अभाज्य गुणनखंडन में \(36^2\times7\) को अंतिम रूप क्यों नहीं माना जाता?
Why is \(36^2\times7\) not considered the final form in prime factorisation?
#prime-factorisation
#composite-base
#hard
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A क्योंकि 36 संयुक्त संख्या है / Because 36 is composite
B क्योंकि 7 संयुक्त संख्या है / Because 7 is composite
C क्योंकि घात लिखना हमेशा गलत है / Because writing powers is always wrong
D क्योंकि गुणन का प्रयोग नहीं किया जाता / Because multiplication is not used
Explanation opens after your attempt
Correct Answer
A. क्योंकि 36 संयुक्त संख्या है / Because 36 is composite
Step 1
Concept
अंतिम अभाज्य गुणनखंडन में आधार अभाज्य होने चाहिए। / In final prime factorisation, the bases should be prime.
Step 2
Why this answer is correct
\(36=2^2\times3^2\), इसलिए \(36^2\) को \(2^4\times3^4\) में बदलना होगा। / \(36=2^2\times3^2\), so \(36^2\) must be changed into \(2^4\times3^4\).
Step 3
Exam Tip
अंतिम उत्तर में 36 जैसा संयुक्त आधार न रखें। / Do not keep a composite base like 36 in the final answer.
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संख्या 5544 का सही अभाज्य गुणनखंडन क्या है?
What is the correct prime factorisation of 5544?
#prime-factorisation
#number-5544
#hard
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A \(2^3\times3^2\times7\times11\)
B \(2^2\times3^3\times7\times11\)
C \(8\times693\)
D \(2^3\times9\times77\)
Explanation opens after your attempt
Correct Answer
A. \(2^3\times3^2\times7\times11\)
Step 1
Concept
\(5544=8\times693\) लिखें। / Write \(5544=8\times693\).
Step 2
Why this answer is correct
\(8=2^3\) और \(693=3^2\times7\times11\), इसलिए \(5544=2^3\times3^2\times7\times11\)। / \(8=2^3\) and \(693=3^2\times7\times11\), so \(5544=2^3\times3^2\times7\times11\).
Step 3
Exam Tip
693 को अंतिम रूप में न छोड़ें। / Do not leave 693 in the final form.
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संख्या 6048 का अभाज्य गुणनखंडन कौन सा है?
Which is the prime factorisation of 6048?
#prime-factorisation
#number-6048
#hard
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A \(2^5\times3^3\times7\)
B \(2^4\times3^3\times7\)
C \(32\times189\)
D \(2^5\times27\times7\)
Explanation opens after your attempt
Correct Answer
A. \(2^5\times3^3\times7\)
Step 1
Concept
\(6048=32\times189\) लिखें। / Write \(6048=32\times189\).
Step 2
Why this answer is correct
\(32=2^5\) और \(189=3^3\times7\), इसलिए \(6048=2^5\times3^3\times7\)। / \(32=2^5\) and \(189=3^3\times7\), so \(6048=2^5\times3^3\times7\).
Step 3
Exam Tip
189 को \(3^3\times7\) में बदलें। / Change 189 into \(3^3\times7\).
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संख्या 6930 का सही अभाज्य गुणनखंडन क्या है?
What is the correct prime factorisation of 6930?
#prime-factorisation
#number-6930
#hard
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A \(2\times3^2\times5\times7\times11\)
B \(2^2\times3\times5\times7\times11\)
C \(63\times110\)
D \(2\times9\times385\)
Explanation opens after your attempt
Correct Answer
A. \(2\times3^2\times5\times7\times11\)
Step 1
Concept
\(6930=63\times110\) लिखें। / Write \(6930=63\times110\).
Step 2
Why this answer is correct
\(63=3^2\times7\) और \(110=2\times5\times11\), इसलिए \(6930=2\times3^2\times5\times7\times11\)। / \(63=3^2\times7\) and \(110=2\times5\times11\), so \(6930=2\times3^2\times5\times7\times11\).
Step 3
Exam Tip
63 और 110 दोनों को अभाज्य रूप दें। / Give prime form to both 63 and 110.
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संख्या 8316 का अभाज्य गुणनखंडन कौन सा है?
Which is the prime factorisation of 8316?
#prime-factorisation
#number-8316
#hard
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A \(2^2\times3^3\times7\times11\)
B \(2^3\times3^2\times7\times11\)
C \(4\times2079\)
D \(36\times231\)
Explanation opens after your attempt
Correct Answer
A. \(2^2\times3^3\times7\times11\)
Step 1
Concept
\(8316=36\times231\) लिखें। / Write \(8316=36\times231\).
Step 2
Why this answer is correct
\(36=2^2\times3^2\) और \(231=3\times7\times11\), इसलिए \(8316=2^2\times3^3\times7\times11\)। / \(36=2^2\times3^2\) and \(231=3\times7\times11\), so \(8316=2^2\times3^3\times7\times11\).
Step 3
Exam Tip
3 की कुल घात 3 बनती है। / The total power of 3 becomes 3.
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संख्या 9240 का सही अभाज्य गुणनखंडन क्या है?
What is the correct prime factorisation of 9240?
#prime-factorisation
#number-9240
#hard
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A \(2^3\times3\times5\times7\times11\)
B \(2^2\times3\times5\times7\times11\)
C \(8\times1155\)
D \(2^3\times105\times11\)
Explanation opens after your attempt
Correct Answer
A. \(2^3\times3\times5\times7\times11\)
Step 1
Concept
\(9240=8\times1155\) लिखें। / Write \(9240=8\times1155\).
