अभाज्य गुणनखंडन में किसी संयुक्त आधार को अंतिम रूप में रखने से क्या समस्या होती है?
What problem occurs if a composite base is kept in the final prime factorisation?
#prime-factorisation
#concept
#medium
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A गुणनखंडन पूरा नहीं माना जाता / The factorisation is not considered complete
B संख्या का मान हमेशा बदल जाता है / The value of the number always changes
C सभी गुणनखंड अभाज्य बन जाते हैं / All factors become prime
D घातों का उपयोग नहीं किया जा सकता / Powers cannot be used
Explanation opens after your attempt
Correct Answer
A. गुणनखंडन पूरा नहीं माना जाता / The factorisation is not considered complete
Step 1
Concept
अंतिम अभाज्य गुणनखंडन में हर आधार अभाज्य होना चाहिए। / In final prime factorisation, every base must be prime.
Step 2
Why this answer is correct
यदि 12 या 21 जैसा संयुक्त आधार बचा है, तो उसे आगे तोड़ना होगा। / If a composite base like 12 or 21 remains, it must be broken further.
Step 3
Exam Tip
परीक्षा में अंतिम रूप देने से पहले हर आधार की जांच करें। / In exams, check every base before writing the final form.
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संख्या 792 का सही अभाज्य गुणनखंडन क्या है?
What is the correct prime factorisation of 792?
#prime-factorisation
#number-792
#medium
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A \(2^3\times3^2\times11\)
B \(2^2\times3^3\times11\)
C \(8\times99\)
D \(2^3\times9\times11\)
Explanation opens after your attempt
Correct Answer
A. \(2^3\times3^2\times11\)
Step 1
Concept
\(792=8\times99\) लिखें। / Write \(792=8\times99\).
Step 2
Why this answer is correct
\(8=2^3\) और \(99=3^2\times11\), इसलिए \(792=2^3\times3^2\times11\)। / \(8=2^3\) and \(99=3^2\times11\), so \(792=2^3\times3^2\times11\).
Step 3
Exam Tip
99 को अंतिम उत्तर में न छोड़ें। / Do not leave 99 in the final answer.
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संख्या 825 का अभाज्य गुणनखंडन कौन सा है?
Which is the prime factorisation of 825?
#prime-factorisation
#number-825
#medium
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A \(3\times5^2\times11\)
B \(3^2\times5\times11\)
C \(25\times33\)
D \(5\times165\)
Explanation opens after your attempt
Correct Answer
A. \(3\times5^2\times11\)
Step 1
Concept
\(825=25\times33\) लिखें। / Write \(825=25\times33\).
Step 2
Why this answer is correct
\(25=5^2\) और \(33=3\times11\), इसलिए \(825=3\times5^2\times11\)। / \(25=5^2\) and \(33=3\times11\), so \(825=3\times5^2\times11\).
Step 3
Exam Tip
25 और 33 दोनों संयुक्त हैं, इसलिए उन्हें अभाज्य रूप में बदलें। / Both 25 and 33 are composite, so convert them into prime form.
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संख्या 864 का सही अभाज्य गुणनखंडन क्या है?
What is the correct prime factorisation of 864?
#prime-factorisation
#number-864
#medium
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A \(2^5\times3^3\)
B \(2^4\times3^3\)
C \(32\times27\)
D \(2^5\times27\)
Explanation opens after your attempt
Correct Answer
A. \(2^5\times3^3\)
Step 1
Concept
\(864=32\times27\) लिखें। / Write \(864=32\times27\).
Step 2
Why this answer is correct
\(32=2^5\) और \(27=3^3\), इसलिए \(864=2^5\times3^3\)। / \(32=2^5\) and \(27=3^3\), so \(864=2^5\times3^3\).
Step 3
Exam Tip
32 और 27 को अभाज्य घातों में बदलना जरूरी है। / It is necessary to convert 32 and 27 into prime powers.
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संख्या 924 का अभाज्य गुणनखंडन कौन सा है?
Which is the prime factorisation of 924?
#prime-factorisation
#number-924
#medium
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A \(2^2\times3\times7\times11\)
B \(2\times3\times7\times22\)
C \(4\times231\)
D \(2^2\times21\times11\)
Explanation opens after your attempt
Correct Answer
A. \(2^2\times3\times7\times11\)
Step 1
Concept
\(924=4\times231\) लिखें। / Write \(924=4\times231\).
Step 2
Why this answer is correct
\(4=2^2\) और \(231=3\times7\times11\), इसलिए \(924=2^2\times3\times7\times11\)। / \(4=2^2\) and \(231=3\times7\times11\), so \(924=2^2\times3\times7\times11\).
Step 3
Exam Tip
231 को पूरा अभाज्य रूप दें। / Give 231 its complete prime form.
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संख्या 1050 का सही अभाज्य गुणनखंडन क्या है?
What is the correct prime factorisation of 1050?
#prime-factorisation
#number-1050
#medium
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A \(2\times3\times5^2\times7\)
B \(2^2\times3\times5\times7\)
C \(105\times10\)
D \(2\times15\times35\)
Explanation opens after your attempt
Correct Answer
A. \(2\times3\times5^2\times7\)
Step 1
Concept
\(1050=105\times10\) लिखें। / Write \(1050=105\times10\).
