Find the HCF of 2^3 * 3^2 * 5 and 2^2 * 3^4 * 7. | Mathematics MCQ | Examoogle

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Find the HCF of 2³ * 3² * 5 and 2² * 3⁴ * 7.

$2^2 \times 3^2$

$2^3 \times 3^2$

$2^2 \times 3^4$

$2^3 \times 3^4$

Correct Answer

📐 MA • Math Prime FactorizationMathematicsMathematics

Option A

$2^2 \times 3^2$

Quick Summary:

To find the HCF, simply select the common prime bases and pick the smallest exponent associated with each base. Ignore any factors that do not appear in both numbers.

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📐MAMath SolutionPrime Factorization

📋 Given

Two numbers are expressed as products of their prime factors: $N_1 = 2^3 imes 3^2 imes 5$ and $N_2 = 2^2 imes 3^4 imes 7$ .

🔢 Formula Used

HCF(a,b)=∏pimin⁡(ai,bi)\text{HCF}(a, b) = \prod p_i^{\min(a_i, b_i)}HCF(a,b)=∏pimin(ai​,bi​)​

⚡ Exam Hall Shortcut / Speed Trick

To find the HCF, simply select the common prime bases and pick the smallest exponent associated with each base. Ignore any factors that do not appear in both numbers.

⚠️ Common Student Trap / Pitfall

Students often pick the highest powers instead of the lowest powers, which would calculate the LCM instead of the HCF.

📊 Diagram / Illustration

🔢 Step-by-Step Solution

Identify common prime bases

Compare the factors of $N_1 = 2^3 imes 3^2 imes 5$ and $N_2 = 2^2 imes 3^4 imes 7$ . The common prime bases present in both numbers are $2$ and $3$ .

2,32, 32,3

Select the minimum exponent for each base

For base $2$ , the powers are $3$ and $2$ , so the minimum is $2$ . For base $3$ , the powers are $2$ and $4$ , so the minimum is $2$ .

min⁡(3,2)=2,min⁡(2,4)=2\min(3, 2) = 2, \min(2, 4) = 2min(3,2)=2,min(2,4)=2

Calculate the HCF

Multiply the prime bases raised to their respective minimum exponents to find the result.

22imes322^2 imes 3^222imes32

✅

A is correct because the product of the lowest powers of common prime factors $2$ and $3$ is $2^2 imes 3^2$ .

Core Concepts Used

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Prime Factorization Highest Common Factor Exponent Laws

💡 EXAM TIP

This concept of taking the minimum exponent for HCF and maximum exponent for LCM is foundational for solving problems involving divisibility and bells/light-flashing intervals in time-based aptitude questions.

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