Fundamental Theorem of Arithmetic

Step 1: The theorem says that prime factorisation of a number greater than 1 is fixed. Step 2: The order may change, but the prime bases and their powers do not change. Step 3: Keep only prime factors in the final answer.

Step 1: Write (443520=40320\times11). Step 2: Since (40320=2^7\times3^2\times5\times7), the complete prime form is (2^7\times3^2\times5\times7\times11). Step 3: 40320 is composite, so it should not remain in the final form.

Step 1: For HCF, take the smaller powers of common prime factors only. Step 2: The common factors are 2 and 3, with smaller powers (2^6) and (3^6). Step 3: (64\times729=46656), so the answer is 46656.

Step 1: For LCM, take the highest powers of all prime factors. Step 2: The highest powers are (2^9), (3^8), (5^4), and (7^3). Step 3: So the correct form is (2^9\times3^8\times5^4\times7^3).

Step 1: For two numbers, product (=) HCF (\times) LCM. Step 2: LCM (=3628800\div1440=2520). Step 3: Use this relation directly for exactly two numbers.

Step 1: The product of two numbers equals the product of their HCF and LCM. Step 2: (420\times18480=7761600). Step 3: Apply this formula directly only when exactly two numbers are involved.

Step 1: (86400=2^7\times3^3\times5^2). Step 2: For a perfect square, all exponents must be even, but powers of 2 and 3 are odd. Step 3: Multiplying by (2\times3=6) makes all powers even.

Step 1: (105840=2^4\times3^3\times5\times7^2). Step 2: For a perfect square, odd powers of 3 and 5 must be reduced. Step 3: Dividing by (3\times5=15) leaves (2^4\times3^2\times7^2).

Step 1: (60480=2^6\times3^3\times5\times7). Step 2: For a perfect cube, every exponent must be a multiple of 3. Step 3: Powers of 5 and 7 are 1, so multiply by (5^2\times7^2=1225).

Step 1: Write (1000188=4\times250047). Step 2: (4=2^2) and (250047=3^6\times7^3), so the prime form is (2^2\times3^6\times7^3). Step 3: Treat 729 and 343 as powers of prime bases and keep prime bases in the final answer.

Step 1: For a perfect square, every exponent must be even. Step 2: Powers of 2 and 5 are odd, while the other powers are even. Step 3: Multiplying by (2\times5=10) makes all powers even.

Step 1: For a perfect cube, exponents must be multiples of 3. Step 2: We must make 11 to 12, 8 to 9, 5 to 6, and 2 to 3. Step 3: Therefore, the smallest multiplier is (2\times3\times5\times7).

Step 1: In multiplication, exponents of the same prime base are added. Step 2: The power of 3 in (a) is 7 and in (b) is 9. Step 3: In (ab), the power of 3 is (7+9=16).

Step 1: In (xy), powers with the same base 5 are added. Step 2: The power of 5 in (x) is 8 and in (y) is 6. Step 3: The total power is (8+6=14).

Step 1: HCF uses the smaller powers of common prime factors only. Step 2: The common factors are 2 and 3, with smaller powers (2^{10}) and (3^9). Step 3: Therefore, the correct form is (2^{10}\times3^9).

Step 1: LCM uses the highest powers of all prime factors. Step 2: The highest powers are (2^{13}), (3^{11}), (5^5), and (7^4). Step 3: So the correct form is (2^{13}\times3^{11}\times5^5\times7^4).

Step 1: Product of the two numbers is (180\times27720=4989600). Step 2: One number is 1260, so the other number is (4989600\div1260=3960). Step 3: As a check, the HCF of 1260 and 3960 is 180.

Step 1: Product of the two numbers is (216\times30240=6531840). Step 2: The other number is (6531840\div1512=4320). Step 3: To check, the HCF of 1512 and 4320 is 216.

Step 1: Write (90720=2592\times35). Step 2: (2592=2^5\times3^4) and (35=5\times7), so (p=2^5\times3^4\times5\times7). Step 3: Comparing gives (a=5).

Step 1: (3628800=2^8\times3^4\times5^2\times7). Step 2: The given power of 2 does not fully match, but the power of 3 is clearly 4. Step 3: Therefore, (b=4).

Step 1: To count with repetition, add the exponents. Step 2: (2^{10}) gives 10, (3^8) gives 8, (5^7) gives 7, and (19^2) gives 2 factors. Step 3: Total (10+8+7+2=27), so the answer is 27.

Step 1: When counting distinct prime factors, exponents are not added. Step 2: The prime bases are 2, 3, 5, 7, and 17. Step 3: Therefore, the number of distinct prime factors is 5.

Step 1: The prime factors of the first number are 2, 3, and 13. Step 2: The prime factors of the second number are 5, 7, and 11. Step 3: There is no common prime factor, so they are co-prime.

Step 1: Co-prime numbers have HCF 1. Step 2: For two numbers, product (=) HCF (\times) LCM. Step 3: Therefore, the LCM will be 12673.

Step 1: Divide 131072 repeatedly by 2. Step 2: (131072=2^{17}). Step 3: (4^9) and (16^5) are not prime factorisations because 4 and 16 are composite.