Fundamental Theorem of Arithmetic

Step 1: Divide 512 repeatedly by 2. Step 2: (512=2^9), so this is the prime factorisation. Step 3: 4 and 16 are composite, so do not keep them in the final prime form.

Step 1: The common prime factors are 2, 3, and 5. Step 2: The smaller powers are (2^1), (3^1), and (5^1), so (2\times3\times5=30). Step 3: Take smaller powers for HCF.

Step 1: For LCM, take the highest powers. Step 2: (2^1\times3^2\times5^2=2\times9\times25=450). Step 3: LCM includes the highest powers of all prime factors.

Step 1: A number divisible by both 2 and 5 is also divisible by 10. Step 2: 70 ends in 0, so it is divisible by 10. Step 3: Divisibility checks help in prime factorisation.

Step 1: Find (2^4=16) and (5^2=25). Step 2: (16\times3\times25=1200). Step 3: In questions with powers, simplify the powers first.

Step 1: The theorem says prime factorisation is unique. Step 2: The order of factors may change, but the prime factors remain the same. Step 3: Understand uniqueness separately from order.

Step 1: Write (462=42\times11). Step 2: (42=2\times3\times7), so (462=2\times3\times7\times11). Step 3: In the final form, all factors must be prime.

Step 1: Multiply the given prime factors. Step 2: (3\times7\times11=231), so the number is 231. Step 3: To get the original number, multiply all prime factors.

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: Write repeated prime factors using powers.

Step 1: Co-prime numbers have HCF 1. Step 2: For two numbers, product (=) HCF (\times) LCM. Step 3: Therefore, the LCM of co-prime numbers is equal to their product.

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.