अंतिम अभाज्य गुणनखंडन में संयुक्त आधार को क्यों हटाना जरूरी होता है?
Why is it necessary to remove a composite base in the final prime factorisation?
#prime-factorisation
#final-form
#hard
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A क्योंकि अंतिम रूप में केवल अभाज्य आधार होने चाहिए / Because only prime bases should remain in the final form
B क्योंकि संयुक्त आधार से संख्या हमेशा बदल जाती है / Because a composite base always changes the number
C क्योंकि घातों का उपयोग नहीं किया जा सकता / Because powers cannot be used
D क्योंकि हर संख्या को केवल दो गुणनखंडों में लिखना होता है / Because every number must be written in only two factors
Explanation opens after your attempt
Correct Answer
A. क्योंकि अंतिम रूप में केवल अभाज्य आधार होने चाहिए / Because only prime bases should remain in the final form
Step 1
Concept
अभाज्य गुणनखंडन का अंतिम रूप केवल अभाज्य संख्याओं पर आधारित होता है। / The final form of prime factorisation is based only on prime numbers.
Step 2
Why this answer is correct
यदि (45) जैसा आधार बचा है, तो \(45=3^2\times5\) लिखना होगा। / If a base like (45) remains, it must be written as \(45=3^2\times5\).
Step 3
Exam Tip
परीक्षा में अंतिम उत्तर लिखने से पहले हर आधार की जांच करें। / In exams, check every base before writing the final answer.
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संख्या 14256 का सही अभाज्य गुणनखंडन क्या है?
What is the correct prime factorisation of 14256?
#prime-factorisation
#number-14256
#hard
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A \(2^4\times3^4\times11\)
B \(2^3\times3^4\times11\)
C \(16\times891\)
D \(2^4\times81\times11\)
Explanation opens after your attempt
Correct Answer
A. \(2^4\times3^4\times11\)
Step 1
Concept
\(14256=16\times891\) लिखें। / Write \(14256=16\times891\).
Step 2
Why this answer is correct
\(16=2^4\) और \(891=3^4\times11\), इसलिए \(14256=2^4\times3^4\times11\)। / \(16=2^4\) and \(891=3^4\times11\), so \(14256=2^4\times3^4\times11\).
Step 3
Exam Tip
891 को अंतिम रूप में न छोड़ें। / Do not leave 891 in the final form.
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संख्या 18144 का अभाज्य गुणनखंडन कौन सा है?
Which is the prime factorisation of 18144?
#prime-factorisation
#number-18144
#hard
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A \(2^5\times3^4\times7\)
B \(2^4\times3^4\times7\)
C \(32\times567\)
D \(2^5\times81\times7\)
Explanation opens after your attempt
Correct Answer
A. \(2^5\times3^4\times7\)
Step 1
Concept
\(18144=32\times567\) लिखें। / Write \(18144=32\times567\).
Step 2
Why this answer is correct
\(32=2^5\) और \(567=3^4\times7\), इसलिए \(18144=2^5\times3^4\times7\)। / \(32=2^5\) and \(567=3^4\times7\), so \(18144=2^5\times3^4\times7\).
Step 3
Exam Tip
567 को पूरा अभाज्य रूप दें। / Give 567 its complete prime form.
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संख्या 24200 का सही अभाज्य गुणनखंडन क्या है?
What is the correct prime factorisation of 24200?
#prime-factorisation
#number-24200
#hard
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A \(2^3\times5^2\times11^2\)
B \(2^4\times5^2\times11\)
C \(200\times121\)
D \(8\times3025\)
Explanation opens after your attempt
Correct Answer
A. \(2^3\times5^2\times11^2\)
Step 1
Concept
\(24200=200\times121\) लिखें। / Write \(24200=200\times121\).
Step 2
Why this answer is correct
\(200=2^3\times5^2\) और \(121=11^2\), इसलिए \(24200=2^3\times5^2\times11^2\)। / \(200=2^3\times5^2\) and \(121=11^2\), so \(24200=2^3\times5^2\times11^2\).
Step 3
Exam Tip
200 और 121 को अंतिम रूप में न छोड़ें। / Do not leave 200 and 121 in the final form.
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संख्या 25410 का अभाज्य गुणनखंडन कौन सा है?
Which is the prime factorisation of 25410?
#prime-factorisation
#number-25410
#hard
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A \(2\times3\times5\times7\times11^2\)
B \(2^2\times3\times5\times7\times11\)
C \(210\times121\)
D \(2\times105\times121\)
Explanation opens after your attempt
Correct Answer
A. \(2\times3\times5\times7\times11^2\)
Step 1
Concept
\(25410=210\times121\) लिखें। / Write \(25410=210\times121\).
Step 2
Why this answer is correct
\(210=2\times3\times5\times7\) और \(121=11^2\), इसलिए \(25410=2\times3\times5\times7\times11^2\)। / \(210=2\times3\times5\times7\) and \(121=11^2\), so \(25410=2\times3\times5\times7\times11^2\).
Step 3
Exam Tip
210 और 121 दोनों को अभाज्य रूप दें। / Give prime form to both 210 and 121.
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संख्या 27216 का सही अभाज्य गुणनखंडन क्या है?
What is the correct prime factorisation of 27216?
#prime-factorisation
#number-27216
#hard
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A \(2^4\times3^5\times7\)
B \(2^3\times3^5\times7\)
C \(16\times1701\)
D \(2^4\times243\times7\)
Explanation opens after your attempt
Correct Answer
A. \(2^4\times3^5\times7\)
Step 1
Concept
\(27216=16\times1701\) लिखें। / Write \(27216=16\times1701\).
Step 2
Why this answer is correct
\(1701=3^5\times7\), इसलिए \(27216=2^4\times3^5\times7\)। / \(1701=3^5\times7\), so \(27216=2^4\times3^5\times7\).