Step 2
Why this answer is correct
\(105=3\times5\times7\) और \(10=2\times5\), इसलिए \(1050=2\times3\times5^2\times7\)। / \(105=3\times5\times7\) and \(10=2\times5\), so \(1050=2\times3\times5^2\times7\).
Step 3
Exam Tip
5 दो बार आता है, इसलिए \(5^2\) लिखें। / Since 5 appears twice, write \(5^2\).
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संख्या 1188 का अभाज्य गुणनखंडन कौन सा है?
Which is the prime factorisation of 1188?
#prime-factorisation
#number-1188
#medium
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A \(2^2\times3^3\times11\)
B \(2^3\times3^2\times11\)
C \(4\times297\)
D \(2^2\times27\times11\)
Explanation opens after your attempt
Correct Answer
A. \(2^2\times3^3\times11\)
Step 1
Concept
\(1188=4\times297\) लिखें। / Write \(1188=4\times297\).
Step 2
Why this answer is correct
\(4=2^2\) और \(297=3^3\times11\), इसलिए \(1188=2^2\times3^3\times11\)। / \(4=2^2\) and \(297=3^3\times11\), so \(1188=2^2\times3^3\times11\).
Step 3
Exam Tip
297 को 3 और 11 के रूप में तोड़ें। / Break 297 into 3 and 11 form.
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संख्या 1260 का सही अभाज्य गुणनखंडन क्या है?
What is the correct prime factorisation of 1260?
#prime-factorisation
#number-1260
#medium
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A \(2^2\times3^2\times5\times7\)
B \(2\times3^2\times5\times7\)
C \(126\times10\)
D \(2^2\times9\times35\)
Explanation opens after your attempt
Correct Answer
A. \(2^2\times3^2\times5\times7\)
Step 1
Concept
\(1260=126\times10\) लिखें। / Write \(1260=126\times10\).
Step 2
Why this answer is correct
\(126=2\times3^2\times7\) और \(10=2\times5\), इसलिए \(1260=2^2\times3^2\times5\times7\)। / \(126=2\times3^2\times7\) and \(10=2\times5\), so \(1260=2^2\times3^2\times5\times7\).
Step 3
Exam Tip
2 की कुल घात 2 है। / The total power of 2 is 2.
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संख्या 1350 का अभाज्य गुणनखंडन कौन सा है?
Which is the prime factorisation of 1350?
#prime-factorisation
#number-1350
#medium
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A \(2\times3^3\times5^2\)
B \(2\times3^2\times5^2\)
C \(27\times50\)
D \(2\times27\times25\)
Explanation opens after your attempt
Correct Answer
A. \(2\times3^3\times5^2\)
Step 1
Concept
\(1350=27\times50\) लिखें। / Write \(1350=27\times50\).
Step 2
Why this answer is correct
\(27=3^3\) और \(50=2\times5^2\), इसलिए \(1350=2\times3^3\times5^2\)। / \(27=3^3\) and \(50=2\times5^2\), so \(1350=2\times3^3\times5^2\).
Step 3
Exam Tip
27 और 50 को अंतिम रूप में न छोड़ें। / Do not leave 27 and 50 in the final form.
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संख्या 1386 का सही अभाज्य गुणनखंडन क्या है?
What is the correct prime factorisation of 1386?
#prime-factorisation
#number-1386
#medium
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A \(2\times3^2\times7\times11\)
B \(2^2\times3\times7\times11\)
C \(2\times693\)
D \(18\times77\)
Explanation opens after your attempt
Correct Answer
A. \(2\times3^2\times7\times11\)
Step 1
Concept
\(1386=18\times77\) लिखें। / Write \(1386=18\times77\).
Step 2
Why this answer is correct
\(18=2\times3^2\) और \(77=7\times11\), इसलिए \(1386=2\times3^2\times7\times11\)। / \(18=2\times3^2\) and \(77=7\times11\), so \(1386=2\times3^2\times7\times11\).
Step 3
Exam Tip
18 और 77 दोनों को अभाज्य रूप दें। / Give prime form to both 18 and 77.
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संख्या 1500 का अभाज्य गुणनखंडन कौन सा है?
Which is the prime factorisation of 1500?
#prime-factorisation
#number-1500
#medium
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A \(2^2\times3\times5^3\)
B \(2^3\times3\times5^2\)
C \(15\times100\)
D \(3\times500\)
Explanation opens after your attempt
Correct Answer
A. \(2^2\times3\times5^3\)
Step 1
Concept
\(1500=15\times100\) लिखें। / Write \(1500=15\times100\).
Step 2
Why this answer is correct
\(15=3\times5\) और \(100=2^2\times5^2\), इसलिए \(1500=2^2\times3\times5^3\)। / \(15=3\times5\) and \(100=2^2\times5^2\), so \(1500=2^2\times3\times5^3\).
Step 3
Exam Tip
5 की कुल घात 3 गिनें। / Count the total power of 5 as 3.
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संख्या 1584 का सही अभाज्य गुणनखंडन क्या है?
What is the correct prime factorisation of 1584?
#prime-factorisation
#number-1584
#medium
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A \(2^4\times3^2\times11\)
B \(2^3\times3^2\times11\)
C \(16\times99\)
D \(2^4\times9\times11\)
Explanation opens after your attempt
Correct Answer
A. \(2^4\times3^2\times11\)
Step 1
Concept
\(1584=16\times99\) लिखें। / Write \(1584=16\times99\).
