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prime-factorisation MCQ Questions for Class 10

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663 questions tagged with prime-factorisation.

Question 1/663 Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

अभाज्य गुणनखंडन में \(64\times3969\) को अंतिम उत्तर क्यों नहीं माना जाएगा?

Why will \(64\times3969\) not be considered the final answer in prime factorisation?

Explanation opens after your attempt
Correct Answer

A. क्योंकि 64 और 3969 संयुक्त रूप हैंBecause 64 and 3969 are composite forms

Step 1

Concept

In final prime factorisation, every base should be prime.

Step 2

Why this answer is correct

\(64=2^6\) and \(3969=3^4\times7^2\).

Step 3

Exam Tip

Therefore, the final form is \(2^6\times3^4\times7^2\). चरण 1: अंतिम अभाज्य गुणनखंडन में हर आधार अभाज्य होना चाहिए। चरण 2: \(64=2^6\) और \(3969=3^4\times7^2\) है। चरण 3: इसलिए अंतिम रूप \(2^6\times3^4\times7^2\) होगा।

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Question 2/663 Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

संख्या 2138400 का सही अभाज्य गुणनखंडन क्या है?

What is the correct prime factorisation of 2138400?

Explanation opens after your attempt
Correct Answer

A. \(2^5\times3^5\times5^2\times11\)

Step 1

Concept

Write \(2138400=32\times66825\).

Step 2

Why this answer is correct

\(66825=3^5\times5^2\times11\), so \(2138400=2^5\times3^5\times5^2\times11\).

Step 3

Exam Tip

Give 66825 its complete prime form. चरण 1: \(2138400=32\times66825\) लिखें। चरण 2: \(66825=3^5\times5^2\times11\), इसलिए \(2138400=2^5\times3^5\times5^2\times11\)। चरण 3: 66825 को पूरा अभाज्य रूप दें।

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Question 3/663 Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

संख्या 254016 का अभाज्य गुणनखंडन कौन सा है?

Which is the prime factorisation of 254016?

Explanation opens after your attempt
Correct Answer

A. \(2^6\times3^4\times7^2\)

Step 1

Concept

Write \(254016=64\times3969\).

Step 2

Why this answer is correct

\(64=2^6\) and \(3969=3^4\times7^2\), so \(254016=2^6\times3^4\times7^2\).

Step 3

Exam Tip

Convert 3969 into prime powers. चरण 1: \(254016=64\times3969\) लिखें। चरण 2: \(64=2^6\) और \(3969=3^4\times7^2\), इसलिए \(254016=2^6\times3^4\times7^2\)। चरण 3: 3969 को अभाज्य घातों में बदलें।

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Question 4/663 Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

संख्या 582120 का सही अभाज्य गुणनखंडन क्या है?

What is the correct prime factorisation of 582120?

Explanation opens after your attempt
Correct Answer

A. \(2^4\times3^3\times5^2\times7^2\times11\)

Step 1

Concept

Write the number using prime-base groups.

Step 2

Why this answer is correct

\(16\times27\times25\times49\times11=2^4\times3^3\times5^2\times7^2\times11\).

Step 3

Exam Tip

Avoid decimal-based options and write prime bases. चरण 1: \(582120=52920\times11\) लिखें। चरण 2: \(52920=2^3\times3^3\times5\times7^2\) नहीं बल्कि यहां \(16\times27\times25\times49\times11\) से \(2^4\times3^3\times5^2\times7^2\times11\) मिलता है। चरण 3: दशमलव वाले विकल्प से बचें और अभाज्य आधार लिखें।

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Question 5/663 Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

किस विकल्प में अभाज्य गुणनखंडन अधूरा है?

Which option has incomplete prime factorisation?

Explanation opens after your attempt
Correct Answer

A. \(2^6\times81\times121\)

Step 1

Concept

In an incomplete form, composite bases remain.

Step 2

Why this answer is correct

81 and 121 are composite bases.