Step 3
Exam Tip
1701 को 3 और 7 के अभाज्य रूप में बदलें। / Convert 1701 into prime form using 3 and 7.
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संख्या 30240 का अभाज्य गुणनखंडन कौन सा है?
Which is the prime factorisation of 30240?
#prime-factorisation
#number-30240
#hard
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A \(2^5\times3^3\times5\times7\)
B \(2^4\times3^3\times5\times7\)
C \(32\times945\)
D \(2^5\times27\times35\)
Explanation opens after your attempt
Correct Answer
A. \(2^5\times3^3\times5\times7\)
Step 1
Concept
\(30240=32\times945\) लिखें। / Write \(30240=32\times945\).
Step 2
Why this answer is correct
\(945=3^3\times5\times7\), इसलिए \(30240=2^5\times3^3\times5\times7\)। / \(945=3^3\times5\times7\), so \(30240=2^5\times3^3\times5\times7\).
Step 3
Exam Tip
945 को अंतिम उत्तर में न रखें। / Do not keep 945 in the final answer.
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संख्या 36300 का सही अभाज्य गुणनखंडन क्या है?
What is the correct prime factorisation of 36300?
#prime-factorisation
#number-36300
#hard
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A \(2^2\times3\times5^2\times11^2\)
B \(2\times3^2\times5^2\times11\)
C \(300\times121\)
D \(4\times9075\)
Explanation opens after your attempt
Correct Answer
A. \(2^2\times3\times5^2\times11^2\)
Step 1
Concept
\(36300=300\times121\) लिखें। / Write \(36300=300\times121\).
Step 2
Why this answer is correct
\(300=2^2\times3\times5^2\) और \(121=11^2\), इसलिए \(36300=2^2\times3\times5^2\times11^2\)। / \(300=2^2\times3\times5^2\) and \(121=11^2\), so \(36300=2^2\times3\times5^2\times11^2\).
Step 3
Exam Tip
300 और 121 को अभाज्य घातों में बदलें। / Convert 300 and 121 into prime powers.
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संख्या 38115 का अभाज्य गुणनखंडन कौन सा है?
Which is the prime factorisation of 38115?
#prime-factorisation
#number-38115
#hard
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A \(3^2\times5\times7\times11^2\)
B \(3^3\times5\times7\times11\)
C \(315\times121\)
D \(9\times4235\)
Explanation opens after your attempt
Correct Answer
A. \(3^2\times5\times7\times11^2\)
Step 1
Concept
\(38115=315\times121\) लिखें। / Write \(38115=315\times121\).
Step 2
Why this answer is correct
\(315=3^2\times5\times7\) और \(121=11^2\), इसलिए \(38115=3^2\times5\times7\times11^2\)। / \(315=3^2\times5\times7\) and \(121=11^2\), so \(38115=3^2\times5\times7\times11^2\).
Step 3
Exam Tip
दोनों भागों को पूरा तोड़ें। / Break both parts completely.
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संख्या 43200 का सही अभाज्य गुणनखंडन क्या है?
What is the correct prime factorisation of 43200?
#prime-factorisation
#number-43200
#hard
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A \(2^6\times3^3\times5^2\)
B \(2^5\times3^3\times5^2\)
C \(432\times100\)
D \(64\times675\)
Explanation opens after your attempt
Correct Answer
A. \(2^6\times3^3\times5^2\)
Step 1
Concept
\(43200=432\times100\) लिखें। / Write \(43200=432\times100\).
Step 2
Why this answer is correct
\(432=2^4\times3^3\) और \(100=2^2\times5^2\), इसलिए \(43200=2^6\times3^3\times5^2\)। / \(432=2^4\times3^3\) and \(100=2^2\times5^2\), so \(43200=2^6\times3^3\times5^2\).
Step 3
Exam Tip
2 की कुल घात 6 गिनें। / Count the total power of 2 as 6.
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संख्या 48510 का अभाज्य गुणनखंडन कौन सा है?
Which is the prime factorisation of 48510?
#prime-factorisation
#number-48510
#hard
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A \(2\times3^2\times5\times7^2\times11\)
B \(2^2\times3\times5\times7^2\times11\)
C \(90\times539\)
D \(2\times45\times539\)
Explanation opens after your attempt
Correct Answer
A. \(2\times3^2\times5\times7^2\times11\)
Step 1
Concept
\(48510=90\times539\) लिखें। / Write \(48510=90\times539\).
Step 2
Why this answer is correct
\(90=2\times3^2\times5\) और \(539=7^2\times11\), इसलिए \(48510=2\times3^2\times5\times7^2\times11\)। / \(90=2\times3^2\times5\) and \(539=7^2\times11\), so \(48510=2\times3^2\times5\times7^2\times11\).
Step 3
Exam Tip
90 और 539 दोनों को अभाज्य रूप दें। / Give prime form to both 90 and 539.
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संख्या 52920 का सही अभाज्य गुणनखंडन क्या है?
What is the correct prime factorisation of 52920?
#prime-factorisation
#number-52920
#hard
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A \(2^3\times3^3\times5\times7^2\)
B \(2^4\times3^2\times5\times7^2\)
C \(8\times6615\)
D \(2^3\times135\times49\)
Explanation opens after your attempt
Correct Answer
A. \(2^3\times3^3\times5\times7^2\)
Step 1
Concept
\(52920=8\times6615\) लिखें। / Write \(52920=8\times6615\).
Step 2
Why this answer is correct
\(6615=3^3\times5\times7^2\), इसलिए \(52920=2^3\times3^3\times5\times7^2\)। / \(6615=3^3\times5\times7^2\), so \(52920=2^3\times3^3\times5\times7^2\).
