If (m=3^2\times5\times7), what is the value of (m)?
Step 1: Calculate (3^2=9). Step 2: (9\times5\times7=315). Step 3: Simplify the factor with a power first.
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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: Calculate (3^2=9). Step 2: (9\times5\times7=315). Step 3: Simplify the factor with a power first.
View question detailsStep 1: In final prime factorisation, every factor must be prime. Step 2: 2, 3, 7, and 11 are all prime. Step 3: 6, 21, and 33 are composite, so they cannot remain in final form.
View question detailsStep 1: Final prime factorisation must not contain a composite factor. Step 2: 6 is composite, so (6\times5^2) is not final form. Step 3: Change 6 into (2\times3).
View question detailsStep 1: Divide 256 repeatedly by 2. Step 2: Eight factors of 2 give (256=2^8). Step 3: 4 and 16 are composite, so do not write them in final prime form.
View question detailsStep 1: Divide 243 repeatedly by 3. Step 2: Five factors of 3 give (243=3^5). Step 3: 9 and 27 are composite, so keep prime base 3.
View question detailsStep 1: 169 is a square number. Step 2: (169=13\times13=13^2). Step 3: Since 13 is prime, (13^2) is the correct prime factorisation.
View question detailsStep 1: (289=17\times17). Step 2: Since 17 is prime, (289=17^2). Step 3: In a square number, the same prime appears twice.
View question detailsStep 1: Divide 512 repeatedly by 2. Step 2: Nine factors of 2 give (512=2^9). Step 3: 8, 16, and 32 are composite, so write the power of 2 in final prime form.
View question detailsStep 1: (625) can be written as (25\times25). Step 2: (25=5^2), so (625=5^4). Step 3: 25 is composite, so write the power of 5 in final form.
View question detailsStep 1: Write (216=8\times27). Step 2: (8=2^3) and (27=3^3), so (216=2^3\times3^3). Step 3: 6, 8, and 27 are composite, so write prime bases in final form.
View question detailsStep 1: Divide 128 repeatedly by 2. Step 2: Seven factors of 2 give (128=2^7). Step 3: 4, 8, and 16 are composite, so write (2^7) in final form.
View question detailsStep 1: Write (540=54\times10). Step 2: (54=2\times3^3) and (10=2\times5), so (540=2^2\times3^3\times5). Step 3: Break 54 completely into prime form.
View question detailsStep 1: Write (588=12\times49). Step 2: (12=2^2\times3) and (49=7^2), so (588=2^2\times3\times7^2). Step 3: Convert 12 and 49 into prime powers.
View question detailsStep 1: Write (660=66\times10). Step 2: (66=2\times3\times11) and (10=2\times5), so (660=2^2\times3\times5\times11). Step 3: Since 2 appears twice, write (2^2).
View question detailsStep 1: Calculate (2^2=4). Step 2: (4\times3\times11=132). Step 3: Multiply all factors to get the number from prime factorisation.
View question detailsStep 1: Calculate (3^2=9) and (5^2=25). Step 2: (9\times25=225). Step 3: Simplify powers first and then multiply.
View question detailsStep 1: All prime factors are given. Step 2: (2\times3\times5\times7=210). Step 3: When no power is written, each prime is taken once.
View question detailsStep 1: Calculate (2^2=4) and (3^2=9). Step 2: (4\times9\times7=252). Step 3: Evaluate prime powers and then multiply.
View question detailsStep 1: Calculate (2^4=16) and (5^2=25). Step 2: (16\times25=400). Step 3: Finding powers first makes the calculation easier.
View question detailsStep 1: A prime number has exactly two factors. Step 2: 1 has only one factor, so 1 is not prime. Step 3: Do not write 1 as a final factor in prime factorisation.
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