Step 2
Why this answer is correct
\(16=2^4\) और \(99=3^2\times11\), इसलिए \(1584=2^4\times3^2\times11\)। / \(16=2^4\) and \(99=3^2\times11\), so \(1584=2^4\times3^2\times11\).
Step 3
Exam Tip
99 को अंतिम रूप में न छोड़ें। / Do not leave 99 in the final form.
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संख्या 1680 का अभाज्य गुणनखंडन कौन सा है?
Which is the prime factorisation of 1680?
#prime-factorisation
#number-1680
#medium
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A \(2^4\times3\times5\times7\)
B \(2^3\times3\times5\times7\)
C \(168\times10\)
D \(16\times105\)
Explanation opens after your attempt
Correct Answer
A. \(2^4\times3\times5\times7\)
Step 1
Concept
\(1680=16\times105\) लिखें। / Write \(1680=16\times105\).
Step 2
Why this answer is correct
\(16=2^4\) और \(105=3\times5\times7\), इसलिए \(1680=2^4\times3\times5\times7\)। / \(16=2^4\) and \(105=3\times5\times7\), so \(1680=2^4\times3\times5\times7\).
Step 3
Exam Tip
105 को अभाज्य गुणनखंडों में बदलें। / Change 105 into prime factors.
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संख्या 1890 का सही अभाज्य गुणनखंडन क्या है?
What is the correct prime factorisation of 1890?
#prime-factorisation
#number-1890
#medium
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A \(2\times3^3\times5\times7\)
B \(2\times3^2\times5\times7\)
C \(27\times70\)
D \(2\times27\times35\)
Explanation opens after your attempt
Correct Answer
A. \(2\times3^3\times5\times7\)
Step 1
Concept
\(1890=27\times70\) लिखें। / Write \(1890=27\times70\).
Step 2
Why this answer is correct
\(27=3^3\) और \(70=2\times5\times7\), इसलिए \(1890=2\times3^3\times5\times7\)। / \(27=3^3\) and \(70=2\times5\times7\), so \(1890=2\times3^3\times5\times7\).
Step 3
Exam Tip
27 को \(3^3\) में बदलें। / Change 27 into \(3^3\).
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संख्या 2100 का अभाज्य गुणनखंडन कौन सा है?
Which is the prime factorisation of 2100?
#prime-factorisation
#number-2100
#medium
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A \(2^2\times3\times5^2\times7\)
B \(2\times3\times5^2\times7\)
C \(21\times100\)
D \(2^2\times3\times25\times7\)
Explanation opens after your attempt
Correct Answer
A. \(2^2\times3\times5^2\times7\)
Step 1
Concept
\(2100=21\times100\) लिखें। / Write \(2100=21\times100\).
Step 2
Why this answer is correct
\(21=3\times7\) और \(100=2^2\times5^2\), इसलिए \(2100=2^2\times3\times5^2\times7\)। / \(21=3\times7\) and \(100=2^2\times5^2\), so \(2100=2^2\times3\times5^2\times7\).
Step 3
Exam Tip
25 या 100 को अंतिम उत्तर में न रखें। / Do not keep 25 or 100 in the final answer.
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संख्या 2310 का सही अभाज्य गुणनखंडन क्या है?
What is the correct prime factorisation of 2310?
#prime-factorisation
#number-2310
#medium
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? Hint Small clue
A \(2\times3\times5\times7\times11\)
B \(21\times110\)
C \(2\times3\times5\times77\)
D \(30\times77\)
Explanation opens after your attempt
Correct Answer
A. \(2\times3\times5\times7\times11\)
Step 1
Concept
\(2310=30\times77\) लिखें। / Write \(2310=30\times77\).
Step 2
Why this answer is correct
\(30=2\times3\times5\) और \(77=7\times11\), इसलिए \(2310=2\times3\times5\times7\times11\)। / \(30=2\times3\times5\) and \(77=7\times11\), so \(2310=2\times3\times5\times7\times11\).
Step 3
Exam Tip
30 और 77 दोनों को पूरा तोड़ें। / Break both 30 and 77 completely.
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संख्या 2520 का अभाज्य गुणनखंडन कौन सा है?
Which is the prime factorisation of 2520?
#prime-factorisation
#number-2520
#medium
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? Hint Small clue
A \(2^3\times3^2\times5\times7\)
B \(2^2\times3^2\times5\times7\)
C \(252\times10\)
D \(8\times315\)
Explanation opens after your attempt
Correct Answer
A. \(2^3\times3^2\times5\times7\)
Step 1
Concept
\(2520=252\times10\) लिखें। / Write \(2520=252\times10\).
Step 2
Why this answer is correct
\(252=2^2\times3^2\times7\) और \(10=2\times5\), इसलिए \(2520=2^3\times3^2\times5\times7\)। / \(252=2^2\times3^2\times7\) and \(10=2\times5\), so \(2520=2^3\times3^2\times5\times7\).
Step 3
Exam Tip
2 की कुल घात को सही गिनें। / Count the total power of 2 correctly.
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संख्या 2772 का सही अभाज्य गुणनखंडन क्या है?
What is the correct prime factorisation of 2772?