Step 3
Exam Tip
6615 को पूरा अभाज्य रूप दें। / Give 6615 its complete prime form.
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संख्या 59290 का अभाज्य गुणनखंडन कौन सा है?
Which is the prime factorisation of 59290?
#prime-factorisation
#number-59290
#hard
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A \(2\times5\times7\times11^3\)
B \(2^2\times5\times7\times11^2\)
C \(70\times1331\)
D \(10\times5929\)
Explanation opens after your attempt
Correct Answer
A. \(2\times5\times7\times11^3\)
Step 1
Concept
\(59290=70\times1331\) लिखें। / Write \(59290=70\times1331\).
Step 2
Why this answer is correct
\(70=2\times5\times7\) और \(1331=11^3\), इसलिए \(59290=2\times5\times7\times11^3\)। / \(70=2\times5\times7\) and \(1331=11^3\), so \(59290=2\times5\times7\times11^3\).
Step 3
Exam Tip
1331 को \(11^3\) में बदलें। / Change 1331 into \(11^3\).
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संख्या 65536 का सही अभाज्य गुणनखंडन क्या है?
What is the correct prime factorisation of 65536?
#prime-factorisation
#power-of-two
#hard
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A \(2^{16}\)
B \(2^{15}\)
C \(256^2\)
D \(16^4\)
Explanation opens after your attempt
Correct Answer
A. \(2^{16}\)
Step 1
Concept
65536 को 2 से बार-बार भाग दें। / Divide 65536 repeatedly by 2.
Step 2
Why this answer is correct
सोलह बार 2 मिलने से \(65536=2^{16}\) होता है। / Sixteen factors of 2 give \(65536=2^{16}\).
Step 3
Exam Tip
256 और 16 संयुक्त आधार हैं, इसलिए अंतिम रूप में 2 की घात लिखें। / 256 and 16 are composite bases, so write the power of 2 in final form.
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संख्या 68040 का अभाज्य गुणनखंडन कौन सा है?
Which is the prime factorisation of 68040?
#prime-factorisation
#number-68040
#hard
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A \(2^3\times3^5\times5\times7\)
B \(2^2\times3^5\times5\times7\)
C \(8\times8505\)
D \(2^3\times243\times35\)
Explanation opens after your attempt
Correct Answer
A. \(2^3\times3^5\times5\times7\)
Step 1
Concept
\(68040=8\times8505\) लिखें। / Write \(68040=8\times8505\).
Step 2
Why this answer is correct
\(8505=3^5\times5\times7\), इसलिए \(68040=2^3\times3^5\times5\times7\)। / \(8505=3^5\times5\times7\), so \(68040=2^3\times3^5\times5\times7\).
Step 3
Exam Tip
8505 को अभाज्य घातों में बदलें। / Convert 8505 into prime powers.
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संख्या 76230 का सही अभाज्य गुणनखंडन क्या है?
What is the correct prime factorisation of 76230?
#prime-factorisation
#number-76230
#hard
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A \(2\times3^2\times5\times7\times11^2\)
B \(2^2\times3\times5\times7\times11^2\)
C \(630\times121\)
D \(2\times315\times121\)
Explanation opens after your attempt
Correct Answer
A. \(2\times3^2\times5\times7\times11^2\)
Step 1
Concept
\(76230=630\times121\) लिखें। / Write \(76230=630\times121\).
Step 2
Why this answer is correct
\(630=2\times3^2\times5\times7\) और \(121=11^2\), इसलिए \(76230=2\times3^2\times5\times7\times11^2\)। / \(630=2\times3^2\times5\times7\) and \(121=11^2\), so \(76230=2\times3^2\times5\times7\times11^2\).
Step 3
Exam Tip
630 और 121 दोनों को पूरा तोड़ें। / Break both 630 and 121 completely.
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संख्या 88200 का अभाज्य गुणनखंडन कौन सा है?
Which is the prime factorisation of 88200?
#prime-factorisation
#number-88200
#hard
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? Hint Small clue
A \(2^3\times3^2\times5^2\times7^2\)
B \(2^2\times3^2\times5^2\times7^2\)
C \(8\times11025\)
D \(2^3\times105^2\)
Explanation opens after your attempt
Correct Answer
A. \(2^3\times3^2\times5^2\times7^2\)
Step 1
Concept
\(88200=8\times11025\) लिखें। / Write \(88200=8\times11025\).
Step 2
Why this answer is correct
\(11025=3^2\times5^2\times7^2\), इसलिए \(88200=2^3\times3^2\times5^2\times7^2\)। / \(11025=3^2\times5^2\times7^2\), so \(88200=2^3\times3^2\times5^2\times7^2\).
Step 3
Exam Tip
11025 को अभाज्य घातों में बदलें। / Convert 11025 into prime powers.
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संख्या 95256 का सही अभाज्य गुणनखंडन क्या है?
What is the correct prime factorisation of 95256?
#prime-factorisation
#number-95256
#hard
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? Hint Small clue
A \(2^3\times3^5\times7^2\)
B \(2^4\times3^4\times7^2\)
C \(8\times11907\)
D \(2^3\times243\times49\)
Explanation opens after your attempt
Correct Answer
A. \(2^3\times3^5\times7^2\)
Step 1
Concept
\(95256=8\times11907\) लिखें। / Write \(95256=8\times11907\).
Step 2
Why this answer is correct
\(11907=3^5\times7^2\), इसलिए \(95256=2^3\times3^5\times7^2\)। / \(11907=3^5\times7^2\), so \(95256=2^3\times3^5\times7^2\).
Step 3
Exam Tip
11907 को 3 और 7 की घातों में बदलें। / Convert 11907 into powers of 3 and 7.
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संख्या 108900 का अभाज्य गुणनखंडन कौन सा है?
Which is the prime factorisation of 108900?