Step 2
Why this answer is correct
\(8=2^3\) और \(1155=3\times5\times7\times11\), इसलिए \(9240=2^3\times3\times5\times7\times11\)। / \(8=2^3\) and \(1155=3\times5\times7\times11\), so \(9240=2^3\times3\times5\times7\times11\).
Step 3
Exam Tip
1155 को पूरा अभाज्य रूप दें। / Give 1155 its complete prime form.
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संख्या 10395 का अभाज्य गुणनखंडन कौन सा है?
Which is the prime factorisation of 10395?
#prime-factorisation
#number-10395
#hard
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A \(3^3\times5\times7\times11\)
B \(3^2\times5\times7\times11\)
C \(27\times385\)
D \(135\times77\)
Explanation opens after your attempt
Correct Answer
A. \(3^3\times5\times7\times11\)
Step 1
Concept
\(10395=27\times385\) लिखें। / Write \(10395=27\times385\).
Step 2
Why this answer is correct
\(27=3^3\) और \(385=5\times7\times11\), इसलिए \(10395=3^3\times5\times7\times11\)। / \(27=3^3\) and \(385=5\times7\times11\), so \(10395=3^3\times5\times7\times11\).
Step 3
Exam Tip
385 को भी अभाज्य रूप में तोड़ें। / Break 385 into prime form too.
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संख्या 11025 का सही अभाज्य गुणनखंडन क्या है?
What is the correct prime factorisation of 11025?
#prime-factorisation
#square-number
#hard
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A \(3^2\times5^2\times7^2\)
B \(3^3\times5^2\times7\)
C \(105\times105\)
D \(9\times1225\)
Explanation opens after your attempt
Correct Answer
A. \(3^2\times5^2\times7^2\)
Step 1
Concept
\(11025=105^2\) पहचाना जा सकता है। / Recognise \(11025=105^2\).
Step 2
Why this answer is correct
\(105=3\times5\times7\), इसलिए \(11025=3^2\times5^2\times7^2\)। / Since \(105=3\times5\times7\), \(11025=3^2\times5^2\times7^2\).
Step 3
Exam Tip
वर्ग संख्या में अभाज्य घातें सम आती हैं। / In a square number, prime exponents are even.
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संख्या 11760 का अभाज्य गुणनखंडन कौन सा है?
Which is the prime factorisation of 11760?
#prime-factorisation
#number-11760
#hard
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A \(2^4\times3\times5\times7^2\)
B \(2^3\times3^2\times5\times7\)
C \(16\times735\)
D \(2^4\times15\times49\)
Explanation opens after your attempt
Correct Answer
A. \(2^4\times3\times5\times7^2\)
Step 1
Concept
\(11760=16\times735\) लिखें। / Write \(11760=16\times735\).
Step 2
Why this answer is correct
\(16=2^4\) और \(735=3\times5\times7^2\), इसलिए \(11760=2^4\times3\times5\times7^2\)। / \(16=2^4\) and \(735=3\times5\times7^2\), so \(11760=2^4\times3\times5\times7^2\).
Step 3
Exam Tip
735 को पूरा अभाज्य रूप दें। / Give 735 its complete prime form.
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संख्या 14112 का सही अभाज्य गुणनखंडन क्या है?
What is the correct prime factorisation of 14112?
#prime-factorisation
#number-14112
#hard
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A \(2^5\times3^2\times7^2\)
B \(2^4\times3^3\times7^2\)
C \(32\times441\)
D \(2^5\times21^2\)
Explanation opens after your attempt
Correct Answer
A. \(2^5\times3^2\times7^2\)
Step 1
Concept
\(14112=32\times441\) लिखें। / Write \(14112=32\times441\).
Step 2
Why this answer is correct
\(32=2^5\) और \(441=3^2\times7^2\), इसलिए \(14112=2^5\times3^2\times7^2\)। / \(32=2^5\) and \(441=3^2\times7^2\), so \(14112=2^5\times3^2\times7^2\).
Step 3
Exam Tip
441 को \(3^2\times7^2\) में बदलें। / Change 441 into \(3^2\times7^2\).
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संख्या 15120 का अभाज्य गुणनखंडन कौन सा है?
Which is the prime factorisation of 15120?
#prime-factorisation
#number-15120
#hard
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A \(2^4\times3^3\times5\times7\)
B \(2^3\times3^3\times5\times7\)
C \(16\times945\)
D \(2^4\times27\times35\)
Explanation opens after your attempt
Correct Answer
A. \(2^4\times3^3\times5\times7\)
Step 1
Concept
\(15120=16\times945\) लिखें। / Write \(15120=16\times945\).
Step 2
Why this answer is correct
\(16=2^4\) और \(945=3^3\times5\times7\), इसलिए \(15120=2^4\times3^3\times5\times7\)। / \(16=2^4\) and \(945=3^3\times5\times7\), so \(15120=2^4\times3^3\times5\times7\).
Step 3
Exam Tip
945 को अंतिम रूप में न छोड़ें। / Do not leave 945 in the final form.
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संख्या 16632 का सही अभाज्य गुणनखंडन क्या है?
What is the correct prime factorisation of 16632?
#prime-factorisation
#number-16632
#hard
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A \(2^3\times3^3\times7\times11\)
B \(2^2\times3^4\times7\times11\)
C \(8\times2079\)
D \(2^3\times27\times77\)
Explanation opens after your attempt
Correct Answer
A. \(2^3\times3^3\times7\times11\)
Step 1
Concept
\(16632=8\times2079\) लिखें। / Write \(16632=8\times2079\).