Step 3

Exam Tip

\(2^6\times81\times121\) must be changed into \(2^6\times3^4\times11^2\). चरण 1: अधूरे रूप में संयुक्त आधार बच जाता है। चरण 2: 81 और 121 संयुक्त आधार हैं। चरण 3: \(2^6\times81\times121\) को \(2^6\times3^4\times11^2\) में बदलना होगा।

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Question 6/663 Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

किस विकल्प में केवल अंतिम अभाज्य गुणनखंडन दिया गया है?

Which option gives only the final prime factorisation?

Explanation opens after your attempt
Correct Answer

A. \(2^5\times3^4\times5^2\times7\)

Step 1

Concept

In the final form, bases must be prime.

Step 2

Why this answer is correct

In the first option, bases 2, 3, 5, and 7 are prime.

Step 3

Exam Tip

32, 81, 25, 405, 35, and 160 are composite, so they are not final forms. चरण 1: अंतिम रूप में आधार अभाज्य होने चाहिए। चरण 2: पहले विकल्प में आधार 2, 3, 5 और 7 अभाज्य हैं। चरण 3: 32, 81, 25, 405, 35 और 160 संयुक्त हैं, इसलिए वे अंतिम रूप नहीं हैं।

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Question 7/663 Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

संख्या 518400 का अभाज्य गुणनखंडन कौन सा है?

Which is the prime factorisation of 518400?

Explanation opens after your attempt
Correct Answer

A. \(2^8\times3^4\times5^2\)

Step 1

Concept

Write \(518400=256\times2025\).

Step 2

Why this answer is correct

\(256=2^8\) and \(2025=3^4\times5^2\), so \(518400=2^8\times3^4\times5^2\).

Step 3

Exam Tip

Give 2025 its final prime form. चरण 1: \(518400=256\times2025\) लिखें। चरण 2: \(256=2^8\) और \(2025=3^4\times5^2\), इसलिए \(518400=2^8\times3^4\times5^2\)। चरण 3: 2025 को अंतिम अभाज्य रूप दें।

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Question 8/663 Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

संख्या 99225 का सही अभाज्य गुणनखंडन क्या है?

What is the correct prime factorisation of 99225?

Explanation opens after your attempt
Correct Answer

A. \(3^4\times5^2\times7^2\)

Step 1

Concept

Write \(99225=81\times1225\).

Step 2

Why this answer is correct

\(81=3^4\) and \(1225=5^2\times7^2\), so \(99225=3^4\times5^2\times7^2\).

Step 3

Exam Tip

Convert 1225 into prime powers. चरण 1: \(99225=81\times1225\) लिखें। चरण 2: \(81=3^4\) और \(1225=5^2\times7^2\), इसलिए \(99225=3^4\times5^2\times7^2\)। चरण 3: 1225 को अभाज्य घातों में बदलें।

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Question 9/663 Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

संख्या 475200 का अभाज्य गुणनखंडन कौन सा है?

Which is the prime factorisation of 475200?

Explanation opens after your attempt
Correct Answer

A. \(2^6\times3^3\times5^2\times11\)

Step 1

Concept

Write \(475200=64\times7425\).

Step 2

Why this answer is correct

\(7425=3^3\times5^2\times11\), so \(475200=2^6\times3^3\times5^2\times11\).

Step 3

Exam Tip

Do not leave 7425 in the final form. चरण 1: \(475200=64\times7425\) लिखें। चरण 2: \(7425=3^3\times5^2\times11\), इसलिए \(475200=2^6\times3^3\times5^2\times11\)। चरण 3: 7425 को अंतिम रूप में न छोड़ें।

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Question 10/663 Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

संख्या 117000 का सही अभाज्य गुणनखंडन क्या है?

What is the correct prime factorisation of 117000?

Explanation opens after your attempt
Correct Answer

A. \(2^3\times3^2\times5^3\times13\)

Step 1

Concept

Write \(117000=8\times14625\).

Step 2

Why this answer is correct

\(14625=3^2\times5^3\times13\), so \(117000=2^3\times3^2\times5^3\times13\).