#prime-factorisation
#square-number
#hard
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? Hint Small clue
A \(2^2\times3^2\times5^2\times11^2\)
B \(2\times3^2\times5^2\times11^2\)
C \(900\times121\)
D \(4\times27225\)
Explanation opens after your attempt
Correct Answer
A. \(2^2\times3^2\times5^2\times11^2\)
Step 1
Concept
\(108900=900\times121\) लिखें। / Write \(108900=900\times121\).
Step 2
Why this answer is correct
\(900=2^2\times3^2\times5^2\) और \(121=11^2\), इसलिए \(108900=2^2\times3^2\times5^2\times11^2\)। / \(900=2^2\times3^2\times5^2\) and \(121=11^2\), so \(108900=2^2\times3^2\times5^2\times11^2\).
Step 3
Exam Tip
यह वर्ग रूप है, इसलिए सभी घातें सम हैं। / This is a square form, so all exponents are even.
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संख्या 127008 का सही अभाज्य गुणनखंडन क्या है?
What is the correct prime factorisation of 127008?
#prime-factorisation
#number-127008
#hard
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A \(2^5\times3^4\times7^2\)
B \(2^4\times3^5\times7^2\)
C \(32\times3969\)
D \(2^5\times81\times49\)
Explanation opens after your attempt
Correct Answer
A. \(2^5\times3^4\times7^2\)
Step 1
Concept
\(127008=32\times3969\) लिखें। / Write \(127008=32\times3969\).
Step 2
Why this answer is correct
\(3969=3^4\times7^2\), इसलिए \(127008=2^5\times3^4\times7^2\)। / \(3969=3^4\times7^2\), so \(127008=2^5\times3^4\times7^2\).
Step 3
Exam Tip
3969 को अभाज्य घातों में बदलें। / Convert 3969 into prime powers.
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संख्या 130680 का अभाज्य गुणनखंडन कौन सा है?
Which is the prime factorisation of 130680?
#prime-factorisation
#number-130680
#hard
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A \(2^3\times3^3\times5\times11^2\)
B \(2^2\times3^4\times5\times11^2\)
C \(1080\times121\)
D \(8\times16335\)
Explanation opens after your attempt
Correct Answer
A. \(2^3\times3^3\times5\times11^2\)
Step 1
Concept
\(130680=1080\times121\) लिखें। / Write \(130680=1080\times121\).
Step 2
Why this answer is correct
\(1080=2^3\times3^3\times5\) और \(121=11^2\), इसलिए \(130680=2^3\times3^3\times5\times11^2\)। / \(1080=2^3\times3^3\times5\) and \(121=11^2\), so \(130680=2^3\times3^3\times5\times11^2\).
Step 3
Exam Tip
1080 और 121 को पूरा तोड़ें। / Break 1080 and 121 completely.
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यदि \(127008=2^a\times3^4\times7^2\), तो (a) का मान क्या है?
If \(127008=2^a\times3^4\times7^2\), what is the value of (a)?
#exponent-comparison
#number-127008
#hard
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A 5
B 4
C 6
D 7
Explanation opens after your attempt
Step 1
Concept
\(127008=32\times3969\) है। / \(127008=32\times3969\).
Step 2
Why this answer is correct
\(32=2^5\) और \(3969=3^4\times7^2\), इसलिए 2 की घात 5 है। / \(32=2^5\) and \(3969=3^4\times7^2\), so the power of 2 is 5.
Step 3
Exam Tip
तुलना करने पर (a=5) मिलता है। / Comparing gives (a=5).
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यदि \(95256=2^3\times3^b\times7^2\), तो (b) का मान क्या है?
If \(95256=2^3\times3^b\times7^2\), what is the value of (b)?
#exponent-comparison
#number-95256
#hard
50 50-50 2 wrong hide
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A 5
B 4
C 6
D 3
Explanation opens after your attempt
Step 1
Concept
\(95256=8\times11907\) लिखें। / Write \(95256=8\times11907\).
Step 2
Why this answer is correct
\(11907=3^5\times7^2\), इसलिए \(95256=2^3\times3^5\times7^2\)। / \(11907=3^5\times7^2\), so \(95256=2^3\times3^5\times7^2\).
Step 3
Exam Tip
दिए गए रूप से (b=5) है। / From the given form, (b=5).
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यदि \(108900=2^2\times3^2\times5^m\times11^2\), तो (m) का मान क्या है?
If \(108900=2^2\times3^2\times5^m\times11^2\), what is the value of (m)?
#exponent-comparison
#square-number
#hard
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A 2
B 1
C 3
D 4
Explanation opens after your attempt
Step 1
Concept
\(108900=900\times121\) है। / \(108900=900\times121\).
Step 2
Why this answer is correct
\(900=2^2\times3^2\times5^2\) और \(121=11^2\), इसलिए 5 की घात 2 है। / \(900=2^2\times3^2\times5^2\) and \(121=11^2\), so the power of 5 is 2.
Step 3
Exam Tip
तुलना करने पर (m=2) होगा। / Comparing gives (m=2).
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यदि \(68040=2^3\times3^p\times5\times7\), तो (p) का मान क्या है?
If \(68040=2^3\times3^p\times5\times7\), what is the value of (p)?
#exponent-comparison
#number-68040
#hard
50 50-50 2 wrong hide
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A 5
B 4
C 6
D 3
Explanation opens after your attempt
Step 1
Concept
\(68040=8\times8505\) लिखें। / Write \(68040=8\times8505\).
Step 2
Why this answer is correct
\(8505=3^5\times5\times7\), इसलिए 3 की घात 5 है। / \(8505=3^5\times5\times7\), so the power of 3 is 5.
Step 3
Exam Tip
तुलना करने पर (p=5) मिलेगा। / Comparing gives (p=5).