Step 2
Why this answer is correct
\(2079=3^3\times7\times11\), इसलिए \(16632=2^3\times3^3\times7\times11\)। / \(2079=3^3\times7\times11\), so \(16632=2^3\times3^3\times7\times11\).
Step 3
Exam Tip
2079 को अभाज्य गुणनखंडों तक तोड़ें। / Break 2079 down to prime factors.
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संख्या 17640 का अभाज्य गुणनखंडन कौन सा है?
Which is the prime factorisation of 17640?
#prime-factorisation
#number-17640
#hard
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A \(2^3\times3^2\times5\times7^2\)
B \(2^4\times3^2\times5\times7\)
C \(8\times2205\)
D \(2^3\times45\times49\)
Explanation opens after your attempt
Correct Answer
A. \(2^3\times3^2\times5\times7^2\)
Step 1
Concept
\(17640=8\times2205\) लिखें। / Write \(17640=8\times2205\).
Step 2
Why this answer is correct
\(2205=3^2\times5\times7^2\), इसलिए \(17640=2^3\times3^2\times5\times7^2\)। / \(2205=3^2\times5\times7^2\), so \(17640=2^3\times3^2\times5\times7^2\).
Step 3
Exam Tip
2205 को पूरा अभाज्य रूप दें। / Give 2205 its complete prime form.
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संख्या 19600 का सही अभाज्य गुणनखंडन क्या है?
What is the correct prime factorisation of 19600?
#prime-factorisation
#square-number
#hard
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A \(2^4\times5^2\times7^2\)
B \(2^3\times5^2\times7^2\)
C \(140\times140\)
D \(16\times1225\)
Explanation opens after your attempt
Correct Answer
A. \(2^4\times5^2\times7^2\)
Step 1
Concept
\(19600=140^2\) पहचाना जा सकता है। / Recognise \(19600=140^2\).
Step 2
Why this answer is correct
\(140=2^2\times5\times7\), इसलिए \(19600=2^4\times5^2\times7^2\)। / Since \(140=2^2\times5\times7\), \(19600=2^4\times5^2\times7^2\).
Step 3
Exam Tip
वर्ग संख्या में घातों की समता जांचें। / Check evenness of exponents in square numbers.
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संख्या 20790 का अभाज्य गुणनखंडन कौन सा है?
Which is the prime factorisation of 20790?
#prime-factorisation
#number-20790
#hard
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A \(2\times3^3\times5\times7\times11\)
B \(2^2\times3^2\times5\times7\times11\)
C \(54\times385\)
D \(2\times27\times385\)
Explanation opens after your attempt
Correct Answer
A. \(2\times3^3\times5\times7\times11\)
Step 1
Concept
\(20790=54\times385\) लिखें। / Write \(20790=54\times385\).
Step 2
Why this answer is correct
\(54=2\times3^3\) और \(385=5\times7\times11\), इसलिए \(20790=2\times3^3\times5\times7\times11\)। / \(54=2\times3^3\) and \(385=5\times7\times11\), so \(20790=2\times3^3\times5\times7\times11\).
Step 3
Exam Tip
54 और 385 दोनों को पूरा तोड़ें। / Break both 54 and 385 completely.
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संख्या 21168 का सही अभाज्य गुणनखंडन क्या है?
What is the correct prime factorisation of 21168?
#prime-factorisation
#number-21168
#hard
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? Hint Small clue
A \(2^4\times3^3\times7^2\)
B \(2^5\times3^2\times7^2\)
C \(16\times1323\)
D \(2^4\times27\times49\)
Explanation opens after your attempt
Correct Answer
A. \(2^4\times3^3\times7^2\)
Step 1
Concept
\(21168=16\times1323\) लिखें। / Write \(21168=16\times1323\).
Step 2
Why this answer is correct
\(1323=3^3\times7^2\), इसलिए \(21168=2^4\times3^3\times7^2\)। / \(1323=3^3\times7^2\), so \(21168=2^4\times3^3\times7^2\).
Step 3
Exam Tip
1323 को 3 और 7 की घातों में बदलें। / Convert 1323 into powers of 3 and 7.
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संख्या 24255 का अभाज्य गुणनखंडन कौन सा है?
Which is the prime factorisation of 24255?
#prime-factorisation
#number-24255
#hard
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? Hint Small clue
A \(3^2\times5\times7^2\times11\)
B \(3^3\times5\times7\times11\)
C \(45\times539\)
D \(9\times2695\)
Explanation opens after your attempt
Correct Answer
A. \(3^2\times5\times7^2\times11\)
Step 1
Concept
\(24255=45\times539\) लिखें। / Write \(24255=45\times539\).
Step 2
Why this answer is correct
\(45=3^2\times5\) और \(539=7^2\times11\), इसलिए \(24255=3^2\times5\times7^2\times11\)। / \(45=3^2\times5\) and \(539=7^2\times11\), so \(24255=3^2\times5\times7^2\times11\).
Step 3
Exam Tip
539 को भी अभाज्य रूप दें। / Give 539 its prime form too.
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संख्या 27720 का सही अभाज्य गुणनखंडन क्या है?
What is the correct prime factorisation of 27720?
#prime-factorisation
#number-27720
#hard
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? Hint Small clue
A \(2^3\times3^2\times5\times7\times11\)
B \(2^2\times3^3\times5\times7\times11\)
C \(8\times3465\)
D \(2^3\times9\times385\)
Explanation opens after your attempt
Correct Answer
A. \(2^3\times3^2\times5\times7\times11\)
Step 1
Concept
\(27720=8\times3465\) लिखें। / Write \(27720=8\times3465\).