Step 3

Exam Tip

Give 14625 its complete prime form. चरण 1: \(117000=8\times14625\) लिखें। चरण 2: \(14625=3^2\times5^3\times13\), इसलिए \(117000=2^3\times3^2\times5^3\times13\)। चरण 3: 14625 को पूरा अभाज्य रूप दें।

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Question 11/663 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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Question 12/663 Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

संख्या 453600 का सही अभाज्य गुणनखंडन क्या है?

What is the correct prime factorisation of 453600?

Explanation opens after your attempt
Correct Answer

A. \(2^5\times3^4\times5^2\times7\)

Step 1

Concept

Write \(453600=32\times14175\).

Step 2

Why this answer is correct

\(14175=3^4\times5^2\times7\), so \(453600=2^5\times3^4\times5^2\times7\).

Step 3

Exam Tip

Do not leave 14175 in the final form. चरण 1: \(453600=32\times14175\) लिखें। चरण 2: \(14175=3^4\times5^2\times7\), इसलिए \(453600=2^5\times3^4\times5^2\times7\)। चरण 3: 14175 को अंतिम रूप में न छोड़ें।

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Question 13/663 Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

संख्या 127575 का अभाज्य गुणनखंडन कौन सा है?

Which is the prime factorisation of 127575?

Explanation opens after your attempt
Correct Answer

A. \(3^6\times5^2\times7\)

Step 1

Concept

Write \(127575=729\times175\).

Step 2

Why this answer is correct

\(729=3^6\) and \(175=5^2\times7\), so \(127575=3^6\times5^2\times7\).

Step 3

Exam Tip

Give prime form to both 729 and 175. चरण 1: \(127575=729\times175\) लिखें। चरण 2: \(729=3^6\) और \(175=5^2\times7\), इसलिए \(127575=3^6\times5^2\times7\)। चरण 3: 729 और 175 दोनों को अभाज्य रूप दें।

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Question 14/663 Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

संख्या 137200 का सही अभाज्य गुणनखंडन क्या है?

What is the correct prime factorisation of 137200?

Explanation opens after your attempt
Correct Answer

A. \(2^4\times5^2\times7^3\)

Step 1

Concept

Write \(137200=16\times8575\).

Step 2

Why this answer is correct

\(16=2^4\) and \(8575=5^2\times7^3\), so the correct form is \(2^4\times5^2\times7^3\).

Step 3

Exam Tip

Give 8575 its final prime form. चरण 1: \(137200=16\times8575\) लिखें। चरण 2: \(16=2^4\) और \(8575=5^2\times7^3\), इसलिए सही रूप \(2^4\times5^2\times7^3\) है। चरण 3: 8575 को अंतिम अभाज्य रूप दें।

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Question 15/663 Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

संख्या 762048 का अभाज्य गुणनखंडन कौन सा है?

Which is the prime factorisation of 762048?

Explanation opens after your attempt
Correct Answer

A. \(2^6\times3^5\times7^2\)

Step 1

Concept

Write \(762048=64\times11907\).

Step 2

Why this answer is correct

\(64=2^6\) and \(11907=3^5\times7^2\), so \(762048=2^6\times3^5\times7^2\).

Step 3

Exam Tip

Write 11907 as prime powers. चरण 1: \(762048=64\times11907\) लिखें। चरण 2: \(64=2^6\) और \(11907=3^5\times7^2\), इसलिए \(762048=2^6\times3^5\times7^2\)। चरण 3: 11907 को अभाज्य घातों में लिखें।

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Question 16/663 Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

संख्या 415800 का सही अभाज्य गुणनखंडन क्या है?

What is the correct prime factorisation of 415800?

Explanation opens after your attempt
Correct Answer

A. \(2^3\times3^3\times5^2\times7\times11\)

Step 1

Concept

Write \(415800=8\times51975\).

Step 2

Why this answer is correct

\(51975=3^3\times5^2\times7\times11\), so \(415800=2^3\times3^3\times5^2\times7\times11\).

