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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
Up to 25 questions from this page. Select your focus, then start.
Step 1: In prime factorisation, add 1 to each exponent and multiply. Step 2: Here the number of factors is ((3+1)(2+1)(1+1)=24). Step 3: In exams, identify the exponents first and then multiply carefully.
Step 1: The bases in prime factorisation are the prime factors. Step 2: Among (2,3,7), the smallest prime number is (2). Step 3: For the smallest prime factor, look at the base, not the exponent.
Step 1: Evaluate the powers first. Step 2: (2^2 \times 3 \times 5^2=4 \times 3 \times 25=300). Step 3: In such questions, solve powers before multiplying.
Step 1: In HCF, take the smaller exponent of common prime factors. Step 2: The exponents of (3) are (2) and (4), so the smaller one is (2). Step 3: For HCF, always choose the minimum exponent.
Step 1: In LCM, take the larger exponent of each prime factor. Step 2: The exponents of (5) are (2) and (4), so the larger one is (4). Step 3: Remember that LCM uses maximum exponents.
Step 1: Write (360=36 \times 10). Step 2: Since (36=2^2 \times 3^2) and (10=2 \times 5), (360=2^3 \times 3^2 \times 5). Step 3: Verify by multiplying back to the original number.
Step 1: Prime factorise (72). Step 2: (72=8 \times 9=2^3 \times 3^2), so (a=3). Step 3: To find an unknown exponent, compare prime factorisations on both sides.
Step 1: Distinct prime factors are counted by their prime bases. Step 2: The bases are (2,3,5), so there are (3) distinct prime factors. Step 3: Do not count exponents as separate factors.
Step 1: (2^4 \times 3 \times 7=16 \times 21=336). Step 2: (336,672,1008,1344) are all divisible by (336), so none is actually not divisible. Step 3: In exams, read negative wording and test every option carefully.
Step 1: Evaluate the power first. Step 2: (2^3=8), so (N=8 \times 3 \times 11=264). Step 3: Group smaller products to calculate faster.
Step 1: The condition requires (2^4 \times 5^2). Step 2: (2^4 \times 5^2=16 \times 25=400), and it has no factor (3). Step 3: Check the exponent of every prime before choosing.
Step 1: Write (840=84 \times 10). Step 2: Since (84=2^2 \times 3 \times 7) and (10=2 \times 5), (840=2^3 \times 3 \times 5 \times 7). The exponent of (7) is (1). Step 3: A prime appearing once has exponent (1).
Step 1: A divisor must not require prime exponents higher than those available. Step 2: (45=3^2 \times 5), which is fully contained in (3^2 \times 5^3). Step 3: Compare exponents to test divisibility.
Step 1: For a square factor, every prime exponent must be even. Step 2: For (2^3), even choices are (0,2); for (3^2), (0,2); for (5^2), (0,2). Total (2 \times 2 \times 2=8). Step 3: Count only even exponent choices for square factors.
Step 1: Write (540=54 \times 10). Step 2: (54=2 \times 3^3) and (10=2 \times 5), so (540=2^2 \times 3^3 \times 5). Step 3: Splitting a number into two easy parts saves time.
Step 1: First calculate (2^2=4). Step 2: (4 \times 3 \times 13=12 \times 13=156). Step 3: Break multiplication into smaller steps.
Step 1: In a perfect cube, every prime exponent must be a multiple of (3). Step 2: (2^6) is fine, but (3^2) needs one more (3) to become (3^3). Step 3: For cubes, make exponents multiples of (3).
Step 1: In a perfect square, all prime exponents are even. Step 2: (2^5) has an odd exponent, so multiplying by (2) makes it (2^6); (7^2) is already even. Step 3: For squares, fix only the odd exponents.
Step 1: One pair of (2) and (5) makes (10), giving a trailing zero. Step 2: Remove one (2) with the (5), leaving (2^3 \times 3^2=72), so the last non-zero digit is (2). Step 3: After removing trailing-zero pairs, check the last digit of the remaining product.
Step 1: A trailing zero comes from a pair (10=2 \times 5). Step 2: The exponent of (2) is (3) and of (5) is (1), so there is (1) pair. Step 3: For trailing zeros, take the smaller exponent of (2) and (5).
Step 1: Write (1260=126 \times 10). Step 2: (126=2 \times 3^2 \times 7) and (10=2 \times 5), so (1260=2^2 \times 3^2 \times 5 \times 7). It does not include (11). Step 3: Match the options with the prime factorisation.
Step 1: Write (900=9 \times 100). Step 2: (9=3^2) and (100=2^2 \times 5^2), so (900=2^2 \times 3^2 \times 5^2). Step 3: Using square numbers makes factorisation easier.
Step 1: In a perfect square, every prime exponent is even. Step 2: Here all exponents are (2), so (N) is a perfect square. Step 3: To identify a square, check whether all exponents are even.
Step 1: In a perfect cube, all prime exponents are multiples of (3). Step 2: Each exponent is (3), so (N) is a perfect cube. Step 3: To identify a cube, check divisibility of exponents by (3).
Step 1: (144=12^2). Step 2: Since (12=2^2 \times 3), (144=2^4 \times 3^2); the sum is (4+2=6). Step 3: When squaring a number, its prime exponents double.
QUIZ COMPLETE