Step 2
Why this answer is correct
\(3465=3^2\times5\times7\times11\), इसलिए \(27720=2^3\times3^2\times5\times7\times11\)। / \(3465=3^2\times5\times7\times11\), so \(27720=2^3\times3^2\times5\times7\times11\).
Step 3
Exam Tip
3465 को पूरी तरह तोड़ें। / Break 3465 completely.
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संख्या 28224 का अभाज्य गुणनखंडन कौन सा है?
Which is the prime factorisation of 28224?
#prime-factorisation
#number-28224
#hard
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? Hint Small clue
A \(2^6\times3^2\times7^2\)
B \(2^5\times3^3\times7^2\)
C \(64\times441\)
D \(2^6\times21^2\)
Explanation opens after your attempt
Correct Answer
A. \(2^6\times3^2\times7^2\)
Step 1
Concept
\(28224=64\times441\) लिखें। / Write \(28224=64\times441\).
Step 2
Why this answer is correct
\(64=2^6\) और \(441=3^2\times7^2\), इसलिए \(28224=2^6\times3^2\times7^2\)। / \(64=2^6\) and \(441=3^2\times7^2\), so \(28224=2^6\times3^2\times7^2\).
Step 3
Exam Tip
441 को अभाज्य घातों में लिखें। / Write 441 as prime powers.
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संख्या 31104 का सही अभाज्य गुणनखंडन क्या है?
What is the correct prime factorisation of 31104?
#prime-factorisation
#number-31104
#hard
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? Hint Small clue
A \(2^7\times3^5\)
B \(2^6\times3^5\)
C \(128\times243\)
D \(2^7\times81\times3\)
Explanation opens after your attempt
Correct Answer
A. \(2^7\times3^5\)
Step 1
Concept
\(31104=128\times243\) लिखें। / Write \(31104=128\times243\).
Step 2
Why this answer is correct
\(128=2^7\) और \(243=3^5\), इसलिए \(31104=2^7\times3^5\)। / \(128=2^7\) and \(243=3^5\), so \(31104=2^7\times3^5\).
Step 3
Exam Tip
128 और 243 संयुक्त हैं, इसलिए अभाज्य आधार रखें। / 128 and 243 are composite, so keep prime bases.
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संख्या 35280 का अभाज्य गुणनखंडन कौन सा है?
Which is the prime factorisation of 35280?
#prime-factorisation
#number-35280
#hard
50 50-50 2 wrong hide
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? Hint Small clue
A \(2^4\times3^2\times5\times7^2\)
B \(2^3\times3^3\times5\times7^2\)
C \(16\times2205\)
D \(2^4\times45\times49\)
Explanation opens after your attempt
Correct Answer
A. \(2^4\times3^2\times5\times7^2\)
Step 1
Concept
\(35280=16\times2205\) लिखें। / Write \(35280=16\times2205\).
Step 2
Why this answer is correct
\(2205=3^2\times5\times7^2\), इसलिए \(35280=2^4\times3^2\times5\times7^2\)। / \(2205=3^2\times5\times7^2\), so \(35280=2^4\times3^2\times5\times7^2\).
Step 3
Exam Tip
2205 को पूरा अभाज्य रूप दें। / Give 2205 its complete prime form.
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यदि \(27720=2^a\times3^2\times5\times7\times11\), तो (a) का मान क्या है?
If \(27720=2^a\times3^2\times5\times7\times11\), what is the value of (a)?
#exponent-comparison
#number-27720
#hard
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? Hint Small clue
A 3
B 2
C 4
D 5
Explanation opens after your attempt
Step 1
Concept
\(27720=8\times3465\) है। / \(27720=8\times3465\).
Step 2
Why this answer is correct
\(8=2^3\) और \(3465=3^2\times5\times7\times11\), इसलिए 2 की घात 3 है। / \(8=2^3\) and \(3465=3^2\times5\times7\times11\), so the power of 2 is 3.
Step 3
Exam Tip
तुलना करने पर (a=3) है। / Comparing gives (a=3).
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यदि \(31104=2^7\times3^b\), तो (b) का मान क्या है?
If \(31104=2^7\times3^b\), what is the value of (b)?
#exponent-comparison
#number-31104
#hard
50 50-50 2 wrong hide
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+10 Time+ 10 sec extra
? Hint Small clue
A 5
B 4
C 6
D 7
Explanation opens after your attempt
Step 1
Concept
\(31104=128\times243\) लिखें। / Write \(31104=128\times243\).
Step 2
Why this answer is correct
\(128=2^7\) और \(243=3^5\), इसलिए \(31104=2^7\times3^5\)। / \(128=2^7\) and \(243=3^5\), so \(31104=2^7\times3^5\).
Step 3
Exam Tip
दिए गए रूप से तुलना करने पर (b=5) है। / Comparing with the given form gives (b=5).
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यदि \(19600=2^4\times5^m\times7^2\), तो (m) का मान क्या है?
If \(19600=2^4\times5^m\times7^2\), what is the value of (m)?
#exponent-comparison
#square-number
#hard
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? Hint Small clue
A 2
B 1
C 3
D 4
Explanation opens after your attempt
Step 1
Concept
\(19600=140^2\) है। / \(19600=140^2\).
Step 2
Why this answer is correct
\(140=2^2\times5\times7\), इसलिए \(19600=2^4\times5^2\times7^2\)। / \(140=2^2\times5\times7\), so \(19600=2^4\times5^2\times7^2\).
