Prime Factorisation

Step 1: Total factors are ((r+1)(2+1)). Step 2: ((r+1)\times3=15), so (r+1=5) and (r=4). Step 3: Write the factor-count formula first, then solve the simple equation.

Step 1: A perfect square needs every exponent even. Step 2: (2^3) and (5^1) have odd powers, so multiply by (2\times5). Step 3: Do not multiply the prime whose exponent is already even.

Step 1: HCF is (2^1\times3^1\times11^1), and LCM is (2^2\times3^3\times11^2). Step 2: Dividing both sides by the HCF gives (1:2^1\times3^2\times11^1). Step 3: Simplifying the ratio is important in such questions.

Step 1: An even factor must contain at least one (2). Step 2: Power of (2) has (2) choices, (1) or (2); each of (3,5,7) has (2) choices. Total (=16). Step 3: For even factors, exclude the case (2^0).

Step 1: Write (990=99\times10). Step 2: (99=3^2\times11) and (10=2\times5), so (990=2\times3^2\times5\times11). Step 3: Split the number into easy parts for faster factorisation.

Step 1: Since (15=3\times5), the factor must contain both (3) and (5). Step 2: Power choices are (2:4) choices, (3:2) choices, (5:2) choices, and (7:2) choices. Total (=4\times2\times2\times2=32). Step 3: Start restricted prime powers from the minimum required value.

Step 1: Total factors are ((a+1)(b+1)). Step 2: For (a=2,\ b=5), we get ((3)(6)=18). The other options do not give (18). Step 3: In option-based questions, substitute values into the rule quickly.

Step 1: HCF uses lower exponents and LCM uses higher exponents. Step 2: Exponent differences give (2^{4-2}\times3^{2-1}\times5^{3-1}=2^2\times3\times5^2). Step 3: For how many times, divide LCM by HCF.

Step 1: A perfect square requires all prime exponents to be even. Step 2: The exponent of (2) is already (3), which is odd, so changing only (x) cannot make the number a perfect square. Step 3: Check the whole prime factorisation, not only the unknown exponent.

Step 1: Total factors are ((x+1)(2+1)(4+1)). Step 2: ((x+1)\times3\times5=45), so (x+1=3) and (x=2). Step 3: Divide the given factor count by the known parts first.

Step 1: Write (1260=126\times10). Step 2: (126=2\times3^2\times7) and (10=2\times5), so (1260=2^2\times3^2\times5\times7). Step 3: Combine repeated prime factors into powers at the end.

Step 1: A square factor must have even exponents for every prime. Step 2: For (2), choices are (0,2); for (3), choices are (0,2); for (5), only (0). Total (=2\times2\times1=4). Step 3: Count only even exponent choices for square factors.

Step 1: A cube factor must have exponents that are multiples of (3). Step 2: For (2), only (0); for (3), (0) or (3); for (5), only (0). Total (=2). Step 3: Remember that (0) is also a multiple of (3).

Step 1: When multiplying powers with the same prime base, add exponents. Step 2: (2^{3+2}\times3^{1+2}\times5^{1+2}=2^5\times3^3\times5^3). Step 3: Add exponents for multiplication with the same base.

Step 1: When dividing powers with the same base, subtract exponents. Step 2: (2^{6-2}\times3^{4-1}\times5^{2-1}=2^4\times3^3\times5). Step 3: In division, subtract the smaller exponent from the larger one.

Step 1: A factor divisible by (5) must have power of (5) at least (1). Step 2: Choices are (4) for (2), (3) for (3), and (4) for (5). Total (=4\times3\times4=48). Step 3: Do not include zero power for the required prime.

Step 1: For total factors, add (1) to each exponent. Step 2: ((2+1)(3+1)(1+1)(2+1)=3\times4\times2\times3=72). Step 3: The same rule works even when many prime factors are present.

Step 1: After division, remaining exponents must be multiples of (3). Step 2: Remove (2^2), remove (3) from (3^4), and remove (5^2) from (5^5). Step 3: Reduce each exponent to the nearest lower multiple of (3).

Step 1: Total factors are ((3+1)(2+1)(x+1)). Step 2: (4\times3\times(x+1)=96), so (x+1=8) and (x=7). Step 3: Multiply known parts first, then solve for the unknown exponent.

Step 1: Product of two numbers (=) HCF (\times) LCM. Step 2: (12=2^2\times3) and (180=2^2\times3^2\times5), so the product is (2^4\times3^3\times5). Step 3: Convert given numbers to prime form before multiplying.