#prime-factorisation
#number-2772
#medium
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? Hint Small clue
A \(2^2\times3^2\times7\times11\)
B \(2\times3^2\times7\times11\)
C \(36\times77\)
D \(4\times693\)
Explanation opens after your attempt
Correct Answer
A. \(2^2\times3^2\times7\times11\)
Step 1
Concept
\(2772=36\times77\) लिखें। / Write \(2772=36\times77\).
Step 2
Why this answer is correct
\(36=2^2\times3^2\) और \(77=7\times11\), इसलिए \(2772=2^2\times3^2\times7\times11\)। / \(36=2^2\times3^2\) and \(77=7\times11\), so \(2772=2^2\times3^2\times7\times11\).
Step 3
Exam Tip
36 और 77 को अभाज्य रूप में लिखें। / Write 36 and 77 in prime form.
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संख्या 3024 का अभाज्य गुणनखंडन कौन सा है?
Which is the prime factorisation of 3024?
#prime-factorisation
#number-3024
#medium
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? Hint Small clue
A \(2^4\times3^3\times7\)
B \(2^3\times3^3\times7\)
C \(16\times189\)
D \(2^4\times27\times7\)
Explanation opens after your attempt
Correct Answer
A. \(2^4\times3^3\times7\)
Step 1
Concept
\(3024=16\times189\) लिखें। / Write \(3024=16\times189\).
Step 2
Why this answer is correct
\(16=2^4\) और \(189=3^3\times7\), इसलिए \(3024=2^4\times3^3\times7\)। / \(16=2^4\) and \(189=3^3\times7\), so \(3024=2^4\times3^3\times7\).
Step 3
Exam Tip
189 को पूरा अभाज्य रूप दें। / Give 189 its complete prime form.
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संख्या 3150 का सही अभाज्य गुणनखंडन क्या है?
What is the correct prime factorisation of 3150?
#prime-factorisation
#number-3150
#medium
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? Hint Small clue
A \(2\times3^2\times5^2\times7\)
B \(2^2\times3^2\times5\times7\)
C \(315\times10\)
D \(2\times9\times25\times7\)
Explanation opens after your attempt
Correct Answer
A. \(2\times3^2\times5^2\times7\)
Step 1
Concept
\(3150=315\times10\) लिखें। / Write \(3150=315\times10\).
Step 2
Why this answer is correct
\(315=3^2\times5\times7\) और \(10=2\times5\), इसलिए \(3150=2\times3^2\times5^2\times7\)। / \(315=3^2\times5\times7\) and \(10=2\times5\), so \(3150=2\times3^2\times5^2\times7\).
Step 3
Exam Tip
5 की घात 2 बनती है। / The power of 5 becomes 2.
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संख्या 3360 का अभाज्य गुणनखंडन कौन सा है?
Which is the prime factorisation of 3360?
#prime-factorisation
#number-3360
#medium
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A \(2^5\times3\times5\times7\)
B \(2^4\times3\times5\times7\)
C \(32\times105\)
D \(2^5\times15\times7\)
Explanation opens after your attempt
Correct Answer
A. \(2^5\times3\times5\times7\)
Step 1
Concept
\(3360=32\times105\) लिखें। / Write \(3360=32\times105\).
Step 2
Why this answer is correct
\(32=2^5\) और \(105=3\times5\times7\), इसलिए \(3360=2^5\times3\times5\times7\)। / \(32=2^5\) and \(105=3\times5\times7\), so \(3360=2^5\times3\times5\times7\).
Step 3
Exam Tip
105 को अंतिम रूप में न रखें। / Do not keep 105 in the final form.
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यदि \(1260=2^a\times3^2\times5\times7\), तो (a) का मान क्या है?
If \(1260=2^a\times3^2\times5\times7\), what is the value of (a)?
#exponent-comparison
#prime-factorisation
#medium
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A 2
B 1
C 3
D 4
Explanation opens after your attempt
Step 1
Concept
\(1260=126\times10\) लिखें। / Write \(1260=126\times10\).
Step 2
Why this answer is correct
\(126=2\times3^2\times7\) और \(10=2\times5\), इसलिए 2 की कुल घात 2 है। / \(126=2\times3^2\times7\) and \(10=2\times5\), so the total power of 2 is 2.
Step 3
Exam Tip
इसलिए (a=2) होगा। / Therefore, (a=2).
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यदि \(1350=2\times3^b\times5^2\), तो (b) का मान क्या है?
If \(1350=2\times3^b\times5^2\), what is the value of (b)?
#exponent-comparison
#number-1350
#medium
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A 3
B 2
C 4
D 1
Explanation opens after your attempt
Step 1
Concept
\(1350=27\times50\) है। / \(1350=27\times50\).
Step 2
Why this answer is correct
\(27=3^3\) और \(50=2\times5^2\), इसलिए \(1350=2\times3^3\times5^2\)। / \(27=3^3\) and \(50=2\times5^2\), so \(1350=2\times3^3\times5^2\).
Step 3
Exam Tip
तुलना करने पर (b=3) है। / Comparing gives (b=3).
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यदि \(1500=2^p\times3\times5^3\), तो (p) का मान क्या है?
If \(1500=2^p\times3\times5^3\), what is the value of (p)?
#exponent-comparison
#number-1500
#medium
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A 2
B 1
C 3
D 4
Explanation opens after your attempt
Step 1
Concept
\(1500=15\times100\) लिखें। / Write \(1500=15\times100\).
