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Question Hard Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

किस संख्या का अभाज्य गुणनखंडन \(2^5\times3^3\times7\) है?

Which number has prime factorisation \(2^5\times3^3\times7\)?

Explanation opens after your attempt
Correct Answer

A. 6048

Step 1

Concept

Calculate \(2^5=32\) and \(3^3=27\).

Step 2

Why this answer is correct

\(32\times27\times7=6048\).

Step 3

Exam Tip

Find the values of higher powers separately. चरण 1: \(2^5=32\) और \(3^3=27\) निकालें। चरण 2: \(32\times27\times7=6048\)। चरण 3: बड़ी घातों का मान अलग से निकालें।

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Question Hard Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

संख्या 6048 का अभाज्य गुणनखंडन कौन सा है?

Which is the prime factorisation of 6048?

Explanation opens after your attempt
Correct Answer

A. \(2^5\times3^3\times7\)

Step 1

Concept

Write \(6048=32\times189\).

Step 2

Why this answer is correct

\(32=2^5\) and \(189=3^3\times7\), so \(6048=2^5\times3^3\times7\).

Step 3

Exam Tip

Change 189 into \(3^3\times7\). चरण 1: \(6048=32\times189\) लिखें। चरण 2: \(32=2^5\) और \(189=3^3\times7\), इसलिए \(6048=2^5\times3^3\times7\)। चरण 3: 189 को \(3^3\times7\) में बदलें।

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