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SubjectsMathematics
अंकगणित का मौलिक प्रमेय
In this Class 10 Mathematics topic from the chapter Real Numbers, students learn that every integer greater than 1 can be expressed as a product of prime numbers, and that this prime factorisation is unique apart from the order of the factors. They practise finding prime factors and use the theorem to understand and determine the HCF and LCM of numbers. The topic builds clear reasoning about the structure of whole numbers and supports later work with divisibility and number relationships.
Correct answer: A. As a product of prime numbers
Explanation: Step 1: This theorem is connected with prime factorisation. Step 2: Every positive integer greater than 1 can be written as a product of prime numbers. Step 3: In exams, remember it through prime factorisation.
Correct answer: A. (2\times3^3)
Explanation: Step 1: Write (54=2\times27). Step 2: (27=3^3), so (54=2\times3^3). Step 3: 6 and 9 are composite, so do not leave them in the final prime form.
Correct answer: A. (2^4\times5)
Explanation: Step 1: (80) can be written as (16\times5). Step 2: (16=2^4), so (80=2^4\times5). Step 3: In the final prime factorisation, keep only prime numbers.
Correct answer: A. The product does not change
Explanation: Step 1: Changing the order in multiplication does not change the product. Step 2: (2\times5\times3) and (3\times2\times5) both give 30. Step 3: In prime factorisation, the set of factors matters, not the order.
Correct answer: B. 4
Explanation: Step 1: Write (112=16\times7). Step 2: (16=2^4), so (112=2^4\times7). Step 3: Count repeated prime factors to write the exponent.
Correct answer: C. 168
Explanation: Step 1: First find (2^3=8). Step 2: (8\times3\times7=168), so the number is 168. Step 3: In expressions with powers, evaluate the power first.
Correct answer: A. (3^2\times11)
Explanation: Step 1: Write (99=9\times11). Step 2: (9=3^2) and 11 is prime, so (99=3^2\times11). Step 3: 9 and 33 are composite, so do not keep them in the final answer.
Correct answer: A. 47
Explanation: Step 1: A prime number has exactly two positive factors. Step 2: 47 is divisible only by 1 and 47, while 51, 57, and 63 are composite. Step 3: Check small numbers by 2, 3, 5, and 7.
Correct answer: C. 91
Explanation: Step 1: A composite number has more than two positive factors. Step 2: (91=7\times13), so it is composite. Step 3: If a number can be written as a product of smaller primes, it is composite.
Correct answer: A. (3\times5\times7)
Explanation: Step 1: Write (105=15\times7). Step 2: (15=3\times5), so (105=3\times5\times7). Step 3: 15 is composite, so do not keep it in the final prime factorisation.
Correct answer: B. 3
Explanation: Step 1: Look at distinct prime factors, not their powers. Step 2: In (180=2^2\times3^2\times5), the distinct primes are 2, 3, and 5. Step 3: Even with a higher power, count the same prime once.
Correct answer: A. (13^2)
Explanation: Step 1: 169 is a perfect square. Step 2: (169=13\times13=13^2). Step 3: In prime factorisation, 1 is not written as a separate factor.
Correct answer: A. 12
Explanation: Step 1: For HCF, take smaller powers of common prime factors. Step 2: The smaller power of (2) is 2 and of (3) is 1, so (2^2\times3=12). Step 3: Use smaller powers for HCF.
Correct answer: C. 144
Explanation: Step 1: For LCM, take the highest powers of all prime factors. Step 2: The highest power of (2) is 4 and of (3) is 2, so (2^4\times3^2=144). Step 3: Use highest powers for LCM.
Correct answer: A. (2^5\times3^2)
Explanation: Step 1: Write (288=32\times9). Step 2: (32=2^5) and (9=3^2), so (288=2^5\times3^2). Step 3: 32 and 9 are composite, so write prime powers in the final form.
Correct answer: C. 135
Explanation: Step 1: First find (3^3=27). Step 2: (27\times5=135), so the number is 135. Step 3: In a form with powers, evaluate the power first.
Correct answer: A. Because 1 is not a prime number
Explanation: Step 1: In prime factorisation, only prime factors are written. Step 2: 1 is not prime because it does not have exactly two positive factors. Step 3: Therefore, (1\times n) is not treated as prime factorisation.
Correct answer: A. (3\times7^2)
Explanation: Step 1: Write (147=3\times49). Step 2: (49=7^2), so (147=3\times7^2). Step 3: 21 and 49 are composite, so do not keep them in the final form.
Correct answer: A. 2 and 31
Explanation: Step 1: Write (124=4\times31). Step 2: (4=2^2) and 31 is prime, so (124=2^2\times31). Step 3: The distinct prime factors are 2 and 31.
Correct answer: A. 1
Explanation: Step 1: Co-prime numbers have no common factor except 1. Step 2: Therefore, their HCF is 1. Step 3: To identify co-prime numbers, check common prime factors.
Correct answer: A. 25 and 36
Explanation: Step 1: (25=5^2) and (36=2^2\times3^2). Step 2: They have no common prime factor, so they are co-prime. Step 3: If a common prime factor appears, the pair is not co-prime.
Correct answer: A. 84
Explanation: Step 1: The common prime factors are 2, 3, and 7. Step 2: The smaller powers are (2^2), (3^1), and (7^1), so (4\times3\times7=84). Step 3: HCF uses smaller powers.
Correct answer: B. 504
Explanation: Step 1: For LCM, take the highest powers of all prime factors. Step 2: (2^3\times3^2\times7=8\times9\times7=504). Step 3: LCM uses highest powers.
Correct answer: A. (2^3\times3^2\times5)
Explanation: Step 1: Write (360=36\times10). Step 2: (36=2^2\times3^2) and (10=2\times5), so (360=2^3\times3^2\times5). Step 3: Combine repeated prime factors using powers.
Correct answer: C. 60
Explanation: Step 1: Find (2^2=4). Step 2: (4\times3\times5=60), so the number is 60. Step 3: To get the number from prime factorisation, multiply all factors.
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