Fundamental Theorem of Arithmetic

Step 1: Divide 243 repeatedly by 3. Step 2: (243=3^5), so this is the prime factorisation. Step 3: 9 and 81 are composite, so do not keep them in the final prime form.

Step 1: A prime number has exactly two positive factors. Step 2: 1 has only one positive factor, 1. Step 3: Therefore, 1 is neither prime nor composite.

Step 1: Look at the given form (625=5^4). Step 2: The exponent written on 5 is 4. Step 3: It is important to identify the base and the exponent separately.

Step 1: Write (182=2\times91). Step 2: (91=7\times13), so (182=2\times7\times13). Step 3: 14 is composite, so (14\times13) is not the final prime factorisation.

Step 1: For two numbers, product (=) HCF (\times) LCM. Step 2: (6\times72=432). Step 3: Apply this relation directly for two numbers.

Step 1: 289 is a perfect square. Step 2: (289=17\times17=17^2). Step 3: 1 is not treated as a separate factor in prime factorisation.

Step 1: Write (231=21\times11). Step 2: (21=3\times7), so (231=3\times7\times11). Step 3: The distinct prime factors are 3, 7, and 11.

Step 1: The common prime factors are 2 and 3. Step 2: The smaller powers are (2^1) and (3^2), so the HCF is (2\times9=18). Step 3: Take smaller powers for HCF.

Step 1: For LCM, take the highest powers. Step 2: (2^2\times3^3=4\times27=108). Step 3: LCM includes the highest powers of all prime factors.

Step 1: Write (375=3\times125). Step 2: (125=5^3), so (375=3\times5^3). Step 3: 15 and 25 are composite, so do not keep them in the final prime form.

Step 1: Find (3^2=9). Step 2: (2\times9\times7=126), so the number is 126. Step 3: To get the number from prime factorisation, simplify powers first.

Step 1: Find (2^6=64). Step 2: (64\times3=192), so the number is 192. Step 3: First evaluate the power, then multiply.

Step 1: In prime factorisation, every factor must be prime. Step 2: In (9\times5), 9 is not prime. Step 3: If a composite factor appears, break it into prime factors.

Step 1: Write (208=16\times13). Step 2: (16=2^4) and 13 is prime, so (208=2^4\times13). Step 3: Since 16 is composite, write it as (2^4) in the final form.

Step 1: (26=2\times13) 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: 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.