Step 2
Why this answer is correct
\(15=3\times5\) और \(100=2^2\times5^2\), इसलिए 2 की घात 2 है। / \(15=3\times5\) and \(100=2^2\times5^2\), so the power of 2 is 2.
Step 3
Exam Tip
तुलना करने पर (p=2) होगा। / Comparing gives (p=2).
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यदि \(2100=2^2\times3\times5^q\times7\), तो (q) का मान क्या है?
If \(2100=2^2\times3\times5^q\times7\), what is the value of (q)?
#exponent-comparison
#number-2100
#medium
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A 2
B 1
C 3
D 4
Explanation opens after your attempt
Step 1
Concept
\(2100=21\times100\) है। / \(2100=21\times100\).
Step 2
Why this answer is correct
\(21=3\times7\) और \(100=2^2\times5^2\), इसलिए 5 की घात 2 है। / \(21=3\times7\) and \(100=2^2\times5^2\), so the power of 5 is 2.
Step 3
Exam Tip
इसलिए (q=2) होगा। / Therefore, (q=2).
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किस संख्या का अभाज्य गुणनखंडन \(2^3\times3^2\times11\) है?
Which number has prime factorisation \(2^3\times3^2\times11\)?
#evaluate-factorisation
#number-792
#medium
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A 792
B 396
C 1584
D 1188
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
\(8\times9\times11=792\)। / \(8\times9\times11=792\).
Step 3
Exam Tip
अभाज्य घातों को पहले हल करें, फिर गुणा करें। / Solve prime powers first, then multiply.
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किस संख्या का अभाज्य गुणनखंडन \(2^4\times3\times5\times7\) है?
Which number has prime factorisation \(2^4\times3\times5\times7\)?
#evaluate-factorisation
#number-1680
#medium
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A 1680
B 840
C 3360
D 1260
Explanation opens after your attempt
Step 1
Concept
\(2^4=16\) निकालें। / Calculate \(2^4=16\).
Step 2
Why this answer is correct
\(16\times3\times5\times7=1680\)। / \(16\times3\times5\times7=1680\).
Step 3
Exam Tip
चार गुणनखंड हों तो क्रम से छोटे गुणन करें। / When there are four factors, multiply smaller products in order.
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किस संख्या का अभाज्य गुणनखंडन \(2\times3^3\times5\times7\) है?
Which number has prime factorisation \(2\times3^3\times5\times7\)?
#evaluate-factorisation
#number-1890
#medium
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A 1890
B 945
C 3780
D 1350
Explanation opens after your attempt
Step 1
Concept
\(3^3=27\) निकालें। / Calculate \(3^3=27\).
Step 2
Why this answer is correct
\(2\times27\times5\times7=1890\)। / \(2\times27\times5\times7=1890\).
Step 3
Exam Tip
घात का मान पहले निकालना गणना को सरल बनाता है। / Finding the value of the power first makes calculation simple.
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किस संख्या का अभाज्य गुणनखंडन \(2^2\times3^2\times7\times11\) है?
Which number has prime factorisation \(2^2\times3^2\times7\times11\)?
#evaluate-factorisation
#number-2772
#medium
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A 2772
B 1386
C 5544
D 1980
Explanation opens after your attempt
Step 1
Concept
\(2^2=4\) और \(3^2=9\) निकालें। / Calculate \(2^2=4\) and \(3^2=9\).
Step 2
Why this answer is correct
\(4\times9\times7\times11=2772\)। / \(4\times9\times7\times11=2772\).
Step 3
Exam Tip
पहले 4 और 9 को गुणा करें, फिर बाकी गुणनखंड जोड़ें। / First multiply 4 and 9, then include the remaining factors.
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किस संख्या का अभाज्य गुणनखंडन \(2^4\times3^3\times7\) है?
Which number has prime factorisation \(2^4\times3^3\times7\)?
#evaluate-factorisation
#number-3024
#medium
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A 3024
B 1512
C 6048
D 2016
Explanation opens after your attempt
Step 1
Concept
\(2^4=16\) और \(3^3=27\) निकालें। / Calculate \(2^4=16\) and \(3^3=27\).
Step 2
Why this answer is correct
\(16\times27\times7=3024\)। / \(16\times27\times7=3024\).
Step 3
Exam Tip
बड़ी घातों को अलग से सरल करके गुणा करें। / Simplify higher powers separately and multiply.
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यदि \(n=2^2\times3\times5^2\times7\), तो (n) का मान क्या है?
If \(n=2^2\times3\times5^2\times7\), what is the value of (n)?
#evaluate-factorisation
#powers
#medium
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A 2100
B 1050
C 4200
D 700
Explanation opens after your attempt
Step 1
Concept
\(2^2=4\) और \(5^2=25\) हैं। / \(2^2=4\) and \(5^2=25\).
Step 2
Why this answer is correct
\(4\times3\times25\times7=2100\)। / \(4\times3\times25\times7=2100\).
Step 3
Exam Tip
दो घातों को पहले हल करने से गुणा आसान होता है। / Solving the powers first makes multiplication easier.
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यदि \(m=2^3\times3^2\times5\times7\), तो (m) का मान क्या है?
If \(m=2^3\times3^2\times5\times7\), what is the value of (m)?
#evaluate-factorisation
#powers
#medium
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? Hint Small clue
A 2520
B 1260
C 5040
D 1680
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
\(8\times9\times5\times7=2520\)। / \(8\times9\times5\times7=2520\).
