Fundamental Theorem of Arithmetic

Step 1: The theorem gives certainty of prime factorisation for numbers greater than 1. Step 2: Prime factors and their powers remain fixed; only their order may change. Step 3: In the final answer, do not leave composite factors.

Step 1: Write (221760=20160\times11). Step 2: Since (20160=2^6\times3^2\times5\times7), the complete prime form is (2^6\times3^2\times5\times7\times11). Step 3: 20160 is composite, so do not keep it in the final answer.

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 (2^5) and (3^5). Step 3: (32\times243=7776), so the answer is 7776.

Step 1: For LCM, take the highest powers of all prime factors. Step 2: The highest powers are (2^8), (3^7), (5^2), and (7^3). Step 3: Their product is 38482790400, so that is the answer.

Step 1: For two numbers, product (=) HCF (\times) LCM. Step 2: LCM (=1814400\div720=2520). Step 3: Use this relation directly only for two numbers.

Step 1: The product of two numbers equals the product of their HCF and LCM. Step 2: (210\times9240=1940400). Step 3: Apply this formula directly only when exactly two numbers are involved.

Step 1: (21600=216\times100=2^5\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: (52920=2^3\times3^3\times5\times7^2). Step 2: For a perfect square, odd powers of 2, 3, and 5 must be reduced. Step 3: Dividing by (2\times3\times5=30) leaves (2^2\times3^2\times7^2).

Step 1: (30240=2^5\times3^3\times5\times7). Step 2: For a perfect cube, exponents must be multiples of 3. Step 3: We need (2), (5^2), and (7^2), so the smallest multiplier is (2\times25\times49=2450).

Step 1: Write (500094=2\times250047). Step 2: (250047=3^3\times7^3\times13), so the full form is (2\times3^3\times7^3\times13). Step 3: Composite bases should not remain in final prime form.

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

Step 1: For a perfect cube, every exponent must be a multiple of 3. Step 2: We must make 10 to 12, 8 to 9, 5 to 6, and 1 to 3. Step 3: So the smallest multiplier is (2^2\times3\times5\times7^2).

Step 1: In multiplication, exponents of the same prime base are added. Step 2: The power of 3 in (a) is 6 and in (b) is 8. Step 3: In (ab), the power of 3 is (6+8=14).

Step 1: In (xy), powers with the same base 5 are added. Step 2: The power of 5 in (x) is 7 and in (y) is 5. Step 3: The total power is (7+5=12).

Step 1: HCF uses the smaller powers of common prime factors only. Step 2: The common factors are 2 and 3, with smaller powers (2^9) and (3^8). Step 3: So the correct form is (2^9\times3^8).

Step 1: LCM uses the highest powers of all prime factors. Step 2: The highest powers are (2^{12}), (3^{10}), (5^4), and (7^5). Step 3: So the correct form is (2^{12}\times3^{10}\times5^4\times7^5).

Step 1: Product of the two numbers is (120\times9240=1108800). Step 2: One number is 840, so the other is (1108800\div840=1320). Step 3: As a check, the HCF of 840 and 1320 is 120.

Step 1: Product of the two numbers is (144\times10080=1451520). Step 2: The other number is (1451520\div720=2016). Step 3: To check, the HCF of 720 and 2016 is 144.

Step 1: Write (136080=3888\times35). Step 2: (3888=2^4\times3^5) and (35=5\times7), so (p=2^4\times3^5\times5\times7). Step 3: Comparing gives (a=4).

Step 1: Write (1814400=18144\times100). Step 2: (18144=2^5\times3^4\times7) and (100=2^2\times5^2), so the actual form is (2^7\times3^4\times5^2\times7). Step 3: The power of 3 is (b=4).

Step 1: To count with repetition, add the exponents. Step 2: (2^9) gives 9, (3^7) gives 7, (5^6) gives 6, and (17^2) gives 2 factors. Step 3: Total (9+7+6+2=24), so the answer is 24.

Step 1: When counting distinct prime factors, exponents are not added. Step 2: The prime bases are 2, 3, 5, 7, and 13. 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 7429.

Step 1: Divide 65536 repeatedly by 2. Step 2: (65536=2^{16}). Step 3: (4^8) and (16^4) can give the value, but 4 and 16 are not prime.