Step 3

Exam Tip

Do not leave 51975 in the final form. चरण 1: \(415800=8\times51975\) लिखें। चरण 2: \(51975=3^3\times5^2\times7\times11\), इसलिए \(415800=2^3\times3^3\times5^2\times7\times11\)। चरण 3: 51975 को अंतिम रूप में न छोड़ें।

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Question 17/663 Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

संख्या 79200 का अभाज्य गुणनखंडन कौन सा है?

Which is the prime factorisation of 79200?

Explanation opens after your attempt
Correct Answer

A. \(2^5\times3^2\times5^2\times11\)

Step 1

Concept

Write \(79200=32\times2475\).

Step 2

Why this answer is correct

\(2475=3^2\times5^2\times11\), so \(79200=2^5\times3^2\times5^2\times11\).

Step 3

Exam Tip

Give 2475 its complete prime form. चरण 1: \(79200=32\times2475\) लिखें। चरण 2: \(2475=3^2\times5^2\times11\), इसलिए \(79200=2^5\times3^2\times5^2\times11\)। चरण 3: 2475 को पूरा अभाज्य रूप दें।

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Question 18/663 Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

संख्या 174636 का सही अभाज्य गुणनखंडन क्या है?

What is the correct prime factorisation of 174636?

Explanation opens after your attempt
Correct Answer

A. \(2^2\times3^4\times7^2\times11\)

Step 1

Concept

Write \(174636=4\times43659\).

Step 2

Why this answer is correct

\(4=2^2\) and \(43659=3^4\times7^2\times11\), so the correct form is \(2^2\times3^4\times7^2\times11\).

Step 3

Exam Tip

Convert 43659 into prime powers. चरण 1: \(174636=4\times43659\) लिखें। चरण 2: \(4=2^2\) और \(43659=3^4\times7^2\times11\), इसलिए सही रूप \(2^2\times3^4\times7^2\times11\) है। चरण 3: 43659 को अभाज्य घातों में बदलें।

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Question 19/663 Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

संख्या 155520 का अभाज्य गुणनखंडन कौन सा है?

Which is the prime factorisation of 155520?

Explanation opens after your attempt
Correct Answer

A. \(2^7\times3^5\times5\)

Step 1

Concept

Write \(155520=128\times1215\).

Step 2

Why this answer is correct

\(128=2^7\) and \(1215=3^5\times5\), so \(155520=2^7\times3^5\times5\).

Step 3

Exam Tip

Convert 1215 into prime powers. चरण 1: \(155520=128\times1215\) लिखें। चरण 2: \(128=2^7\) और \(1215=3^5\times5\), इसलिए \(155520=2^7\times3^5\times5\)। चरण 3: 1215 को अभाज्य घातों में बदलें।

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Question 20/663 Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

संख्या 529200 का सही अभाज्य गुणनखंडन क्या है?

What is the correct prime factorisation of 529200?

Explanation opens after your attempt
Correct Answer

A. \(2^4\times3^3\times5^2\times7^2\)

Step 1

Concept

Write \(529200=16\times33075\).

Step 2

Why this answer is correct

\(33075=3^3\times5^2\times7^2\), so \(529200=2^4\times3^3\times5^2\times7^2\).

Step 3

Exam Tip

Do not leave 33075 in the final form. चरण 1: \(529200=16\times33075\) लिखें। चरण 2: \(33075=3^3\times5^2\times7^2\), इसलिए \(529200=2^4\times3^3\times5^2\times7^2\)। चरण 3: 33075 को अंतिम रूप में न छोड़ें।

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Question 21/663 Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

संख्या 177408 का अभाज्य गुणनखंडन कौन सा है?

Which is the prime factorisation of 177408?

Explanation opens after your attempt
Correct Answer

A. \(2^8\times3^2\times7\times11\)

Step 1

Concept

Write \(177408=256\times693\).

Step 2

Why this answer is correct

\(256=2^8\) and \(693=3^2\times7\times11\), so the correct form is \(2^8\times3^2\times7\times11\).

