Fundamental Theorem of Arithmetic

Step 1: 121 is a perfect square. Step 2: (121=11\times11=11^2). Step 3: 1 is not written as a factor in prime factorisation.

Step 1: Write (198=2\times99). Step 2: (99=9\times11=3^2\times11), so the distinct prime factors are 2, 3, and 11. Step 3: Write repeated 3 only once in the distinct list.

Step 1: The common prime factors are 2 and 3. Step 2: Their product is (2\times3=6), so the HCF is 6. Step 3: For HCF, take only common prime factors.

Step 1: For LCM, take all required prime factors. Step 2: (2\times3\times5\times7=210). Step 3: LCM includes all necessary prime factors from both numbers.

Step 1: Write (250=2\times125). Step 2: (125=5^3), so (250=2\times5^3). Step 3: 25 and 10 are composite, so do not keep them in the final prime form.

Step 1: Find (2^3=8). Step 2: (8\times7=56), so the number is 56. Step 3: To get the number from prime factorisation, simplify powers first.

Step 1: Find (2^5=32). Step 2: (32\times3=96), so the number is 96. Step 3: First evaluate the power, then multiply.

Step 1: In prime factorisation, every factor must be prime. Step 2: In (6\times5), 6 is not prime. Step 3: If a composite factor appears, break it further into prime factors.

Step 1: Write (176=16\times11). Step 2: (16=2^4) and 11 is prime, so (176=2^4\times11). Step 3: Since 16 is composite, write it as (2^4) in the final form.

Step 1: (22=2\times11) and (35=5\times7). Step 2: They have no common prime factor, so they are co-prime. Step 3: Co-prime numbers have HCF 1.

Step 1: Divide 256 repeatedly by 2. Step 2: (256=2^8), so this is the prime factorisation. Step 3: 4 and 16 are composite, so (4^4) or (16^2) are not final prime forms.

Step 1: The common prime factors are 2, 3, and 7. Step 2: The smaller powers are (2^1), (3^1), and (7^1), so (2\times3\times7=42). Step 3: Take smaller powers for HCF.

Step 1: For LCM, take the highest powers. Step 2: (2^2\times3^2\times7=4\times9\times7=252). Step 3: LCM includes the highest powers of all prime factors.

Step 1: A number divisible by both 3 and 5 is divisible by 15. Step 2: The digit sum of 45 is 9, so it is divisible by 3, and it ends in 5, so it is also divisible by 5. Step 3: Small divisibility rules help in factorisation.

Step 1: Find (2^3=8) and (5^2=25). Step 2: (8\times3\times25=600). Step 3: In questions with powers, simplify the powers first.

Step 1: This theorem is stated for positive integers greater than 1. Step 2: Every such number can be prime factorised. Step 3: 1 is not included in the usual prime factorisation statement.

Step 1: Write (330=33\times10). Step 2: (33=3\times11) and (10=2\times5), so (330=2\times3\times5\times11). Step 3: In the final form, all factors must be prime.

Step 1: Multiply the given prime factors. Step 2: (2\times5\times11=110), so the number is 110. Step 3: To get the original number, multiply all prime factors.

Step 1: Write (270=27\times10). Step 2: (27=3^3) and (10=2\times5), so (270=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.