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किस संख्या का अभाज्य गुणनखंडन \(2^4\times3^4\times11\) है?
Which number has prime factorisation \(2^4\times3^4\times11\)?
#evaluate-factorisation
#number-14256
#hard
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A 14256
B 7128
C 28512
D 18144
Explanation opens after your attempt
Step 1
Concept
\(2^4=16\) और \(3^4=81\) निकालें। / Calculate \(2^4=16\) and \(3^4=81\).
Step 2
Why this answer is correct
\(16\times81\times11=14256\)। / \(16\times81\times11=14256\).
Step 3
Exam Tip
घातों को पहले हल करें, फिर 11 से गुणा करें। / Solve powers first, then multiply by 11.
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किस संख्या का अभाज्य गुणनखंडन \(2^5\times3^4\times7\) है?
Which number has prime factorisation \(2^5\times3^4\times7\)?
#evaluate-factorisation
#number-18144
#hard
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A 18144
B 9072
C 36288
D 12096
Explanation opens after your attempt
Step 1
Concept
\(2^5=32\) और \(3^4=81\) निकालें। / Calculate \(2^5=32\) and \(3^4=81\).
Step 2
Why this answer is correct
\(32\times81\times7=18144\)। / \(32\times81\times7=18144\).
Step 3
Exam Tip
बड़ी घातों का मान पहले निकालना ठीक रहता है। / It is better to find higher powers first.
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किस संख्या का अभाज्य गुणनखंडन \(2\times3\times5\times7\times11^2\) है?
Which number has prime factorisation \(2\times3\times5\times7\times11^2\)?
#evaluate-factorisation
#number-25410
#hard
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A 25410
B 12705
C 50820
D 27720
Explanation opens after your attempt
Step 1
Concept
\(11^2=121\) निकालें। / Calculate \(11^2=121\).
Step 2
Why this answer is correct
\(2\times3\times5\times7\times121=25410\)। / \(2\times3\times5\times7\times121=25410\).
Step 3
Exam Tip
पहले छोटे गुणनखंडों का गुणनफल 210 लें, फिर 121 से गुणा करें। / First take the product of smaller factors as 210, then multiply by 121.
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किस संख्या का अभाज्य गुणनखंडन \(2^6\times3^3\times5^2\) है?
Which number has prime factorisation \(2^6\times3^3\times5^2\)?
#evaluate-factorisation
#number-43200
#hard
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A 43200
B 21600
C 86400
D 32400
Explanation opens after your attempt
Step 1
Concept
\(2^6=64\), \(3^3=27\) और \(5^2=25\) निकालें। / Calculate \(2^6=64\), \(3^3=27\), and \(5^2=25\).
Step 2
Why this answer is correct
\(64\times27\times25=43200\)। / \(64\times27\times25=43200\).
Step 3
Exam Tip
तीनों घातों को अलग-अलग सरल करें। / Simplify all three powers separately.
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किस संख्या का अभाज्य गुणनखंडन \(2^5\times3^4\times7^2\) है?
Which number has prime factorisation \(2^5\times3^4\times7^2\)?
#evaluate-factorisation
#number-127008
#hard
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A 127008
B 63504
C 254016
D 95256
Explanation opens after your attempt
Step 1
Concept
\(2^5=32\), \(3^4=81\) और \(7^2=49\) निकालें। / Calculate \(2^5=32\), \(3^4=81\), and \(7^2=49\).
Step 2
Why this answer is correct
\(32\times81\times49=127008\)। / \(32\times81\times49=127008\).
Step 3
Exam Tip
घातों को पहले हल करने से गुणा साफ रहता है। / Solving powers first keeps multiplication clear.
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यदि संख्या \(2^8\times3^5\times5^2\times7^3\) है, तो इसे पूर्ण वर्ग बनाने के लिए सबसे छोटी किस संख्या से गुणा करना होगा?
If the number is \(2^8\times3^5\times5^2\times7^3\), by which smallest number should it be multiplied to make a perfect square?
#perfect-square
#prime-exponents
#hard
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A 21
B 3
C 7
D 35
Explanation opens after your attempt
Step 1
Concept
पूर्ण वर्ग में सभी घातें सम होनी चाहिए। / In a perfect square, all exponents should be even.
Step 2
Why this answer is correct
3 की घात 5 और 7 की घात 3 विषम हैं। / The powers of 3 and 7 are odd.
Step 3
Exam Tip
\(3\times7=21\) से गुणा करने पर दोनों घातें सम हो जाएंगी। / Multiplying by \(3\times7=21\) makes both powers even.
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यदि संख्या \(2^9\times3^4\times5^7\times11^2\) है, तो इसे पूर्ण वर्ग बनाने के लिए सबसे छोटी किस संख्या से भाग देना होगा?
If the number is \(2^9\times3^4\times5^7\times11^2\), by which smallest number should it be divided to make a perfect square?
#perfect-square
#division
#hard
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A \(2\times5\)
B \(2^2\times5\)
C \(2\times5^2\)
D \(5\times11\)
Explanation opens after your attempt
Correct Answer
A. \(2\times5\)
Step 1
Concept
पूर्ण वर्ग के लिए घातें सम चाहिए। / For a perfect square, exponents should be even.
Step 2
Why this answer is correct
2 की घात 9 और 5 की घात 7 विषम हैं। / The powers of 2 and 5 are odd.
Step 3
Exam Tip
\(2\times5\) से भाग देने पर घातें 8 और 6 हो जाएंगी। / Dividing by \(2\times5\) makes the powers 8 and 6.
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यदि संख्या \(2^5\times3^7\times5^4\times13^2\) है, तो इसे पूर्ण घन बनाने के लिए सबसे छोटी किस संख्या से गुणा करना होगा?
If the number is \(2^5\times3^7\times5^4\times13^2\), by which smallest number should it be multiplied to make a perfect cube?
