Prime Factorisation

Step 1: Since (25=5^2), the factor must contain at least (5^2). Step 2: Choices for (2): (5), for (3): (3), for (5): (2) or (3), giving (2) choices. Total (=5\times3\times2=30). Step 3: Treat (25) as (5^2) before counting.

Step 1: (12=2^2\times3), so the factor needs power of (2) at least (2) and power of (3) at least (1). Step 2: Choices are (4) for (2), (3) for (3), and (3) for (5). Total (=4\times3\times3=36). Step 3: First write the divisor in prime form, then set exponent limits.

Step 1: In division, exponents of the same base are subtracted. Step 2: (2^{2-a}=2^0) gives (a=2), (3^{3-b}=3^2) gives (b=1), and (5^{2-c}=5^1) gives (c=1). Step 3: If a prime is not visible in the result, treat its exponent as (0).

Step 1: Add (1) to each exponent and multiply to get total factors. Step 2: In option A, ((4+1)(2+1)(3+1)=5\times3\times4=60). Step 3: In option-based questions, test the exponents of each option.

Step 1: Since (9=3^2), power of (3) must be at least (2). Not divisible by (5) means power of (5) must be (0). Step 2: Choices are (4) for (2), (3) for (3), and (1) for (5). Total (=12). Step 3: Convert each condition into exponent restrictions.

Step 1: If (d=2^a\times3^b\times7^c), then (d^2=2^{2a}\times3^{2b}\times7^{2c}). Step 2: Conditions are (2a\le6), (2b\le2), (2c\le2), so choices are (4,2,2). Total (=16). Step 3: For square divisibility, double the exponents and compare.

Step 1: Let (d=2^a\times3^b\times5^c), so (d^3=2^{3a}\times3^{3b}\times5^{3c}). Step 2: (3a\le5), (3b\le4), and (3c\le3), giving (2) choices each. Total (=8). Step 3: For cube divisibility, triple the exponents and compare.

Step 1: To be divisible by neither (2) nor (3), powers of (2) and (3) must both be (0). Step 2: Power of (5) can be (0,1,2), giving (3) factors. Step 3: For neither-nor conditions, set both restricted prime powers to zero.

Step 1: HCF uses the smaller exponents. Step 2: HCF (=2^3\times3^2). Its number of factors is ((3+1)(2+1)=12). Step 3: First find the HCF, then count its factors.

Step 1: LCM takes the highest exponent of every prime factor. Step 2: LCM (=2^4\times3^2\times5^3). Total factors (=(4+1)(2+1)(3+1)=5\times3\times4=60). Step 3: After finding the LCM, apply the factor-count rule.

Step 1: The final form of prime factorisation is based only on prime numbers. Step 2: If a base like (45) remains, it must be written as (45=3^2\times5). Step 3: In exams, check every base before writing the final answer.

Step 1: Write (14256=16\times891). Step 2: (16=2^4) and (891=3^4\times11), so (14256=2^4\times3^4\times11). Step 3: Do not leave 891 in the final form.

Step 1: Write (18144=32\times567). Step 2: (32=2^5) and (567=3^4\times7), so (18144=2^5\times3^4\times7). Step 3: Give 567 its complete prime form.

Step 1: Write (24200=200\times121). Step 2: (200=2^3\times5^2) and (121=11^2), so (24200=2^3\times5^2\times11^2). Step 3: Do not leave 200 and 121 in the final form.

Step 1: Write (25410=210\times121). Step 2: (210=2\times3\times5\times7) and (121=11^2), so (25410=2\times3\times5\times7\times11^2). Step 3: Give prime form to both 210 and 121.

Step 1: Write (27216=16\times1701). Step 2: (1701=3^5\times7), so (27216=2^4\times3^5\times7). Step 3: Convert 1701 into prime form using 3 and 7.

Step 1: Write (30240=32\times945). Step 2: (945=3^3\times5\times7), so (30240=2^5\times3^3\times5\times7). Step 3: Do not keep 945 in the final answer.

Step 1: Write (36300=300\times121). Step 2: (300=2^2\times3\times5^2) and (121=11^2), so (36300=2^2\times3\times5^2\times11^2). Step 3: Convert 300 and 121 into prime powers.

Step 1: Write (38115=315\times121). Step 2: (315=3^2\times5\times7) and (121=11^2), so (38115=3^2\times5\times7\times11^2). Step 3: Break both parts completely.

Step 1: Write (43200=432\times100). Step 2: (432=2^4\times3^3) and (100=2^2\times5^2), so (43200=2^6\times3^3\times5^2). Step 3: Count the total power of 2 as 6.