Find the H.C.F. of $2^3 \times 3^2 \times 5 \times 7^4$, $2^2 \times 3^5 \times 5^2 \times 7^3$, $2^3 \times 5^3 \times 7^2$.

### Concept & Formula The Highest Common Factor (H.C.F.) of numbers expressed in prime factorization form is the product of the **lowest powers** of all **common** prime factors. If a prime number does not appear in every factorization, it is not common and cannot be included in the H.C.F. ### Step-by-Step Solution **Step 1: Identify the prime factors in each given number.** * First number: Base primes are 2, 3, 5, 7. * Second number: Base primes are 2, 3, 5, 7. * Third number: Base primes are 2, 5, 7. (Notice 3 is missing here). **Step 2: Find the common prime factors.** * The prime numbers common to all three given expressions are 2, 5, and 7. **Step 3: Determine the lowest power of each common prime factor.** * For prime 2: The powers are $2^3$, $2^2$, and $2^3$. The lowest is $2^2$. * For prime 5: The powers are $5^1$, $5^2$, and $5^3$. The lowest is $5^1$. * For prime 7: The powers are $7^4$, $7^3$, and $7^2$. The lowest is $7^2$. **Step 4: Multiply these lowest powers to find the H.C.F.** H.C.F. = $2^2 \times 5 \times 7^2$ H.C.F. = 4 $\times$ 5 $\times$ 49 H.C.F. = 20 $\times$ 49 = 980. ### Exam Strategy & Shortcut In competitive exams, immediately scan for missing prime bases across the numbers. Here, the third term lacks the base 3. You can instantly eliminate any multiple-choice option that is divisible by 3 (by checking if the sum of its digits is a multiple of 3). 980 (9+8+0 = 17) is not divisible by 3, which aligns with our logic. ### Common Pitfall A standard mistake is confusing H.C.F. with L.C.M. when looking at exponents. Students often take the highest powers ($2^3$, $3^5$, etc.) instead of the lowest, or they mistakenly include non-common factors (like $3^2$). Remember: H.C.F. requires the **minimum** intersection. ### Final Answer **Therefore, the correct answer is 980.**