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If (N=2^5\times3^4\times5^3), how many factors (d) are there such that (d^3) divides (N)?
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Answer and explanation
Correct answer: (8)
Step 1: Let (d=2^a\times3^b\times5^c), so (d^3=2^{3a}\times3^{3b}\times5^{3c}). Step 2: (3a\le5), (3b\le4), and (3c\le3), giving (2) choices each. Total (=8). Step 3: For cube divisibility, triple the exponents and compare.
Frequently asked questions
What is the correct answer to this question?
(8)
Why is this the correct answer?
Step 1: Let (d=2^a\times3^b\times5^c), so (d^3=2^{3a}\times3^{3b}\times5^{3c}). Step 2: (3a\le5), (3b\le4), and (3c\le3), giving (2) choices each. Total (=8). Step 3: For cube divisibility, triple the exponents and compare.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Real Numbers. Topic: Prime Factorisation.
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