Fundamental Theorem of Arithmetic

Step 1: For LCM, take the highest powers. Step 2: The highest powers are (2^5) and (3^4). Step 3: (32\times81=2592), so the answer is 2592.

Step 1: Write (1155=105\times11). Step 2: (105=3\times5\times7), so (1155=3\times5\times7\times11). Step 3: The distinct prime factors are 3, 5, 7, and 11.

Step 1: For two numbers, product (=) HCF (\times) LCM. Step 2: (21\times420=8820). Step 3: Use the HCF-LCM relation in the correct situation.

Step 1: Write (2025=45\times45). Step 2: (45=3^2\times5), so (2025=3^4\times5^2). Step 3: In perfect squares, the powers of the base factors get doubled.

Step 1: A prime number must have exactly two positive factors. Step 2: 1 has only one positive factor, so it is not prime. Step 3: A composite number needs more than two factors, so 1 is not composite either.

Step 1: The common prime factors are 2 and 5. Step 2: The smaller powers are (2^2) and (5^1). Step 3: (2^2\times5=4\times5=20), so the HCF is 20.

Step 1: For LCM, take the highest powers of all prime factors. Step 2: The highest powers are (2^3), (3), (5^2), and (7). Step 3: (8\times3\times25\times7=4200), so the answer is 4200.

Step 1: Write (1848=168\times11). Step 2: (168=2^3\times3\times7), so (1848=2^3\times3\times7\times11). Step 3: 168 is composite, so do not keep it in the final form.

Step 1: The main idea of the theorem is prime factorisation. Step 2: Every composite number greater than 1 can be written in a fixed way as prime factors. Step 3: The order may change, but the prime factors do not change.

Step 1: Calculate (2^6=64) and (5^2=25). Step 2: (64\times3\times25=4800). Step 3: Simplifying powers first makes multiplication easier.

Step 1: Having no common prime factor means the numbers are co-prime. Step 2: Co-prime numbers have HCF 1. Step 3: Prime factorisation helps identify co-primality quickly.

Step 1: Calculate (2^4=16). Step 2: (16\times3\times7=336). Step 3: To get the number from prime factorisation, multiply all factors.

Step 1: Write (1080=108\times10). Step 2: (108=2^2\times3^3) and (10=2\times5), so (1080=2^3\times3^3\times5). Step 3: Comparing with the given form gives (a=3).

Step 1: In multiplication, powers with the same base are added. Step 2: The power of 3 in (a) is 2 and in (b) is 1. Step 3: In (ab), the power of 3 will be (2+1=3).

Step 1: While multiplying, add the exponents of the same prime base. Step 2: The power of 5 in (a) is 2 and in (b) is 3. Step 3: In (ab), the power of 5 will be (2+3=5).

Step 1: (2^3) and (5^2) contain both factors 2 and 5. Step 2: The product of 2 and 5 is 10, so the number must be divisible by 10. Step 3: Divisibility can be quickly identified from prime factors.

Step 1: Write (1323=27\times49). Step 2: (27=3^3) and (49=7^2), so (1323=3^3\times7^2). Step 3: 27 and 49 are composite, so write prime powers in the final form.

Step 1: In (xy), the powers of the same base 2 are added. Step 2: The power of 2 in (x) is 3 and in (y) is 2. Step 3: So the power of 2 in (xy) is (3+2=5).

Step 1: For HCF, take the smaller powers of common prime factors. Step 2: The smaller powers are (2^2), (3^1), and (5^1). Step 3: (4\times3\times5=60), so the HCF is 60.

Step 1: The theorem tells us that prime factorisation is fixed. Step 2: The order of factors may change but the prime factors remain the same. Step 3: In exams, do not treat changed order as a new factorisation.