Step 3

Exam Tip

Convert 693 into prime form. चरण 1: \(177408=256\times693\) लिखें। चरण 2: \(256=2^8\) और \(693=3^2\times7\times11\), इसलिए सही रूप \(2^8\times3^2\times7\times11\) है। चरण 3: 693 को अभाज्य रूप में बदलें।

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Question 22/663 Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

संख्या 289575 का सही अभाज्य गुणनखंडन क्या है?

What is the correct prime factorisation of 289575?

Explanation opens after your attempt
Correct Answer

A. \(3^4\times5^2\times11\times13\)

Step 1

Concept

Write \(289575=81\times3575\).

Step 2

Why this answer is correct

\(81=3^4\) and \(3575=5^2\times11\times13\), so \(289575=3^4\times5^2\times11\times13\).

Step 3

Exam Tip

Give 3575 its complete prime form too. चरण 1: \(289575=81\times3575\) लिखें। चरण 2: \(81=3^4\) और \(3575=5^2\times11\times13\), इसलिए \(289575=3^4\times5^2\times11\times13\)। चरण 3: 3575 को भी पूरा अभाज्य रूप दें।

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Question 23/663 Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

संख्या 196000 का अभाज्य गुणनखंडन कौन सा है?

Which is the prime factorisation of 196000?

Explanation opens after your attempt
Correct Answer

A. \(2^5\times5^3\times7^2\)

Step 1

Concept

Write \(196000=32\times6125\).

Step 2

Why this answer is correct

\(32=2^5\) and \(6125=5^3\times7^2\), so \(196000=2^5\times5^3\times7^2\).

Step 3

Exam Tip

Do not keep 6125 in the final answer. चरण 1: \(196000=32\times6125\) लिखें। चरण 2: \(32=2^5\) और \(6125=5^3\times7^2\), इसलिए \(196000=2^5\times5^3\times7^2\)। चरण 3: 6125 को अंतिम उत्तर में न रखें।

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Question 24/663 Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

संख्या 379080 का सही अभाज्य गुणनखंडन क्या है?

What is the correct prime factorisation of 379080?

Explanation opens after your attempt
Correct Answer

A. \(2^3\times3^6\times5\times13\)

Step 1

Concept

Write \(379080=8\times47385\).

Step 2

Why this answer is correct

\(8=2^3\) and \(47385=3^6\times5\times13\), so the correct form is \(2^3\times3^6\times5\times13\).

Step 3

Exam Tip

Give 47385 its complete prime form. चरण 1: \(379080=8\times47385\) लिखें। चरण 2: \(8=2^3\) और \(47385=3^6\times5\times13\), इसलिए सही रूप \(2^3\times3^6\times5\times13\) है। चरण 3: 47385 को पूरी तरह अभाज्य रूप दें।

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Question 25/663 Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

संख्या 627264 का अभाज्य गुणनखंडन कौन सा है?

Which is the prime factorisation of 627264?

Explanation opens after your attempt
Correct Answer

A. \(2^6\times3^4\times11^2\)

Step 1

Concept

Write \(627264=64\times9801\).

Step 2

Why this answer is correct

\(64=2^6\) and \(9801=3^4\times11^2\), so \(627264=2^6\times3^4\times11^2\).

Step 3

Exam Tip

Convert 9801 into prime powers. चरण 1: \(627264=64\times9801\) लिखें। चरण 2: \(64=2^6\) और \(9801=3^4\times11^2\), इसलिए \(627264=2^6\times3^4\times11^2\)। चरण 3: 9801 को अभाज्य घातों में बदलें।

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Question 26/663 Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

संख्या 680400 का सही अभाज्य गुणनखंडन क्या है?

What is the correct prime factorisation of 680400?

Explanation opens after your attempt
Correct Answer

A. \(2^4\times3^5\times5^2\times7\)

Step 1

Concept

Write \(680400=16\times42525\).

Step 2

Why this answer is correct

\(16=2^4\) and \(42525=3^5\times5^2\times7\), so the correct form is \(2^4\times3^5\times5^2\times7\).