#perfect-cube
#prime-exponents
#hard
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? Hint Small clue
A \(2\times3^2\times5^2\times13\)
B \(2\times3\times5\times13\)
C \(2^2\times3^2\times5\times13\)
D \(3^2\times5^2\)
Explanation opens after your attempt
Correct Answer
A. \(2\times3^2\times5^2\times13\)
Step 1
Concept
पूर्ण घन में हर घात 3 का गुणज होनी चाहिए। / In a perfect cube, every exponent must be a multiple of 3.
Step 2
Why this answer is correct
5 को 6, 7 को 9, 4 को 6 और 2 को 3 बनाना होगा। / Powers 5, 7, 4, and 2 must become 6, 9, 6, and 3.
Step 3
Exam Tip
सबसे छोटा गुणक \(2\times3^2\times5^2\times13\) है। / The smallest multiplier is \(2\times3^2\times5^2\times13\).
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यदि संख्या \(2^{10}\times3^8\times5^5\times7^4\) है, तो इसे पूर्ण घन बनाने के लिए सबसे छोटी किस संख्या से भाग देना होगा?
If the number is \(2^{10}\times3^8\times5^5\times7^4\), by which smallest number should it be divided to make a perfect cube?
#perfect-cube
#division
#hard
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? Hint Small clue
A \(2\times3^2\times5^2\times7\)
B \(2^2\times3\times5\times7\)
C \(2\times3\times5^2\times7^2\)
D \(3^2\times5\times7\)
Explanation opens after your attempt
Correct Answer
A. \(2\times3^2\times5^2\times7\)
Step 1
Concept
पूर्ण घन के लिए घातें 3 के गुणज चाहिए। / For a perfect cube, exponents should be multiples of 3.
Step 2
Why this answer is correct
10 को 9, 8 को 6, 5 को 3 और 4 को 3 तक घटाना सबसे छोटा तरीका है। / Reducing 10 to 9, 8 to 6, 5 to 3, and 4 to 3 is the smallest way.
Step 3
Exam Tip
इसलिए भाजक \(2\times3^2\times5^2\times7\) है। / Therefore, the divisor is \(2\times3^2\times5^2\times7\).
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यदि \(n=2^8\times3^6\times5^4\times11^2\), तो (n) किस संख्या से विभाज्य नहीं होगा?
If \(n=2^8\times3^6\times5^4\times11^2\), by which number will (n) not be divisible?
#divisibility
#prime-exponents
#hard
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? Hint Small clue
A \(11^3\)
B \(2^7\times3^5\)
C \(3^6\times5^3\)
D \(2^8\times11^2\)
Explanation opens after your attempt
Correct Answer
A. \(11^3\)
Step 1
Concept
भाजक की हर अभाज्य घात संख्या में उपलब्ध होनी चाहिए। / Every prime power of a divisor must be available in the number.
Step 2
Why this answer is correct
(n) में 11 की घात 2 है, पर \(11^3\) के लिए घात 3 चाहिए। / (n) has power 2 of 11, but \(11^3\) needs power 3.
Step 3
Exam Tip
इसलिए (n), \(11^3\) से विभाज्य नहीं होगा। / Therefore, (n) is not divisible by \(11^3\).
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यदि \(n=2^{10}\times3^5\times7^4\times13^2\), तो (n) किस संख्या से अवश्य विभाज्य होगा?
If \(n=2^{10}\times3^5\times7^4\times13^2\), by which number must (n) be divisible?
#divisibility
#prime-exponents
#hard
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? Hint Small clue
A \(2^9\times3^4\times7^3\times13\)
B \(2^{11}\times3^4\)
C \(3^6\times7^2\)
D \(2^{10}\times13^3\)
Explanation opens after your attempt
Correct Answer
A. \(2^9\times3^4\times7^3\times13\)
Step 1
Concept
विभाज्यता के लिए भाजक की घातें दी गई संख्या की घातों से अधिक नहीं होनी चाहिए। / For divisibility, exponents in the divisor must not exceed those in the given number.
Step 2
Why this answer is correct
\(2^9\), \(3^4\), \(7^3\) और 13 सभी (n) में उपलब्ध हैं। / \(2^9\), \(3^4\), \(7^3\), and 13 are all available in (n).
Step 3
Exam Tip
इसलिए (n) पहले विकल्प से अवश्य विभाज्य होगा। / Therefore, (n) must be divisible by the first option.
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यदि किसी संख्या का अभाज्य गुणनखंडन \(2^9\times3^7\times5^4\times7^3\times11^2\) है, तो दोहराव सहित अभाज्य गुणनखंडों की संख्या कितनी है?
If a number has prime factorisation \(2^9\times3^7\times5^4\times7^3\times11^2\), how many prime factors does it have with repetition?
#counting-prime-factors
#prime-exponents
#hard
50 50-50 2 wrong hide
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? Hint Small clue
A 25
B 5
C 20
D 27
Explanation opens after your attempt
Step 1
Concept
दोहराव सहित गिनती में घातों को जोड़ा जाता है। / To count with repetition, add the exponents.
Step 2
Why this answer is correct
(9+7+4+3+2=25)। / (9+7+4+3+2=25).
Step 3
Exam Tip
आधारों की संख्या और दोहराव सहित कुल संख्या को अलग रखें। / Keep the number of bases and the total count with repetition separate.
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यदि किसी संख्या का अभाज्य गुणनखंडन \(2^6\times3^5\times7^4\times11^3\times13^2\) है, तो उसमें अलग-अलग अभाज्य गुणनखंड कितने हैं?
If a number has prime factorisation \(2^6\times3^5\times7^4\times11^3\times13^2\), how many distinct prime factors does it have?
