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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: The theorem says that prime factorisation of a number greater than 1 is fixed. Step 2: Changing the order does not change the product, such as (2\times3\times5) and (5\times2\times3). Step 3: In exams, do not treat changed order as a different factorisation.
Step 1: Write (540=54\times10). Step 2: (54=2\times3^3) and (10=2\times5), so (540=2^2\times3^3\times5). Step 3: Do not leave composite factors like 54 and 10 in the final form.
Step 1: Write (756=84\times9). Step 2: (84=2^2\times3\times7) and (9=3^2), so (756=2^2\times3^3\times7). Step 3: Combine repeated prime factors using powers.
Step 1: Write (980=98\times10). Step 2: (98=2\times7^2) and (10=2\times5), so (980=2^2\times5\times7^2). Step 3: Composite factors like 98 and 10 should not remain in the final prime form.
Step 1: For HCF, take the smaller powers of common prime factors. Step 2: The common factors are 2 and 3, with smaller powers (2^2) and (3^2). Step 3: (2^2\times3^2=4\times9=36), so the answer is 36.
Step 1: For LCM, take the highest powers of all prime factors. Step 2: The highest powers are (2^4), (3^3), (5), and (7). Step 3: (16\times27\times5\times7=15120), so the answer is 15120.
Step 1: For two numbers, product (=) HCF (\times) LCM. Step 2: (1512=18\times) LCM, so LCM (=1512\div18=84). Step 3: Use this relation directly only for two numbers.
Step 1: For two numbers, product (=) HCF (\times) LCM. Step 2: (15\times210=3150). Step 3: When the question has two numbers, this formula gives the answer quickly.
Step 1: First calculate (2^3=8) and (3^2=9). Step 2: (8\times9\times7=504). Step 3: To get the number from prime factorisation, simplify powers first.
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: The number of times 3 appears as a factor is its power.
Step 1: Write (945=9\times105). Step 2: (9=3^2) and (105=3\times5\times7), so (945=3^3\times5\times7). Step 3: Do not keep composite factors like 9 or 105 in the final answer.
Step 1: Calculate (2^3=8) and (3^2=9). Step 2: (8\times9\times5=360). Step 3: When converting prime factorisation into a number, multiply all factors.
Step 1: Look at distinct prime factors, not powers. Step 2: In (m), the distinct prime factors are 2, 3, and 5. Step 3: In (2^4), count 2 only once in the distinct list.
Step 1: (16=2^4) and (45=3^2\times5). Step 2: They have no common prime factor, so they are co-prime. Step 3: Co-prime numbers have HCF 1.
Step 1: Co-prime numbers have HCF 1. Step 2: For two numbers, product (=) HCF (\times) LCM. Step 3: Therefore, the LCM is (221).
Step 1: In prime factorisation, every factor must be prime. Step 2: (2), (3), and (11) are prime, so (2^2\times3\times11) is correct. Step 3: 4, 6, and 12 are composite, so they should not 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), so (8\times3=24). Step 3: For HCF, take smaller powers of only common factors.
Step 1: For LCM, take the highest powers of all prime factors. Step 2: The highest powers are (2^3), (3^2), and (5^1). Step 3: (8\times9\times5=360), so the answer is 360.
Step 1: Write (1260=126\times10). Step 2: (126=2\times3^2\times7) and (10=2\times5). Step 3: Therefore, (1260=2^2\times3^2\times5\times7).
Step 1: Calculate (2^5=32) and (3^2=9). Step 2: (32\times9=288). Step 3: In prime factorisation with powers, evaluate powers first.
Step 1: When repetition is counted, add the exponents. Step 2: (2^4) gives 4 factors, (3^2) gives 2 factors, and (5) gives 1 factor. Step 3: Total (4+2+1=7), so the answer is 7.
Step 1: Prime factorise 108. Step 2: (108=4\times27=2^2\times3^3), so the given form (2^a\times3^2) does not match 108. Step 3: The power of 3 should also be 3; therefore no value of (a) alone can make it correct.
Step 1: Write 216 as (8\times27). Step 2: (8=2^3) and (27=3^3), so (216=2^3\times3^3). Step 3: Therefore, (a=3).
Step 1: The prime factors of the first number are 2 and 5. Step 2: The prime factors of the second number are 3 and 7. Step 3: There is no common prime factor, so they are co-prime.
Step 1: (18=2\times3^2) and (35=5\times7), so they are co-prime. Step 2: The LCM of co-prime numbers equals their product. Step 3: (18\times35=630), so the answer is 630.
QUIZ COMPLETE