Step 3
Exam Tip
तुलना करने पर (m=2) होगा। / Comparing gives (m=2).
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यदि \(20790=2\times3^p\times5\times7\times11\), तो (p) का मान क्या है?
If \(20790=2\times3^p\times5\times7\times11\), what is the value of (p)?
#exponent-comparison
#number-20790
#hard
50 50-50 2 wrong hide
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? Hint Small clue
A 3
B 2
C 4
D 5
Explanation opens after your attempt
Step 1
Concept
\(20790=54\times385\) लिखें। / Write \(20790=54\times385\).
Step 2
Why this answer is correct
\(54=2\times3^3\) और \(385=5\times7\times11\), इसलिए 3 की घात 3 है। / \(54=2\times3^3\) and \(385=5\times7\times11\), so the power of 3 is 3.
Step 3
Exam Tip
तुलना करने पर (p=3) मिलेगा। / Comparing gives (p=3).
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किस संख्या का अभाज्य गुणनखंडन \(2^3\times3^2\times7\times11\) है?
Which number has prime factorisation \(2^3\times3^2\times7\times11\)?
#evaluate-factorisation
#number-5544
#hard
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? Hint Small clue
A 5544
B 2772
C 11088
D 8316
Explanation opens after your attempt
Step 1
Concept
\(2^3=8\) और \(3^2=9\) निकालें। / Calculate \(2^3=8\) and \(3^2=9\).
Step 2
Why this answer is correct
\(8\times9\times7\times11=5544\)। / \(8\times9\times7\times11=5544\).
Step 3
Exam Tip
घातों को पहले सरल करके गुणा करें। / Simplify powers first and then multiply.
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किस संख्या का अभाज्य गुणनखंडन \(2^5\times3^3\times7\) है?
Which number has prime factorisation \(2^5\times3^3\times7\)?
#evaluate-factorisation
#number-6048
#hard
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A 6048
B 3024
C 12096
D 8064
Explanation opens after your attempt
Step 1
Concept
\(2^5=32\) और \(3^3=27\) निकालें। / Calculate \(2^5=32\) and \(3^3=27\).
Step 2
Why this answer is correct
\(32\times27\times7=6048\)। / \(32\times27\times7=6048\).
Step 3
Exam Tip
बड़ी घातों का मान अलग से निकालें। / Find the values of higher powers separately.
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किस संख्या का अभाज्य गुणनखंडन \(3^2\times5^2\times7^2\) है?
Which number has prime factorisation \(3^2\times5^2\times7^2\)?
#evaluate-factorisation
#square-number
#hard
50 50-50 2 wrong hide
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? Hint Small clue
A 11025
B 3675
C 22050
D 1225
Explanation opens after your attempt
Step 1
Concept
\(3^2=9\), \(5^2=25\) और \(7^2=49\) निकालें। / Calculate \(3^2=9\), \(5^2=25\), and \(7^2=49\).
Step 2
Why this answer is correct
\(9\times25\times49=11025\)। / \(9\times25\times49=11025\).
Step 3
Exam Tip
वर्ग रूप में सभी घातें 2 हैं, इसलिए गुणा ध्यान से करें। / In this square form, all exponents are 2, so multiply carefully.
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किस संख्या का अभाज्य गुणनखंडन \(2^4\times3\times5\times7^2\) है?
Which number has prime factorisation \(2^4\times3\times5\times7^2\)?
#evaluate-factorisation
#number-11760
#hard
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A 11760
B 5880
C 23520
D 17640
Explanation opens after your attempt
Step 1
Concept
\(2^4=16\) और \(7^2=49\) निकालें। / Calculate \(2^4=16\) and \(7^2=49\).
Step 2
Why this answer is correct
\(16\times3\times5\times49=11760\)। / \(16\times3\times5\times49=11760\).
Step 3
Exam Tip
घातों का मान पहले निकालने से गणना आसान होती है। / Calculating powers first makes the work easier.
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किस संख्या का अभाज्य गुणनखंडन \(2^7\times3^5\) है?
Which number has prime factorisation \(2^7\times3^5\)?
#evaluate-factorisation
#number-31104
#hard
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? Hint Small clue
A 31104
B 15552
C 62208
D 20736
Explanation opens after your attempt
Step 1
Concept
\(2^7=128\) और \(3^5=243\) निकालें। / Calculate \(2^7=128\) and \(3^5=243\).
Step 2
Why this answer is correct
\(128\times243=31104\)। / \(128\times243=31104\).
Step 3
Exam Tip
दोनों घातों को पहले सरल करना सही तरीका है। / Simplifying both powers first is the correct method.
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यदि संख्या \(2^5\times3^4\times5^2\times7\) है, तो इसे पूर्ण वर्ग बनाने के लिए सबसे छोटी किस संख्या से गुणा करना होगा?
If the number is \(2^5\times3^4\times5^2\times7\), by which smallest number should it be multiplied to make a perfect square?
#perfect-square
#prime-exponents
#hard
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? Hint Small clue
A 14
B 7
C 2
D 28
Explanation opens after your attempt
Step 1
Concept
पूर्ण वर्ग के लिए सभी घातें सम होनी चाहिए। / For a perfect square, all exponents must be even.
Step 2
Why this answer is correct
2 की घात 5 और 7 की घात 1 विषम हैं। / The powers of 2 and 7 are odd.
Step 3
Exam Tip
\(2\times7=14\) से गुणा करने पर सभी घातें सम हो जाएंगी। / Multiplying by \(2\times7=14\) makes all exponents even.