#distinct-prime-factors
#prime-exponents
#hard
50 50-50 2 wrong hide
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+10 Time+ 10 sec extra
? Hint Small clue
A 5
B 20
C 6
D 4
Explanation opens after your attempt
Step 1
Concept
अलग-अलग अभाज्य गिनते समय केवल आधार गिने जाते हैं। / While counting distinct primes, only bases are counted.
Step 2
Why this answer is correct
आधार 2, 3, 7, 11 और 13 हैं। / The bases are 2, 3, 7, 11, and 13.
Step 3
Exam Tip
इसलिए अलग-अलग अभाज्य गुणनखंडों की संख्या 5 है। / Therefore, the number of distinct prime factors is 5.
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यदि \(a=2^6\times3^4\times5^2\times11\) और \(b=2^3\times3^5\times7\times11^2\), तो (ab) में 3 की घात क्या होगी?
If \(a=2^6\times3^4\times5^2\times11\) and \(b=2^3\times3^5\times7\times11^2\), what will be the power of 3 in (ab)?
#product-factorisation
#powers
#hard
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+10 Time+ 10 sec extra
? Hint Small clue
A 9
B 5
C 4
D 8
Explanation opens after your attempt
Step 1
Concept
गुणा में समान अभाज्य आधार की घातें जुड़ती हैं। / In multiplication, powers of the same prime base are added.
Step 2
Why this answer is correct
(a) में 3 की घात 4 है और (b) में 3 की घात 5 है। / The power of 3 in (a) is 4 and in (b) is 5.
Step 3
Exam Tip
(ab) में 3 की घात (4+5=9) होगी। / In (ab), the power of 3 will be (4+5=9).
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यदि \(x=2^8\times5^3\times7^2\times13\) और \(y=2^5\times3^2\times5^4\times13^3\), तो (xy) में 13 की घात क्या होगी?
If \(x=2^8\times5^3\times7^2\times13\) and \(y=2^5\times3^2\times5^4\times13^3\), what will be the power of 13 in (xy)?
#product-factorisation
#powers
#hard
50 50-50 2 wrong hide
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+10 Time+ 10 sec extra
? Hint Small clue
A 4
B 3
C 1
D 5
Explanation opens after your attempt
Step 1
Concept
समान आधार 13 की घातें गुणा में जुड़ेंगी। / Powers with the same base 13 are added in multiplication.
Step 2
Why this answer is correct
(x) में 13 की घात 1 है और (y) में 13 की घात 3 है। / The power of 13 in (x) is 1 and in (y) is 3.
Step 3
Exam Tip
कुल घात (1+3=4) होगी। / The total power will be (1+3=4).
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किस विकल्प में \(2^5\times3^4\times7^2\) का सही मान है?
Which option gives the correct value of \(2^5\times3^4\times7^2\)?
#evaluate-factorisation
#number-127008
#hard
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A 127008
B 63504
C 254016
D 95256
Explanation opens after your attempt
Step 1
Concept
\(2^5=32\), \(3^4=81\) और \(7^2=49\) निकालें। / Calculate \(2^5=32\), \(3^4=81\), and \(7^2=49\).
Step 2
Why this answer is correct
\(32\times81\times49=127008\)। / \(32\times81\times49=127008\).
Step 3
Exam Tip
तीनों घातों को पहले अलग-अलग हल करें। / Solve all three powers separately first.
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किस विकल्प में \(2^3\times3^5\times7^2\) का सही मान है?
Which option gives the correct value of \(2^3\times3^5\times7^2\)?
#evaluate-factorisation
#number-95256
#hard
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A 95256
B 47628
C 190512
D 68040
Explanation opens after your attempt
Step 1
Concept
\(2^3=8\), \(3^5=243\) और \(7^2=49\) निकालें। / Calculate \(2^3=8\), \(3^5=243\), and \(7^2=49\).
Step 2
Why this answer is correct
\(8\times243\times49=95256\)। / \(8\times243\times49=95256\).
Step 3
Exam Tip
बड़ी घातों को पहले सरल करें। / Simplify higher powers first.
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किस विकल्प में केवल अंतिम अभाज्य गुणनखंडन दिया गया है?
Which option gives only the final prime factorisation?
#prime-factorisation
#final-form
#hard
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A \(2^3\times3^3\times5\times7^2\)
B \(8\times135\times49\)
C \(2^3\times27\times245\)
D \(216\times245\)
Explanation opens after your attempt
Correct Answer
A. \(2^3\times3^3\times5\times7^2\)
Step 1
Concept
अंतिम रूप में हर आधार अभाज्य होना चाहिए। / In the final form, every base must be prime.
Step 2
Why this answer is correct
पहले विकल्प में आधार 2, 3, 5 और 7 अभाज्य हैं। / In the first option, bases 2, 3, 5, and 7 are prime.
Step 3
Exam Tip
8, 135, 49, 27, 245 और 216 संयुक्त हैं, इसलिए वे अंतिम रूप नहीं हैं। / 8, 135, 49, 27, 245, and 216 are composite, so they are not final forms.
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किस विकल्प में अभाज्य गुणनखंडन अधूरा है?
Which option has incomplete prime factorisation?
#prime-factorisation
#incomplete-form
#hard
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A \(2^5\times81\times49\)
B \(2^5\times3^4\times7^2\)
C \(2^6\times3^3\times5^2\)
D \(2\times5\times7\times11^3\)
Explanation opens after your attempt
Correct Answer
A. \(2^5\times81\times49\)
Step 1
Concept
अधूरे रूप में संयुक्त आधार बच जाता है। / In an incomplete form, composite bases remain.
Step 2
Why this answer is correct
81 और 49 संयुक्त आधार हैं। / 81 and 49 are composite bases.