Step 3
Exam Tip
पहले घातों को हल करें, फिर बाकी गुणनखंडों से गुणा करें। / First solve powers, then multiply by the remaining factors.
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किस विकल्प में केवल अभाज्य गुणनखंडों का अंतिम रूप दिया गया है?
Which option gives the final form using only prime factors?
#prime-factors
#concept-check
#medium
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A \(2^4\times3^2\times11\)
B \(16\times9\times11\)
C \(2^4\times99\)
D \(8\times18\times11\)
Explanation opens after your attempt
Correct Answer
A. \(2^4\times3^2\times11\)
Step 1
Concept
अंतिम रूप में आधार केवल अभाज्य होने चाहिए। / In the final form, bases should be prime only.
Step 2
Why this answer is correct
पहले विकल्प में आधार 2, 3 और 11 अभाज्य हैं। / In the first option, bases 2, 3, and 11 are prime.
Step 3
Exam Tip
16, 9, 99 और 18 संयुक्त हैं, इसलिए वे अंतिम रूप नहीं हैं। / 16, 9, 99, and 18 are composite, so they are not final forms.
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किस विकल्प में अंतिम अभाज्य गुणनखंडन नहीं दिया गया है?
Which option is not a final prime factorisation?
#prime-factorisation
#not-final-form
#medium
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A \(2^2\times25\times7\)
B \(2^2\times5^2\times7\)
C \(2\times3^2\times5\times7\)
D \(2^3\times3\times7\)
Explanation opens after your attempt
Correct Answer
A. \(2^2\times25\times7\)
Step 1
Concept
अंतिम अभाज्य गुणनखंडन में 25 जैसा संयुक्त गुणनखंड नहीं होना चाहिए। / A final prime factorisation should not contain a composite factor like 25.
Step 2
Why this answer is correct
\(25=5^2\), इसलिए \(2^2\times25\times7\) अंतिम रूप नहीं है। / Since \(25=5^2\), \(2^2\times25\times7\) is not final form.
Step 3
Exam Tip
25 को \(5^2\) में बदलें। / Change 25 into \(5^2\).
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संख्या 4096 का अभाज्य गुणनखंडन क्या है?
What is the prime factorisation of 4096?
#prime-factorisation
#power-of-two
#medium
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A \(2^{12}\)
B \(2^{11}\)
C \(64^2\)
D \(16^3\)
Explanation opens after your attempt
Correct Answer
A. \(2^{12}\)
Step 1
Concept
4096 को बार-बार 2 से भाग दें। / Divide 4096 repeatedly by 2.
Step 2
Why this answer is correct
बारह बार 2 मिलने से \(4096=2^{12}\) होता है। / Twelve factors of 2 give \(4096=2^{12}\).
Step 3
Exam Tip
64 और 16 संयुक्त हैं, इसलिए अंतिम अभाज्य रूप में \(2^{12}\) लिखें। / 64 and 16 are composite, so write \(2^{12}\) as the final prime form.
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संख्या 6561 का सही अभाज्य गुणनखंडन क्या है?
What is the correct prime factorisation of 6561?
#prime-factorisation
#power-of-three
#medium
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A \(3^8\)
B \(3^7\)
C \(81^2\)
D \(27^3\)
Explanation opens after your attempt
Correct Answer
A. \(3^8\)
Step 1
Concept
6561 को 3 से बार-बार भाग दें। / Divide 6561 repeatedly by 3.
Step 2
Why this answer is correct
आठ बार 3 मिलने से \(6561=3^8\)। / Eight factors of 3 give \(6561=3^8\).
Step 3
Exam Tip
81 और 27 संयुक्त हैं, इसलिए अंतिम रूप में 3 की घात लिखें। / 81 and 27 are composite, so write a power of 3 in the final form.
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संख्या 2197 का अभाज्य गुणनखंडन कौन सा है?
Which is the prime factorisation of 2197?
#prime-factorisation
#cube-number
#medium
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A \(13^3\)
B \(13^2\)
C \(169\times13\)
D \(2197\times1\)
Explanation opens after your attempt
Correct Answer
A. \(13^3\)
Step 1
Concept
\(2197=13\times169\) लिखें। / Write \(2197=13\times169\).
Step 2
Why this answer is correct
\(169=13^2\), इसलिए \(2197=13^3\)। / \(169=13^2\), so \(2197=13^3\).
Step 3
Exam Tip
169 संयुक्त है, इसलिए अंतिम रूप में \(13^3\) लिखें। / Since 169 is composite, write \(13^3\) in the final form.
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संख्या 6250 का सही अभाज्य गुणनखंडन क्या है?
What is the correct prime factorisation of 6250?
#prime-factorisation
#power-of-five
#medium
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A \(2\times5^5\)
B \(2^2\times5^4\)
C \(625\times10\)
D \(2\times25^3\)
Explanation opens after your attempt
Correct Answer
A. \(2\times5^5\)
Step 1
Concept
\(6250=625\times10\) लिखें। / Write \(6250=625\times10\).
Step 2
Why this answer is correct
\(625=5^4\) और \(10=2\times5\), इसलिए \(6250=2\times5^5\)। / \(625=5^4\) and \(10=2\times5\), so \(6250=2\times5^5\).
Step 3
Exam Tip
5 की कुल घात 5 गिनें। / Count the total power of 5 as 5.