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यदि संख्या \(2^4\times3^5\times5^2\) है, तो इसे पूर्ण वर्ग बनाने के लिए सबसे छोटी किस संख्या से भाग देना होगा?
If the number is \(2^4\times3^5\times5^2\), by which smallest number should it be divided to make a perfect square?
#perfect-square
#division
#hard
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? Hint Small clue
A 3
B 9
C 15
D 5
Explanation opens after your attempt
Step 1
Concept
पूर्ण वर्ग में सभी घातें सम होती हैं। / In a perfect square, all exponents are even.
Step 2
Why this answer is correct
केवल 3 की घात 5 विषम है। / Only the power of 3 is odd.
Step 3
Exam Tip
3 से भाग देने पर घात 4 हो जाएगी और संख्या पूर्ण वर्ग बन जाएगी। / Dividing by 3 makes the power 4 and the number becomes a perfect square.
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यदि संख्या \(2^4\times3^2\times5^3\times7\) है, तो इसे पूर्ण घन बनाने के लिए सबसे छोटी किस संख्या से गुणा करना होगा?
If the number is \(2^4\times3^2\times5^3\times7\), 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 588
B 147
C 84
D 441
Explanation opens after your attempt
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
2 की घात 4 को 6, 3 की घात 2 को 3 और 7 की घात 1 को 3 बनाना होगा। / Powers 4, 2, and 1 must become 6, 3, and 3 respectively.
Step 3
Exam Tip
गुणक \(2^2\times3\times7^2=588\) होगा। / The multiplier is \(2^2\times3\times7^2=588\).
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यदि संख्या \(2^8\times3^5\times5^4\times11^2\) है, तो इसे पूर्ण घन बनाने के लिए किस सबसे छोटी संख्या से भाग देना होगा?
If the number is \(2^8\times3^5\times5^4\times11^2\), 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 \(2^2\times3^2\times5\times11^2\)
B \(2\times3\times5^2\times11\)
C \(2^2\times3\times5\times11\)
D \(3^2\times5^2\times11^2\)
Explanation opens after your attempt
Correct Answer
A. \(2^2\times3^2\times5\times11^2\)
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
8 को 6, 5 को 3, 4 को 3 और 2 को 0 तक घटाना सबसे छोटा तरीका है। / Reduce 8 to 6, 5 to 3, 4 to 3, and 2 to 0 for the smallest divisor.
Step 3
Exam Tip
इसलिए भाजक \(2^2\times3^2\times5\times11^2\) है। / So the divisor is \(2^2\times3^2\times5\times11^2\).
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यदि \(n=2^6\times3^4\times5^2\times7\), तो (n) किस संख्या से विभाज्य नहीं होगा?
If \(n=2^6\times3^4\times5^2\times7\), by which number will (n) not be divisible?
#divisibility
#prime-exponents
#hard
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? Hint Small clue
A \(3^5\)
B \(2^5\times3^4\)
C \(2^6\times5^2\times7\)
D \(3^2\times5\times7\)
Explanation opens after your attempt
Correct Answer
A. \(3^5\)
Step 1
Concept
किसी भाजक के लिए उसकी हर अभाज्य घात संख्या में पर्याप्त होनी चाहिए। / For divisibility, every prime power of the divisor must be available in the number.
Step 2
Why this answer is correct
(n) में 3 की घात 4 है, पर \(3^5\) के लिए घात 5 चाहिए। / (n) has power 4 of 3, but \(3^5\) needs power 5.
Step 3
Exam Tip
इसलिए (n), \(3^5\) से विभाज्य नहीं होगा। / Therefore, (n) is not divisible by \(3^5\).
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यदि \(n=2^8\times3^3\times5^5\), तो (n) किस संख्या से अवश्य विभाज्य होगा?
If \(n=2^8\times3^3\times5^5\), by which number must (n) be divisible?
#divisibility
#prime-exponents
#hard
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? Hint Small clue
A \(2^7\times3^2\times5^4\)
B \(2^9\times3^2\)
C \(3^4\times5\)
D \(2^8\times5^6\)
Explanation opens after your attempt
Correct Answer
A. \(2^7\times3^2\times5^4\)
Step 1
Concept
भाज्य होने के लिए भाजक की हर घात दी गई संख्या में बराबर या कम होनी चाहिए। / For divisibility, each exponent in the divisor must be less than or equal to the corresponding exponent in the number.
Step 2
Why this answer is correct
\(2^7\), \(3^2\) और \(5^4\) सभी (n) में उपलब्ध हैं। / \(2^7\), \(3^2\), and \(5^4\) are all available in (n).
Step 3
Exam Tip
इसलिए (n), \(2^7\times3^2\times5^4\) से अवश्य विभाज्य होगा। / Therefore, (n) must be divisible by \(2^7\times3^2\times5^4\).
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यदि किसी संख्या का अभाज्य गुणनखंडन \(2^4\times3^2\times5^2\times7^2\) है, तो उसके कुल अभाज्य गुणनखंडों की संख्या कितनी है, यदि दोहराव गिना जाए?
If a number has prime factorisation \(2^4\times3^2\times5^2\times7^2\), how many prime factors does it have if repetition is counted?
#counting-prime-factors
#prime-exponents
#hard
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? Hint Small clue
A 10
B 4
C 8
D 12
Explanation opens after your attempt
Step 1
Concept
दोहराव सहित गिनने के लिए घातों को जोड़ते हैं। / To count with repetition, add the exponents.
Step 2
Why this answer is correct
(4+2+2+2=10)। / (4+2+2+2=10).