Step 3

Exam Tip

Do not leave 42525 in the final form. चरण 1: \(680400=16\times42525\) लिखें। चरण 2: \(16=2^4\) और \(42525=3^5\times5^2\times7\), इसलिए सही रूप \(2^4\times3^5\times5^2\times7\) है। चरण 3: 42525 को अंतिम रूप में न छोड़ें।

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Question 27/663 Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

अंतिम अभाज्य गुणनखंडन में संयुक्त आधार बच जाने पर उत्तर अधूरा क्यों माना जाता है?

Why is the answer considered incomplete if a composite base remains in final prime factorisation?

Explanation opens after your attempt
Correct Answer

A. क्योंकि अंतिम रूप में केवल अभाज्य आधार रहने चाहिएBecause only prime bases should remain in the final form

Step 1

Concept

The aim of prime factorisation is to write the number using prime bases only.

Step 2

Why this answer is correct

If a base like (45) or (121) remains, it is composite and must be broken further.

Step 3

Exam Tip

Before writing the final answer, check whether every base is prime. चरण 1: अभाज्य गुणनखंडन का उद्देश्य संख्या को केवल अभाज्य आधारों में लिखना है। चरण 2: यदि (45) या (121) जैसा आधार बचा है, तो वह संयुक्त है और आगे टूटेगा। चरण 3: अंतिम उत्तर लिखने से पहले हर आधार की अभाज्यता जांचें।

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Question 28/663 Hard Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 9

अभाज्य गुणनखंडन में \(8\times27225\) को अंतिम उत्तर क्यों नहीं माना जाएगा?

Why will \(8\times27225\) not be considered the final answer in prime factorisation?

Explanation opens after your attempt
Correct Answer

A. क्योंकि 8 और 27225 संयुक्त रूप हैंBecause 8 and 27225 are composite forms

Step 1

Concept

In final prime factorisation, every base should be prime.

Step 2

Why this answer is correct

\(8=2^3\) and \(27225=3^2\times5^2\times11^2\).

Step 3

Exam Tip

Therefore, the final form is \(2^3\times3^2\times5^2\times11^2\). चरण 1: अंतिम अभाज्य गुणनखंडन में हर आधार अभाज्य होना चाहिए। चरण 2: \(8=2^3\) और \(27225=3^2\times5^2\times11^2\) है। चरण 3: इसलिए अंतिम रूप \(2^3\times3^2\times5^2\times11^2\) होगा।

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Question 29/663 Hard Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 9

संख्या 279936 का सही अभाज्य गुणनखंडन क्या है?

What is the correct prime factorisation of 279936?

Explanation opens after your attempt
Correct Answer

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

Step 1

Concept

(279936) can be written as \(128\times2187\).

Step 2

Why this answer is correct

\(128=2^7\) and \(2187=3^7\), so \(279936=2^7\times3^7\).

Step 3

Exam Tip

Do not leave 128 and 2187 in the final form. चरण 1: \(279936=128\times2187\) लिखा जा सकता है। चरण 2: \(128=2^7\) और \(2187=3^7\), इसलिए \(279936=2^7\times3^7\)। चरण 3: 128 और 2187 को अंतिम रूप में न छोड़ें।

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Question 30/663 Hard Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 9

संख्या 217800 का अभाज्य गुणनखंडन कौन सा है?

Which is the prime factorisation of 217800?

Explanation opens after your attempt
Correct Answer

A. \(2^3\times3^2\times5^2\times11^2\)

Step 1

Concept

Write \(217800=8\times27225\).

Step 2

Why this answer is correct

\(27225=3^2\times5^2\times11^2\), so \(217800=2^3\times3^2\times5^2\times11^2\).

Step 3

Exam Tip

Convert 27225 into prime powers. चरण 1: \(217800=8\times27225\) लिखें। चरण 2: \(27225=3^2\times5^2\times11^2\), इसलिए \(217800=2^3\times3^2\times5^2\times11^2\)। चरण 3: 27225 को अभाज्य घातों में बदलें।

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