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If (N) is the smallest number divisible by both (2^5\times3^2\times7) and (2^3\times3^4\times5), and (M) is the HCF of these two numbers, what is (\frac{N}{M})?
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Answer and explanation
Correct answer: (2^2\times3^2\times5\times7)
Step 1: The smallest divisible number (N) is the LCM, and (M) is the HCF. Step 2: (N=2^5\times3^4\times5\times7) and (M=2^3\times3^2), so (\frac{N}{M}=2^2\times3^2\times5\times7). Step 3: While dividing, subtract powers of the same base.
Frequently asked questions
What is the correct answer to this question?
(2^2\times3^2\times5\times7)
Why is this the correct answer?
Step 1: The smallest divisible number (N) is the LCM, and (M) is the HCF. Step 2: (N=2^5\times3^4\times5\times7) and (M=2^3\times3^2), so (\frac{N}{M}=2^2\times3^2\times5\times7). Step 3: While dividing, subtract powers of the same base.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Real Numbers. Topic: HCF and LCM using prime factorisation.
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