Step 3
Exam Tip
अलग-अलग अभाज्य गिनने और दोहराव सहित गिनने में अंतर रखें। / Keep the difference between distinct prime count and repeated prime count clear.
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यदि किसी संख्या का अभाज्य गुणनखंडन \(2^7\times3^5\times11^2\times13\) है, तो उसमें अलग-अलग अभाज्य गुणनखंड कितने हैं?
If a number has prime factorisation \(2^7\times3^5\times11^2\times13\), how many distinct prime factors does it have?
#distinct-prime-factors
#prime-exponents
#hard
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? Hint Small clue
A 4
B 15
C 7
D 5
Explanation opens after your attempt
Step 1
Concept
अलग-अलग अभाज्य गुणनखंड गिनते समय घात नहीं जोड़ते। / While counting distinct prime factors, do not add exponents.
Step 2
Why this answer is correct
अभाज्य आधार 2, 3, 11 और 13 हैं। / The prime bases are 2, 3, 11, and 13.
Step 3
Exam Tip
इसलिए अलग-अलग अभाज्य गुणनखंडों की संख्या 4 है। / Therefore, the number of distinct prime factors is 4.
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यदि \(a=2^5\times3^2\times7\) और \(b=2^3\times3^4\times5\), तो (ab) में 3 की घात क्या होगी?
If \(a=2^5\times3^2\times7\) and \(b=2^3\times3^4\times5\), what will be the power of 3 in (ab)?
#product-factorisation
#powers
#hard
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? Hint Small clue
A 6
B 4
C 2
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 2 है और (b) में 3 की घात 4 है। / The power of 3 in (a) is 2 and in (b) is 4.
Step 3
Exam Tip
(ab) में 3 की घात (2+4=6) होगी। / In (ab), the power of 3 will be (2+4=6).
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यदि \(x=2^8\times5^3\times11\) और \(y=2^4\times3^2\times5^5\), तो (xy) में 5 की घात क्या होगी?
If \(x=2^8\times5^3\times11\) and \(y=2^4\times3^2\times5^5\), what will be the power of 5 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 8
B 5
C 3
D 10
Explanation opens after your attempt
Step 1
Concept
समान आधार 5 की घातें गुणा में जुड़ेंगी। / Powers with the same base 5 are added in multiplication.
Step 2
Why this answer is correct
(x) में 5 की घात 3 है और (y) में 5 की घात 5 है। / The power of 5 in (x) is 3 and in (y) is 5.
Step 3
Exam Tip
कुल घात (3+5=8) होगी। / The total power will be (3+5=8).
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किस विकल्प में \(2^4\times3^3\times7^2\) का सही मान है?
Which option gives the correct value of \(2^4\times3^3\times7^2\)?
#evaluate-factorisation
#number-21168
#hard
50 50-50 2 wrong hide
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+10 Time+ 10 sec extra
? Hint Small clue
A 21168
B 10584
C 28224
D 14112
Explanation opens after your attempt
Step 1
Concept
\(2^4=16\), \(3^3=27\) और \(7^2=49\) निकालें। / Calculate \(2^4=16\), \(3^3=27\), and \(7^2=49\).
Step 2
Why this answer is correct
\(16\times27\times49=21168\)। / \(16\times27\times49=21168\).
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 \(2^6\times3^2\times7^2\) का सही मान है?
Which option gives the correct value of \(2^6\times3^2\times7^2\)?
#evaluate-factorisation
#number-28224
#hard
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A 28224
B 14112
C 56448
D 21168
Explanation opens after your attempt
Step 1
Concept
\(2^6=64\), \(3^2=9\) और \(7^2=49\) निकालें। / Calculate \(2^6=64\), \(3^2=9\), and \(7^2=49\).
Step 2
Why this answer is correct
\(64\times9\times49=28224\)। / \(64\times9\times49=28224\).
Step 3
Exam Tip
गुणा को चरणों में करें ताकि गलती न हो। / Do multiplication step by step to avoid mistakes.
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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 \(2^4\times3^2\times5\times7^2\)
B \(16\times45\times49\)
C \(2^4\times45\times49\)
D \(8\times4410\)
Explanation opens after your attempt
Correct Answer
A. \(2^4\times3^2\times5\times7^2\)
Step 1
Concept
अंतिम रूप में आधार अभाज्य होने चाहिए। / In the final form, bases must be prime.
Step 2
Why this answer is correct
पहले विकल्प में आधार 2, 3, 5 और 7 हैं, जो अभाज्य हैं। / In the first option, bases 2, 3, 5, and 7 are prime.
Step 3
Exam Tip
16, 45, 49 और 4410 संयुक्त हैं, इसलिए वे अंतिम रूप नहीं हैं। / 16, 45, 49, and 4410 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 \(2^5\times21^2\)
B \(2^5\times3^2\times7^2\)
C \(2^3\times3^2\times5\times7\times11\)
D \(3^2\times5^2\times7^2\)
Explanation opens after your attempt
Correct Answer
A. \(2^5\times21^2\)
Step 1
Concept
अधूरे रूप में कोई संयुक्त आधार बचा रहता है। / In an incomplete form, a composite base remains.
Step 2
Why this answer is correct
21 संयुक्त है और \(21=3\times7\), इसलिए \(2^5\times21^2\) अंतिम नहीं है। / 21 is composite and \(21=3\times7\), so \(2^5\times21^2\) is not final.
Step 3
Exam Tip
इसे \(2^5\times3^2\times7^2\) में बदलें। / Change it into \(2^5\times3^2\times7^2\).
