Fundamental Theorem of Arithmetic

Step 1: Having no common prime factor means their only common factor is 1. Step 2: Such numbers are called co-prime numbers. Step 3: Prime factorisation helps identify co-primality quickly.

Step 1: Calculate (2^5=32). Step 2: (32\times3\times7=672). Step 3: To get the number from prime factorisation, multiply all factors.

Step 1: Write (6480=648\times10). Step 2: (648=2^3\times3^4) and (10=2\times5), so (6480=2^4\times3^4\times5). Step 3: Comparing gives (a=4).

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

Step 1: Powers with the same base are added in multiplication. Step 2: The power of 7 in (a) is 2 and in (b) is 1. Step 3: In (ab), the power of 7 will be (2+1=3).

Step 1: The factorisation contains 2, 3, and 5. Step 2: (2\times3\times5=30), so the number must be divisible by 30. Step 3: Identify divisibility quickly from prime factors.

Step 1: Write (3087=63\times49). Step 2: (63=3^2\times7) and (49=7^2). Step 3: Therefore, (3087=3^2\times7^3).

Step 1: In (xy), the powers of the same base 2 are added. Step 2: The power of 2 in (x) is 5 and in (y) is 3. Step 3: So the power of 2 in (xy) is (5+3=8).

Step 1: For HCF, take the smaller powers of common prime factors. Step 2: The smaller powers are (2^3), (3^2), and (5^1). Step 3: (8\times9\times5=360), so the HCF is 360.

Step 1: The Fundamental Theorem of Arithmetic states uniqueness of prime factorisation. Step 2: The order of prime factors may change, but the prime factors themselves do not change. Step 3: In exams, do not treat a change of order as a different factorisation.

Step 1: Write (27720=2520\times11). Step 2: Since (2520=2^3\times3^2\times5\times7), the full factorisation is (2^3\times3^2\times5\times7\times11). Step 3: Do not leave a composite factor like 2520 in the final answer.

Step 1: For HCF, take the smaller powers of only the common prime factors. Step 2: The common prime factors are 2 and 3, with smaller powers (2^2) and (3^3). Step 3: (2^2\times3^3=4\times27=108), so the answer is 108.

Step 1: For LCM, take the highest powers of all prime factors. Step 2: The highest powers are (2^4), (3^5), (5^2), and (7). Step 3: (16\times243\times25\times7=680400), so the answer is 680400.

Step 1: For two numbers, product (=) HCF (\times) LCM. Step 2: Therefore, LCM (=60480\div72=840). Step 3: Use this relation directly only for two numbers.

Step 1: For two numbers, product (=) HCF (\times) LCM. Step 2: (45\times1260=56700). Step 3: In such questions, first notice that exactly two numbers are involved.

Step 1: (1800=18\times100=2^3\times3^2\times5^2). Step 2: For a perfect square, all exponents must be even, but the exponent of 2 is 3. Step 3: Multiplying by 2 makes it 4, so the smallest number is 2.

Step 1: (8820=2^2\times3^2\times5\times7^2). Step 2: For a perfect square, all exponents must be even, but the exponent of 5 is 1. Step 3: Dividing by 5 makes all remaining exponents even, so the smallest number is 5.

Step 1: (5400=54\times100=2^3\times3^3\times5^2). Step 2: For a perfect cube, each prime exponent must be a multiple of 3. Step 3: The exponent of 5 is 2, so one more 5 is needed; the smallest number is 5.

Step 1: Recognise (9261=21^3). Step 2: Since (21=3\times7), (21^3=3^3\times7^3). Step 3: (21^3) gives the value, but 21 is not prime, so write the final prime form separately.

Step 1: For a perfect square, every exponent must be even. Step 2: The exponents of (2^5) and (7^1) are odd, while the others are even. Step 3: Multiplying by (2\times7=14) makes all exponents even.