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संख्या 5103 का अभाज्य गुणनखंडन कौन सा है?
Which is the prime factorisation of 5103?
#prime-factorisation
#number-5103
#medium
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? Hint Small clue
A \(3^6\times7\)
B \(3^5\times7^2\)
C \(729\times7\)
D \(27\times189\)
Explanation opens after your attempt
Correct Answer
A. \(3^6\times7\)
Step 1
Concept
\(5103=729\times7\) लिखें। / Write \(5103=729\times7\).
Step 2
Why this answer is correct
\(729=3^6\), इसलिए \(5103=3^6\times7\)। / \(729=3^6\), so \(5103=3^6\times7\).
Step 3
Exam Tip
729 को अंतिम रूप में न छोड़कर 3 की घात लिखें। / Do not leave 729 in the final form; write it as a power of 3.
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संख्या 5488 का सही अभाज्य गुणनखंडन क्या है?
What is the correct prime factorisation of 5488?
#prime-factorisation
#number-5488
#medium
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? Hint Small clue
A \(2^4\times7^3\)
B \(2^3\times7^3\)
C \(16\times343\)
D \(8\times686\)
Explanation opens after your attempt
Correct Answer
A. \(2^4\times7^3\)
Step 1
Concept
\(5488=16\times343\) लिखें। / Write \(5488=16\times343\).
Step 2
Why this answer is correct
\(16=2^4\) और \(343=7^3\), इसलिए \(5488=2^4\times7^3\)। / \(16=2^4\) and \(343=7^3\), so \(5488=2^4\times7^3\).
Step 3
Exam Tip
16 और 343 को अभाज्य घातों में बदलें। / Convert 16 and 343 into prime powers.
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यदि \(2^4\times3^2\times11\) से संख्या बनाई जाए, तो वह कौन सी होगी?
If a number is formed from \(2^4\times3^2\times11\), which number will it be?
#evaluate-factorisation
#mcq-pattern
#medium
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? Hint Small clue
A 1584
B 792
C 3168
D 1188
Explanation opens after your attempt
Step 1
Concept
\(2^4=16\) और \(3^2=9\) निकालें। / Calculate \(2^4=16\) and \(3^2=9\).
Step 2
Why this answer is correct
\(16\times9\times11=1584\)। / \(16\times9\times11=1584\).
Step 3
Exam Tip
घातों को पहले हल करके गुणा करें। / Solve powers first and then multiply.
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यदि किसी संख्या का अभाज्य गुणनखंडन \(2^2\times3^3\times11\) है, तो वह संख्या कौन सी है?
If the prime factorisation of a number is \(2^2\times3^3\times11\), which number is it?
#evaluate-factorisation
#number-1188
#medium
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? Hint Small clue
A 1188
B 594
C 2376
D 1320
Explanation opens after your attempt
Step 1
Concept
\(2^2=4\) और \(3^3=27\) निकालें। / Calculate \(2^2=4\) and \(3^3=27\).
Step 2
Why this answer is correct
\(4\times27\times11=1188\)। / \(4\times27\times11=1188\).
Step 3
Exam Tip
पहले घातों का मान निकालना ठीक तरीका है। / Finding the value of powers first is the right method.
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यदि किसी संख्या का अभाज्य गुणनखंडन \(2\times3^3\times5^2\) है, तो वह संख्या कौन सी है?
If the prime factorisation of a number is \(2\times3^3\times5^2\), which number is it?
#evaluate-factorisation
#number-1350
#medium
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+10 Time+ 10 sec extra
? Hint Small clue
A 1350
B 675
C 2700
D 1500
Explanation opens after your attempt
Step 1
Concept
\(3^3=27\) और \(5^2=25\) निकालें। / Calculate \(3^3=27\) and \(5^2=25\).
Step 2
Why this answer is correct
\(2\times27\times25=1350\)। / \(2\times27\times25=1350\).
Step 3
Exam Tip
दो घातों को पहले सरल करने से गुणा आसान होता है। / Simplifying the two powers first makes multiplication easy.
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यदि \(2^2\times3^2\times7\times11\) किसी संख्या का अभाज्य गुणनखंडन है, तो संख्या क्या है?
If \(2^2\times3^2\times7\times11\) is the prime factorisation of a number, what is the number?
#evaluate-factorisation
#number-2772
#medium
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A 2772
B 1386
C 5544
D 1980
Explanation opens after your attempt
Step 1
Concept
\(2^2=4\) और \(3^2=9\) निकालें। / Calculate \(2^2=4\) and \(3^2=9\).
Step 2
Why this answer is correct
\(4\times9\times7\times11=2772\)। / \(4\times9\times7\times11=2772\).
Step 3
Exam Tip
गुणा करते समय चरणों में आगे बढ़ें। / Move step by step while multiplying.
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संख्या 4200 का सही अभाज्य गुणनखंडन क्या है?
What is the correct prime factorisation of 4200?
#prime-factorisation
#number-4200
#medium
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A \(2^3\times3\times5^2\times7\)
B \(2^2\times3\times5^2\times7\)
C \(42\times100\)
D \(4\times1050\)
Explanation opens after your attempt
Correct Answer
A. \(2^3\times3\times5^2\times7\)
Step 1
Concept
\(4200=42\times100\) लिखें। / Write \(4200=42\times100\).
