Prime Factorisation

Step 1: For prime factorisation (p^a q^b r^c), total factors are ((a+1)(b+1)(c+1)). Step 2: Here the powers are (3,2,1), so factors (=(4)(3)(2)=24). Step 3: In exams, always add (1) to each exponent before multiplying.

Step 1: In a perfect square, every prime factor must have an even exponent. Step 2: (2^4) and (5^2) are already even powers, but (3^3) is odd. Dividing by (3) leaves (3^2). Step 3: Fix only the prime factor with an odd exponent.

Step 1: In a perfect cube, every prime exponent must be a multiple of (3). Step 2: (2^5) needs one more (2), (3^2) needs one more (3), and (7^1) needs (7^2). Step 3: Complete each exponent to the next multiple of (3).

Step 1: HCF contains only common prime factors. Step 2: The smaller power of (2) is (2), and the smaller power of (3) is (2). So HCF (=2^2\times3^2). Step 3: For HCF, choose the smaller exponent of common primes.

Step 1: LCM includes every prime factor with its highest exponent. Step 2: Highest powers are (2^3), (3^2), and (5^2). Step 3: For LCM, do not miss a prime factor that appears in either number.

Step 1: LCM uses higher exponents and HCF uses lower exponents. Step 2: In the ratio, subtract exponents: (2^{4-2}\times3^{3-1}\times5^{2-1}=2^2\times3^2\times5). Step 3: Use exponent difference instead of calculating both full numbers.

Step 1: Multiply the prime factors according to their powers. Step 2: (2^2=4) and (7^2=49), so (4\times3\times49=588). Step 3: First evaluate powers, then multiply carefully.

Step 1: Total factors are ((a+1)(2+1)(1+1)). Step 2: ((a+1)\times3\times2=24), so (a+1=4) and (a=3). Step 3: For an unknown exponent, apply the factor-count rule directly.

Step 1: A factor divisible by (2) must contain (2^1) at least. Step 2: Powers of (2) can be (1,2,3), giving (3) choices; powers of (3) give (3) choices; (5) and (7) give (2) choices each. Total (=3\times3\times2\times2=36). Step 3: For conditional factors, adjust only the restricted prime exponent.

Step 1: An odd factor must not contain the prime (2). Step 2: Power of (2) is only (0); power of (3) has (3) choices and power of (5) has (2) choices. Total (=3\times2=6). Step 3: When counting odd factors, ignore the power choices of (2) except zero.

Step 1: Coprime numbers do not have any common factor greater than (1). Step 2: Therefore, their HCF is always (1), even if their product is large. Step 3: When you see coprime, think about common factors first.

Step 1: The fundamental theorem of arithmetic says every integer greater than (1) has a unique prime factorisation except for order. Step 2: Option A states this correctly. Step 3: Unique means the same prime factors appear, only their order may change.

Step 1: In square root, each prime exponent becomes half. Step 2: (2^6) becomes (2^3), (3^4) becomes (3^2), and (5^2) becomes (5). Step 3: This direct method works when all exponents are even.

Step 1: For cube root, divide each prime exponent by (3). Step 2: (9\div3=3), (6\div3=2), and (3\div3=1), so the cube root is (2^3\times3^2\times7). Step 3: In a perfect cube, all exponents are multiples of (3).

Step 1: A perfect square needs all exponents to be even. Step 2: (2^2) is already fine, (3^3) needs one (3), and (5^1) needs one (5). Step 3: Make only the odd exponents even.

Step 1: Division reduces exponents, and a perfect cube needs remaining exponents as multiples of (3). Step 2: Divide (2^4) by (2), (3^5) by (3^2), and (11^2) by (11^2). Step 3: In division questions, reduce each exponent to the nearest lower multiple of (3).

Step 1: Product of two numbers equals HCF multiplied by LCM. Step 2: Multiply (2^2\times3) with (2^5\times3^3\times5); add exponents to get (2^7\times3^4\times5). Step 3: When multiplying same bases, add exponents.

Step 1: Write (756) as (4\times189). Step 2: (4=2^2) and (189=27\times7=3^3\times7), so (756=2^2\times3^3\times7). Step 3: Break a large number into convenient factors first.

Step 1: Since (10=2\times5), the factor must contain at least one (2) and one (5). Step 2: Powers of (2): (1,2,3) give (3) choices; powers of (3): (0,1,2) give (3) choices; powers of (5): (1,2) give (2) choices. Total (=18). Step 3: For divisibility, check the required minimum prime powers.

Step 1: Total factors are ((5+1)(4+1)(1+1)=60). Step 2: Factors divisible by (6=2\times3) must have power of (2) at least (1) and power of (3) at least (1), so (5\times4\times2=40). Step 3: Not divisible by (6) means total minus divisible factors, (60-40=20).