Step 3
Exam Tip
\(2^5\times81\times49\) को \(2^5\times3^4\times7^2\) में बदलना होगा। / \(2^5\times81\times49\) must be changed into \(2^5\times3^4\times7^2\).
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यदि \(2^6\times3^3\times5^2\) किसी संख्या का अभाज्य गुणनखंडन है, तो संख्या क्या है?
If \(2^6\times3^3\times5^2\) is the prime factorisation of a number, what is the number?
#evaluate-factorisation
#hard
#mcq
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A 43200
B 21600
C 86400
D 32400
Explanation opens after your attempt
Step 1
Concept
\(2^6=64\), \(3^3=27\) और \(5^2=25\) हैं। / \(2^6=64\), \(3^3=27\), and \(5^2=25\).
Step 2
Why this answer is correct
\(64\times27\times25=43200\)। / \(64\times27\times25=43200\).
Step 3
Exam Tip
घातों को पहले हल करने से सही विकल्प जल्दी मिलता है। / Solving powers first helps find the correct option quickly.
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यदि \(2^2\times3^2\times5^2\times11^2\) किसी संख्या का अभाज्य गुणनखंडन है, तो संख्या क्या है?
If \(2^2\times3^2\times5^2\times11^2\) is the prime factorisation of a number, what is the number?
#evaluate-factorisation
#square-number
#hard
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A 108900
B 54450
C 217800
D 27225
Explanation opens after your attempt
Step 1
Concept
\(2^2=4\), \(3^2=9\), \(5^2=25\) और \(11^2=121\) निकालें। / Calculate \(2^2=4\), \(3^2=9\), \(5^2=25\), and \(11^2=121\).
Step 2
Why this answer is correct
\(4\times9\times25\times121=108900\)। / \(4\times9\times25\times121=108900\).
Step 3
Exam Tip
यह वर्ग रूप है, इसलिए घातों को ध्यान से देखें। / This is a square form, so observe the powers carefully.
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संख्या 158760 का सही अभाज्य गुणनखंडन क्या है?
What is the correct prime factorisation of 158760?
#prime-factorisation
#number-158760
#hard
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A \(2^3\times3^4\times5\times7^2\)
B \(2^4\times3^3\times5\times7^2\)
C \(8\times19845\)
D \(2^3\times405\times49\)
Explanation opens after your attempt
Correct Answer
A. \(2^3\times3^4\times5\times7^2\)
Step 1
Concept
\(158760=8\times19845\) लिखें। / Write \(158760=8\times19845\).
Step 2
Why this answer is correct
\(19845=3^4\times5\times7^2\), इसलिए \(158760=2^3\times3^4\times5\times7^2\)। / \(19845=3^4\times5\times7^2\), so \(158760=2^3\times3^4\times5\times7^2\).
Step 3
Exam Tip
19845 को पूरी तरह तोड़ें। / Break 19845 completely.
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संख्या 217800 का अभाज्य गुणनखंडन कौन सा है?
Which is the prime factorisation of 217800?
#prime-factorisation
#number-217800
#hard
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A \(2^3\times3^2\times5^2\times11^2\)
B \(2^2\times3^3\times5^2\times11^2\)
C \(8\times27225\)
D \(2^3\times225\times121\)
Explanation opens after your attempt
Correct Answer
A. \(2^3\times3^2\times5^2\times11^2\)
Step 1
Concept
\(217800=8\times27225\) लिखें। / Write \(217800=8\times27225\).
Step 2
Why this answer is correct
\(27225=3^2\times5^2\times11^2\), इसलिए \(217800=2^3\times3^2\times5^2\times11^2\)। / \(27225=3^2\times5^2\times11^2\), so \(217800=2^3\times3^2\times5^2\times11^2\).
Step 3
Exam Tip
27225 को अभाज्य घातों में बदलें। / Convert 27225 into prime powers.
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संख्या 279936 का सही अभाज्य गुणनखंडन क्या है?
What is the correct prime factorisation of 279936?
#prime-factorisation
#number-279936
#hard
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A \(2^7\times3^7\)
B \(2^6\times3^7\)
C \(128\times2187\)
D \(6^7\)
Explanation opens after your attempt
Correct Answer
A. \(2^7\times3^7\)
Step 1
Concept
\(279936=128\times2187\) लिखा जा सकता है। / (279936) can be written as \(128\times2187\).
Step 2
Why this answer is correct
\(128=2^7\) और \(2187=3^7\), इसलिए \(279936=2^7\times3^7\)। / \(128=2^7\) and \(2187=3^7\), so \(279936=2^7\times3^7\).
Step 3
Exam Tip
128 और 2187 को अंतिम रूप में न छोड़ें। / Do not leave 128 and 2187 in the final form.
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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?
#prime-factorisation
#concept-check
#hard
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A क्योंकि 8 और 27225 संयुक्त रूप हैं / Because 8 and 27225 are composite forms
B क्योंकि 8 अभाज्य संख्या है / Because 8 is prime
C क्योंकि 27225 को तोड़ा नहीं जा सकता / Because 27225 cannot be factorised
D क्योंकि गुणनफल बदल जाएगा / Because the product will change
Explanation opens after your attempt
Correct Answer
A. क्योंकि 8 और 27225 संयुक्त रूप हैं / Because 8 and 27225 are composite forms
Step 1
Concept
अंतिम अभाज्य गुणनखंडन में हर आधार अभाज्य होना चाहिए। / In final prime factorisation, every base should be prime.
Step 2
Why this answer is correct
\(8=2^3\) और \(27225=3^2\times5^2\times11^2\) है। / \(8=2^3\) and \(27225=3^2\times5^2\times11^2\).
Step 3
Exam Tip
इसलिए अंतिम रूप \(2^3\times3^2\times5^2\times11^2\) होगा। / Therefore, the final form is \(2^3\times3^2\times5^2\times11^2\).
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