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2 results found for "number-139392" in Class 10.

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

किस संख्या का अभाज्य गुणनखंडन \(2^7\times3^2\times11^2\) है?

Which number has prime factorisation \(2^7\times3^2\times11^2\)?

Explanation opens after your attempt
Correct Answer

A. 139392

Step 1

Concept

Calculate \(2^7=128\), \(3^2=9\), and \(11^2=121\).

Step 2

Why this answer is correct

\(128\times9\times121=139392\).

Step 3

Exam Tip

Find the values of powers first, then multiply. चरण 1: \(2^7=128\), \(3^2=9\) और \(11^2=121\) निकालें। चरण 2: \(128\times9\times121=139392\)। चरण 3: पहले घातों का मान निकालें, फिर गुणा करें।

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

संख्या 139392 का अभाज्य गुणनखंडन कौन सा है?

Which is the prime factorisation of 139392?

Explanation opens after your attempt
Correct Answer

A. \(2^7\times3^2\times11^2\)

Step 1

Concept

Write \(139392=128\times1089\).

Step 2

Why this answer is correct

\(128=2^7\) and \(1089=3^2\times11^2\), so \(139392=2^7\times3^2\times11^2\).

Step 3

Exam Tip

Convert 1089 into prime powers. चरण 1: \(139392=128\times1089\) लिखें। चरण 2: \(128=2^7\) और \(1089=3^2\times11^2\), इसलिए \(139392=2^7\times3^2\times11^2\)। चरण 3: 1089 को अभाज्य घातों में बदलें।

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