Prime Factorisation

Step 1: A factor divisible by (3) must have the exponent of (3) at least (1). Step 2: The exponent of (2) has (5) choices from (0) to (4), and the exponent of (3) has (3) choices (1,2,3). Total (5 \times 3=15). Step 3: In conditional factor counting, adjust exponent limits.

Step 1: A factor divisible by (10) must contain both (2) and (5). Step 2: The exponent of (2) has (4) choices from (1) to (4), and the exponent of (5) has (3) choices from (1) to (3). Total (4 \times 3=12). Step 3: Divisibility by (10) needs both primes.

Step 1: In a square factor, every prime exponent must be even. Step 2: For (2), choices are (0,2,4), giving (3) choices; for (3), choices are (0,2), giving (2); for (7), only (0), giving (1). Total (6). Step 3: Count even exponent choices separately.

Step 1: (729=27 \times 27). Step 2: Since (27=3^3), (729=3^3 \times 3^3=3^6). Step 3: For larger powers, split the number into familiar cubes.

Step 1: In prime factorisation, every base must be prime. Step 2: (21) is not prime because (21=3 \times 7), so the third option is not prime factorisation. Step 3: Always identify hidden composite numbers in options.

Step 1: In a perfect square, all prime exponents are even. Step 2: Here the exponents are (4,2,2), and all are even, so (N) is a perfect square. Step 3: To identify a perfect square, check whether the exponents are even.

Step 1: In a perfect cube, all prime exponents are multiples of (3). Step 2: Both (3) and (6) are multiples of (3), so (N) is a perfect cube. Step 3: For a perfect cube, check exponents using (3).

Step 1: An odd factor must not contain (2). Step 2: Removing (2^3) leaves (3^2 \times 5^2=9 \times 25=225). Step 3: To get the greatest odd factor, remove all powers of (2).

Step 1: The given number has no (2), so it is odd. Step 2: Multiplying by just one (2) gives the smallest even multiple. Step 3: When the smallest multiple is asked, do not increase exponents unnecessarily.

Step 1: A square root is an integer only when all prime exponents are even. Step 2: In (2^3 \times 3^4), the exponent of (2) is (3), which is odd. Step 3: In square-root questions, check the evenness of each exponent.

Step 1: In a cube root, divide prime exponents by (3). Step 2: (2^9) becomes (2^3), and (3^3) becomes (3). Step 3: In cube roots, bases stay the same and only exponents change.

Step 1: When taking a square root, halve the prime exponents. Step 2: (2^6) becomes (2^3), and (5^2) becomes (5). Step 3: In square roots, the base does not change; the exponent is halved.

Step 1: (18=2 \times 3^2). Step 2: (2^5) can supply (2) five times, but (3^3) can supply (3^2) only once. So the answer is (1). Step 3: For a composite divisor, the most limiting prime exponent decides the answer.

Step 1: (10=2 \times 5). Step 2: The exponent of (2) is (7), and the exponent of (5) is (3), so (3) complete pairs of (10) can be formed. Step 3: The number of divisions by (10) is decided by the smaller exponent.

Step 1: A divisor's prime exponents must not exceed the given number's exponents. Step 2: (144=2^4 \times 3^2), which is fully contained in (2^4 \times 3^3). Step 3: For divisibility, match each prime exponent separately.

Step 1: Write (1250) as (125 \times 10). Step 2: (125=5^3) and (10=2 \times 5), so (1250=2 \times 5^4). Step 3: Do not keep (10) in the final form because it is not prime.

Step 1: Look at the exponent of (5) in each option. Step 2: The exponents are (1,2,3,2), and the greatest is (3). Step 3: For comparison, you do not need to calculate the full value.

Step 1: Compare all prime exponents. Step 2: The exponent of (2) is (3), of (3) is (2), and of (7) is (2). The greatest exponent is (3), attached to (2). Step 3: Read the base and exponent separately while comparing.

Step 1: (2^2=4) and (7^2=49). Step 2: (4 \times 49=196), so (N=196). Step 3: Knowing squares helps solve such calculations quickly.

Step 1: Recognise (1024) as a power of (2). Step 2: (1024=2^{10}). Step 3: (4^5) gives the value, but prime factorisation must use base (2).