What is the correct prime factorisation of 720?
Step 1: Write (720=72\times10). Step 2: (72=2^3\times3^2) and (10=2\times5), so (720=2^4\times3^2\times5). Step 3: Do not keep composite factors in the final answer.
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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.
TOPIC PRACTICE
Up to 25 questions from this page. Select your focus, then start.
Step 1: Write (720=72\times10). Step 2: (72=2^3\times3^2) and (10=2\times5), so (720=2^4\times3^2\times5). Step 3: Do not keep composite factors in the final answer.
Step 1: Write (924=84\times11). Step 2: (84=2^2\times3\times7), so (924=2^2\times3\times7\times11). Step 3: 84 is composite, so do not leave it in the final prime form.
Step 1: Write (1176=24\times49). Step 2: (24=2^3\times3) and (49=7^2), so (1176=2^3\times3\times7^2). Step 3: 24 and 49 are composite, so write prime powers in the final form.
Step 1: For HCF, take the smaller powers of common prime factors. Step 2: The smaller powers are (2^3), (3^1), and (5^1). Step 3: (8\times3\times5=120), so the answer is 120.
Step 1: For LCM, take the highest powers of all prime factors. Step 2: The highest powers are (2^5), (3^2), (5^2), and (7). Step 3: (32\times9\times25\times7=50400), so the answer is 50400.
Step 1: For two numbers, product (=) HCF (\times) LCM. Step 2: LCM (=2520\div21=120). Step 3: Use this formula directly only for two numbers.
Step 1: For two numbers, product (=) HCF (\times) LCM. Step 2: (24\times360=8640). Step 3: Apply the HCF-LCM relation carefully.
Step 1: First calculate (2^4=16) and (3^2=9). Step 2: (16\times9\times7=1008). Step 3: Simplify powers first while finding the number.
Step 1: Write (2520=252\times10). Step 2: (252=2^2\times3^2\times7) and (10=2\times5), so (2520=2^3\times3^2\times5\times7). Step 3: The number of times 2 appears is its power.
Step 1: Write (1575=15\times105). Step 2: (15=3\times5) and (105=3\times5\times7), so (1575=3^2\times5^2\times7). Step 3: Do not keep composite factors in the final answer.
Step 1: Calculate (2^4=16) and (3^2=9). Step 2: (16\times9\times5=720). Step 3: To get the number from prime factorisation, multiply all factors.
Step 1: Look at distinct prime factors, not powers. Step 2: In (m), the distinct primes are 2, 3, 7, and 11. Step 3: Count 2 and 3 only once each despite their powers.
Step 1: (27=3^3) and (64=2^6). Step 2: They have no common prime factor. Step 3: Therefore, they are co-prime and their HCF is 1.
Step 1: Co-prime numbers have HCF 1. Step 2: For two numbers, product (=) HCF (\times) LCM. Step 3: Therefore, the LCM will be 391.
Step 1: In prime factorisation, every factor must be prime. Step 2: (2), (3), and (13) are prime, so (2^3\times3\times13) is correct. Step 3: 8, 12, and 24 are composite, so they cannot remain in the final form.
Step 1: The common prime factors are 2 and 3. Step 2: The smaller powers are (2^3) and (3^1). Step 3: (8\times3=24), so the HCF is 24.
Step 1: For LCM, take the highest powers. Step 2: The highest powers are (2^4), (3^2), and (7). Step 3: (16\times9\times7=1008), so the answer is 1008.
Step 1: Write (1890=189\times10). Step 2: (189=3^3\times7) and (10=2\times5). Step 3: Therefore, (1890=2\times3^3\times5\times7).
Step 1: Calculate (2^6=64) and (3^2=9). Step 2: (64\times9=576). Step 3: In a form with powers, it is easier to evaluate powers first.
Step 1: To count with repetition, add the exponents. Step 2: (2^5) gives 5, (3^3) gives 3, and (7) gives 1 factor. Step 3: Total (5+3+1=9), so the answer is 9.
Step 1: Prime factorise 180. Step 2: (180=18\times10=2^2\times3^2\times5). Step 3: Comparing with the given form gives (a=2).
Step 1: Write (1512=8\times189). Step 2: (8=2^3) and (189=3^3\times7), so (1512=2^3\times3^3\times7). Step 3: Comparing gives (b=3).
Step 1: The prime factors of the first number are 2 and 7. Step 2: The prime factors of the second number are 3 and 5. Step 3: There is no common prime factor, so they are co-prime.
Step 1: (28=2^2\times7) and (45=3^2\times5), so they are co-prime. Step 2: The LCM of co-prime numbers equals their product. Step 3: (28\times45=1260), so the answer is 1260.
Step 1: Divide 2048 repeatedly by 2. Step 2: (2048=2^{11}). Step 3: 4 and 32 are composite, so do not use them in the final prime form.
QUIZ COMPLETE