Fundamental Theorem of Arithmetic

Step 1: Write (110880=10080\times11). Step 2: Since (10080=2^5\times3^2\times5\times7), the full form is (2^5\times3^2\times5\times7\times11). Step 3: Do not leave a composite factor such as 10080 in the final answer.

Step 1: For HCF, take the smaller powers of common prime factors. Step 2: The common factors are 2 and 3, with smaller powers (2^4) and (3^4). Step 3: (2^4\times3^4=16\times81=1296), so the answer is 1296.

Step 1: For LCM, take the highest powers of all prime factors. Step 2: The highest powers are (2^7), (3^6), (5^3), and (7^2). Step 3: (128\times729\times125\times49=714420000), so the answer is 714420000.

Step 1: For two numbers, product (=) HCF (\times) LCM. Step 2: LCM (=907200\div360=2520). Step 3: Use this relation directly only for two numbers.

Step 1: For two numbers, the product equals HCF multiplied by LCM. Step 2: (126\times4620=582120). Step 3: In such questions, first confirm that exactly two numbers are involved.

Step 1: (43200=432\times100=2^6\times3^3\times5^2). Step 2: For a perfect square, all exponents must be even, but the exponent of 3 is 3. Step 3: Multiplying by 3 makes it 4, so the smallest number is 3.

Step 1: (26460=2^2\times3^3\times5\times7^2). Step 2: For a perfect square, the odd powers of 3 and 5 must be reduced. Step 3: Dividing by (3\times5=15) leaves (2^2\times3^2\times7^2), so the smallest number is 15.

Step 1: (15120=2^4\times3^3\times5\times7). Step 2: For a perfect cube, exponents must be multiples of 3. Step 3: Multiplying by (2^2\times5^2\times7^2=2450) makes the powers 6, 3, 3, and 3.

Step 1: A convenient form is (250047=21^3\times13). Step 2: Since (21=3\times7), (21^3\times13=3^3\times7^3\times13). Step 3: Since 21 is composite, write 3 and 7 in the final prime form.

Step 1: For a perfect square, every exponent must be even. Step 2: The powers of 2, 5, and 11 are 9, 5, and 3, which are odd. Step 3: Multiplying by (2\times5\times11=110) makes all powers even.

Step 1: For a perfect cube, every exponent must be a multiple of 3. Step 2: We must make 8 to 9, 7 to 9, 4 to 6, and 2 to 3. Step 3: So the smallest multiplier is (2\times3^2\times5^2\times7).

Step 1: In multiplication, exponents of the same prime base are added. Step 2: The power of 3 in (a) is 5 and in (b) is 7. Step 3: In (ab), the power of 3 is (5+7=12).

Step 1: In (xy), powers with the same base 5 are added. Step 2: The power of 5 in (x) is 6 and in (y) is 4. Step 3: The total power is (6+4=10).

Step 1: HCF uses the smaller powers of common prime factors only. Step 2: The common factors are 2 and 3, with smaller powers (2^8) and (3^7). Step 3: Therefore, the HCF is (2^8\times3^7).

Step 1: LCM uses the highest powers of all prime factors. Step 2: The highest powers are (2^{11}), (3^9), (5^3), and (7^4). Step 3: So the correct form is (2^{11}\times3^9\times5^3\times7^4).

Step 1: Product of the two numbers is (84\times4620=388080). Step 2: One number is 420, so the other is (388080\div420=924). Step 3: As a check, the HCF of 420 and 924 is 84.

Step 1: Product of the two numbers is (96\times6720=645120). Step 2: The other number is (645120\div480=1344). Step 3: To check, the HCF of 480 and 1344 is 96.

Step 1: Write (45360=648\times70). Step 2: (648=2^3\times3^4) and (70=2\times5\times7), so (45360=2^4\times3^4\times5\times7). Step 3: Comparing gives (a=4).

Step 1: Write (604800=6048\times100). Step 2: (6048=2^5\times3^3\times7) and (100=2^2\times5^2), so the actual form is (2^7\times3^3\times5^2\times7). Step 3: The given power of 2 does not match, but the power of 3 is (b=3).

Step 1: To count with repetition, add the exponents. Step 2: (2^8) gives 8, (3^6) gives 6, (5^5) gives 5, and (13^2) gives 2 factors. Step 3: Total (8+6+5+2=21), so the answer is 21.