किस संख्या का अभाज्य गुणनखंडन \(2^3\times3^2\times5^2\) है?

Which number has prime factorisation \(2^3\times3^2\times5^2\)?

Explanation opens after your attempt
Correct Answer

A. 1800

Step 1

Concept

Calculate \(2^3=8\), \(3^2=9\), and \(5^2=25\).

Step 2

Why this answer is correct

\(8\times9\times25=1800\).

Step 3

Exam Tip

Simplify powers first, then multiply. चरण 1: \(2^3=8\), \(3^2=9\) और \(5^2=25\) निकालें। चरण 2: \(8\times9\times25=1800\)। चरण 3: घातों को पहले सरल करें, फिर गुणा करें।

Question me issue ya doubt hai?

Answer, explanation, typing mistake ya suggestion directly hamari team ko bhejein. 📱Helpline (Call / WhatsApp): +91 7272824365

Related Mathematics Questions

FAQs

Mathematics Answer, Explanation and Revision Hints

किस संख्या का अभाज्य गुणनखंडन \(2^3\times3^2\times5^2\) है? / Which number has prime factorisation \(2^3\times3^2\times5^2\)?

Correct Answer: A. 1800. Explanation: चरण 1: \(2^3=8\), \(3^2=9\) और \(5^2=25\) निकालें। चरण 2: \(8\times9\times25=1800\)। चरण 3: घातों को पहले सरल करें, फिर गुणा करें। / Step 1: Calculate \(2^3=8\), \(3^2=9\), and \(5^2=25\). Step 2: \(8\times9\times25=1800\). Step 3: Simplify powers first, then multiply.

Which concept should I revise for this Mathematics MCQ?

Calculate \(2^3=8\), \(3^2=9\), and \(5^2=25\).

What exam hint can help solve this Mathematics question?

Simplify powers first, then multiply. चरण 1: \(2^3=8\), \(3^2=9\) और \(5^2=25\) निकालें। चरण 2: \(8\times9\times25=1800\)। चरण 3: घातों को पहले सरल करें, फिर गुणा करें।