If (n=2^5\times3\times7), what is the value of (n)?
Step 1: (2^5=32). Step 2: (32\times3\times7=672). Step 3: Multiply all factors to convert prime factorisation into the number.
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SubjectsMathematics
अभाज्य गुणनखंडन
In Class 10 Mathematics, under the Real Numbers chapter, Prime Factorisation teaches students to express a composite number as a product of prime numbers. Students practise identifying prime factors and writing factorisations using multiplication and exponents. This foundational skill supports the Fundamental Theorem of Arithmetic and later work with HCF and LCM.
TOPIC PRACTICE
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Step 1: (2^5=32). Step 2: (32\times3\times7=672). Step 3: Multiply all factors to convert prime factorisation into the number.
View question detailsStep 1: (3^3=27). Step 2: (27\times5\times7=945). Step 3: It is better to simplify the power part first.
View question detailsStep 1: In prime factorisation, bases must be prime. Step 2: In the first option, bases 2, 3, 5, and 7 are prime. Step 3: 8, 9, 35, and 4 are composite, so the other forms are not final.
View question detailsStep 1: Final prime factorisation must not contain a composite factor. Step 2: 21 is composite, so (2^3\times21\times5) is not final form. Step 3: Change 21 into (3\times7).
View question detailsStep 1: Divide 2048 repeatedly by 2. Step 2: Eleven factors of 2 give (2048=2^{11}). Step 3: 4, 32, and 64 are composite, so write (2^{11}) in final form.
View question detailsStep 1: Divide 729 repeatedly by 3. Step 2: Six factors of 3 give (729=3^6). Step 3: 9 and 27 are composite, so write a power of 3 in final form.
View question detailsStep 1: Write (1331=11\times121). Step 2: (121=11^2), so (1331=11^3). Step 3: Since 121 is composite, write (11^3) in the final form.
View question detailsStep 1: (2401) can be written as (49\times49). Step 2: Since (49=7^2), (2401=7^4). Step 3: Since 49 is composite, write the power of 7 in final prime form.
View question detailsStep 1: Write (2500=25\times100). Step 2: (25=5^2) and (100=2^2\times5^2), so (2500=2^2\times5^4). Step 3: Count the total power of 5 as 4.
View question detailsStep 1: Recognise (2744=14^3). Step 2: Since (14=2\times7), (2744=2^3\times7^3). Step 3: Since 14 is composite, write prime bases in final form.
View question detailsStep 1: Calculate (2^4=16) and (3^2=9). Step 2: (16\times9\times5=720). Step 3: Solving powers first helps find the option quickly.
View question detailsStep 1: Calculate (2^3=8), (3^2=9), and (5^2=25). Step 2: (8\times9\times25=1800). Step 3: Find each power separately and then multiply.
View question detailsStep 1: (2^4=16) and (3^4=81). Step 2: (16\times81=1296). Step 3: In such calculations, simplifying powers first is safer.
View question detailsStep 1: Calculate (2^2=4) and (5^4=625). Step 2: (4\times625=2500). Step 3: Finding the higher power first makes the answer easier.
View question detailsStep 1: Write (2700=27\times100). Step 2: (27=3^3) and (100=2^2\times5^2), so (2700=2^2\times3^3\times5^2). Step 3: Convert both 27 and 100 into prime powers.
View question detailsStep 1: Write (3080=8\times385). Step 2: (8=2^3) and (385=5\times7\times11), so (3080=2^3\times5\times7\times11). Step 3: Break 385 into prime form too.
View question detailsStep 1: Write (3960=36\times110). Step 2: (36=2^2\times3^2) and (110=2\times5\times11), so (3960=2^3\times3^2\times5\times11). Step 3: Count the total power of 2 carefully.
View question detailsStep 1: Calculate (2^3=8). Step 2: (8\times5\times7\times11=3080). Step 3: Whatever the order, the product remains the same.
View question detailsStep 1: Divide 4096 repeatedly by 2. Step 2: Twelve factors of 2 give (4096=2^{12}). Step 3: 64 and 16 are composite, so they are not final prime forms.
View question detailsStep 1: In final prime factorisation, the base must be prime. Step 2: 12 is composite and (12=2^2\times3). Step 3: Therefore, (12^2) must be changed further into (2^4\times3^2).
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