Step 2
Why this answer is correct
\(42=2\times3\times7\) और \(100=2^2\times5^2\), इसलिए \(4200=2^3\times3\times5^2\times7\)। / \(42=2\times3\times7\) and \(100=2^2\times5^2\), so \(4200=2^3\times3\times5^2\times7\).
Step 3
Exam Tip
2 और 5 की घातें ध्यान से गिनें। / Count the powers of 2 and 5 carefully.
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संख्या 4620 का अभाज्य गुणनखंडन कौन सा है?
Which is the prime factorisation of 4620?
#prime-factorisation
#number-4620
#medium
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A \(2^2\times3\times5\times7\times11\)
B \(2\times3\times5\times7\times11\)
C \(42\times110\)
D \(4\times1155\)
Explanation opens after your attempt
Correct Answer
A. \(2^2\times3\times5\times7\times11\)
Step 1
Concept
\(4620=42\times110\) लिखें। / Write \(4620=42\times110\).
Step 2
Why this answer is correct
\(42=2\times3\times7\) और \(110=2\times5\times11\), इसलिए \(4620=2^2\times3\times5\times7\times11\)। / \(42=2\times3\times7\) and \(110=2\times5\times11\), so \(4620=2^2\times3\times5\times7\times11\).
Step 3
Exam Tip
2 दो बार आता है, इसलिए \(2^2\) लिखें। / Since 2 appears twice, write \(2^2\).
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संख्या 5400 का सही अभाज्य गुणनखंडन क्या है?
What is the correct prime factorisation of 5400?
#prime-factorisation
#number-5400
#medium
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A \(2^3\times3^3\times5^2\)
B \(2^2\times3^3\times5^2\)
C \(54\times100\)
D \(2^3\times27\times25\)
Explanation opens after your attempt
Correct Answer
A. \(2^3\times3^3\times5^2\)
Step 1
Concept
\(5400=54\times100\) लिखें। / Write \(5400=54\times100\).
Step 2
Why this answer is correct
\(54=2\times3^3\) और \(100=2^2\times5^2\), इसलिए \(5400=2^3\times3^3\times5^2\)। / \(54=2\times3^3\) and \(100=2^2\times5^2\), so \(5400=2^3\times3^3\times5^2\).
Step 3
Exam Tip
54 और 100 दोनों को पूरा तोड़ें। / Break both 54 and 100 completely.
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किस संख्या का अभाज्य गुणनखंडन \(2^2\times3\times5\times7\times11\) है?
Which number has prime factorisation \(2^2\times3\times5\times7\times11\)?
#evaluate-factorisation
#number-4620
#medium
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A 4620
B 2310
C 9240
D 3850
Explanation opens after your attempt
Step 1
Concept
\(2^2=4\) निकालें। / Calculate \(2^2=4\).
Step 2
Why this answer is correct
\(4\times3\times5\times7\times11=4620\)। / \(4\times3\times5\times7\times11=4620\).
Step 3
Exam Tip
कई गुणनखंड हों तो जोड़े बनाकर गुणा करें। / When there are many factors, multiply in pairs.
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संख्या 8192 का अभाज्य गुणनखंडन क्या है?
What is the prime factorisation of 8192?
#prime-factorisation
#power-of-two
#medium
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A \(2^{13}\)
B \(2^{12}\)
C \(64^2\times2\)
D \(16^3\times2\)
Explanation opens after your attempt
Correct Answer
A. \(2^{13}\)
Step 1
Concept
8192 को बार-बार 2 से भाग दें। / Divide 8192 repeatedly by 2.
Step 2
Why this answer is correct
तेरह बार 2 मिलने से \(8192=2^{13}\)। / Thirteen factors of 2 give \(8192=2^{13}\).
Step 3
Exam Tip
64 और 16 संयुक्त हैं, इसलिए अंतिम अभाज्य रूप में 2 की घात लिखें। / 64 and 16 are composite, so write the power of 2 in final prime form.
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अभाज्य गुणनखंडन में \(18\times100\) को अंतिम उत्तर क्यों नहीं माना जाता?
Why is \(18\times100\) not considered a final answer in prime factorisation?
#prime-factorisation
#concept-check
#medium
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A क्योंकि 18 और 100 संयुक्त संख्याएं हैं / Because 18 and 100 are composite numbers
B क्योंकि गुणा करना हमेशा गलत है / Because multiplication is always wrong
C क्योंकि 18 अभाज्य और 100 अभाज्य हैं / Because 18 and 100 are prime
D क्योंकि घातों का उपयोग नहीं किया जा सकता / Because powers cannot be used
Explanation opens after your attempt
Correct Answer
A. क्योंकि 18 और 100 संयुक्त संख्याएं हैं / Because 18 and 100 are composite numbers
Step 1
Concept
अभाज्य गुणनखंडन में हर अंतिम गुणनखंड अभाज्य होना चाहिए। / In prime factorisation, every final factor must be prime.
Step 2
Why this answer is correct
\(18=2\times3^2\) और \(100=2^2\times5^2\), इसलिए दोनों संयुक्त हैं। / \(18=2\times3^2\) and \(100=2^2\times5^2\), so both are composite.
Step 3
Exam Tip
\(18\times100\) को आगे \(2^3\times3^2\times5^2\) में बदलना होगा। / \(18\times100\) must be changed further into \(2^3\times3^2\times5^2\).
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