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यदि \(2^6\times3^2\times7^2\) किसी संख्या का अभाज्य गुणनखंडन है, तो संख्या क्या है?
If \(2^6\times3^2\times7^2\) is the prime factorisation of a number, what is the number?
#evaluate-factorisation
#hard
#mcq
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A 28224
B 14112
C 56448
D 21168
Explanation opens after your attempt
Step 1
Concept
\(2^6=64\), \(3^2=9\) और \(7^2=49\) हैं। / \(2^6=64\), \(3^2=9\), and \(7^2=49\).
Step 2
Why this answer is correct
\(64\times9\times49=28224\)। / \(64\times9\times49=28224\).
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 \(2^4\times5^2\times7^2\) किसी संख्या का अभाज्य गुणनखंडन है, तो संख्या क्या है?
If \(2^4\times5^2\times7^2\) is the prime factorisation of a number, what is the number?
#evaluate-factorisation
#square-number
#hard
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A 19600
B 9800
C 39200
D 4900
Explanation opens after your attempt
Step 1
Concept
\(2^4=16\), \(5^2=25\) और \(7^2=49\) निकालें। / Calculate \(2^4=16\), \(5^2=25\), and \(7^2=49\).
Step 2
Why this answer is correct
\(16\times25\times49=19600\)। / \(16\times25\times49=19600\).
Step 3
Exam Tip
तीनों घातों का मान अलग-अलग निकालना सुरक्षित रहता है। / Finding all three powers separately is safer.
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संख्या 39200 का सही अभाज्य गुणनखंडन क्या है?
What is the correct prime factorisation of 39200?
#prime-factorisation
#number-39200
#hard
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A \(2^5\times5^2\times7^2\)
B \(2^4\times5^2\times7^2\)
C \(32\times1225\)
D \(2^5\times35^2\)
Explanation opens after your attempt
Correct Answer
A. \(2^5\times5^2\times7^2\)
Step 1
Concept
\(39200=32\times1225\) लिखें। / Write \(39200=32\times1225\).
Step 2
Why this answer is correct
\(32=2^5\) और \(1225=5^2\times7^2\), इसलिए \(39200=2^5\times5^2\times7^2\)। / \(32=2^5\) and \(1225=5^2\times7^2\), so \(39200=2^5\times5^2\times7^2\).
Step 3
Exam Tip
1225 को अभाज्य घातों में बदलें। / Convert 1225 into prime powers.
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संख्या 41580 का अभाज्य गुणनखंडन कौन सा है?
Which is the prime factorisation of 41580?
#prime-factorisation
#number-41580
#hard
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A \(2^2\times3^3\times5\times7\times11\)
B \(2\times3^3\times5\times7\times11\)
C \(4\times10395\)
D \(36\times1155\)
Explanation opens after your attempt
Correct Answer
A. \(2^2\times3^3\times5\times7\times11\)
Step 1
Concept
\(41580=4\times10395\) लिखें। / Write \(41580=4\times10395\).
Step 2
Why this answer is correct
\(10395=3^3\times5\times7\times11\), इसलिए \(41580=2^2\times3^3\times5\times7\times11\)। / \(10395=3^3\times5\times7\times11\), so \(41580=2^2\times3^3\times5\times7\times11\).
Step 3
Exam Tip
10395 को पूरा अभाज्य रूप दें। / Give 10395 its complete prime form.
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संख्या 62208 का सही अभाज्य गुणनखंडन क्या है?
What is the correct prime factorisation of 62208?
#prime-factorisation
#number-62208
#hard
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A \(2^8\times3^5\)
B \(2^7\times3^5\)
C \(256\times243\)
D \(2^8\times81\times3\)
Explanation opens after your attempt
Correct Answer
A. \(2^8\times3^5\)
Step 1
Concept
\(62208=256\times243\) लिखें। / Write \(62208=256\times243\).
Step 2
Why this answer is correct
\(256=2^8\) और \(243=3^5\), इसलिए \(62208=2^8\times3^5\)। / \(256=2^8\) and \(243=3^5\), so \(62208=2^8\times3^5\).
Step 3
Exam Tip
256 और 243 को अभाज्य घातों में लिखें। / Write 256 and 243 as prime powers.
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अभाज्य गुणनखंडन में \(8\times3465\) को अंतिम उत्तर क्यों नहीं माना जाएगा?
Why will \(8\times3465\) not be considered the final answer in prime factorisation?
#prime-factorisation
#concept-check
#hard
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A क्योंकि 8 और 3465 संयुक्त रूप हैं / Because 8 and 3465 are composite forms
B क्योंकि 8 अभाज्य है / Because 8 is prime
C क्योंकि 3465 को तोड़ा नहीं जा सकता / Because 3465 cannot be factorised
D क्योंकि गुणनफल बदल जाता है / Because the product changes
Explanation opens after your attempt
Correct Answer
A. क्योंकि 8 और 3465 संयुक्त रूप हैं / Because 8 and 3465 are composite forms
Step 1
Concept
अभाज्य गुणनखंडन में हर अंतिम गुणनखंड अभाज्य आधार में होना चाहिए। / In prime factorisation, every final factor should be in prime-base form.
Step 2
Why this answer is correct
\(8=2^3\) और \(3465=3^2\times5\times7\times11\) है। / \(8=2^3\) and \(3465=3^2\times5\times7\times11\).
Step 3
Exam Tip
इसलिए अंतिम रूप \(2^3\times3^2\times5\times7\times11\) होगा। / Therefore, the final form is \(2^3\times3^2\times5\times7\times11\).
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