximum exponent of each prime among the three numbers.

Using the known H.C.F. and L.C.M., we deduce the prime exponents of the third number N so that both conditions are satisfied.

Step-by-Step Solution:
Step 1: Prime factorise 3240 and 3600. 3240 = 2^3 * 3^4 * 5^1. 3600 = 2^4 * 3^2 * 5^2. Step 2: HCF(3240, 3600) = 2^3 * 3^2 * 5^1 = 360. Step 3: HCF(3240, 3600, N) = 36 = 2^2 * 3^2. → So N must have at least 2^2 and 3^2 and must not have factor 5 (otherwise HCF would include 5). Step 4: LCM of all three is given as 2^4 * 3^5 * 5^2 * 7^2. From 3240 and 3600, maximum exponents so far: 2^4, 3^4, 5^2. Step 5: To reach the LCM exponents, N must contribute exponents: For 3: need 3^5, so N has 3^5. For 7: need 7^2, so N has 7^2. For 2: max exponent given is already 4 from 3600, so N can have 2^a with a ≤ 4 but at least 2 (for HCF). Step 6: To keep LCM exponent of 2 at 4, we choose smallest valid a = 2. Step 7: Therefore, N = 2^2 * 3^5 * 7^2. Step 8: Compute N: 2^2 = 4, 3^5 = 243, 7^2 = 49. Step 9: First 4 * 243 = 972. Step 10: Then 972 * 49 = 47628.

Verification / Alternative check:
Check H.C.F. condition: gcd(3240, 3600, 47628) is 36, and the L.C.M. computed from these three numbers matches 2^4 * 3^5 * 5^2 * 7^2. This confirms that N = 47628 is consistent with the given information.

Why Other Options Are Wrong:
49874, 24157, 42146, 95256: These numbers do not have the exact prime factor structure required to produce both the specified H.C.F. and L.C.M. with 3240 and 3600. In particular, they either introduce unwanted factors, omit needed powers of 3 or 7, or disturb the H.C.F. from 36.

Common Pitfalls:
Many learners confuse how to use exponents for H.C.F. and L.C.M., sometimes taking maximum exponents for H.C.F. or minimum exponents for L.C.M., which is reversed. Others may not fully factorise the given numbers, leading to mistakes in deducing the third number. Careful factorisation and systematic reasoning are crucial here.

Final Answer:
The value of the third number is 47628.