Prime Factorisation

Step 1: In a cube root, divide prime exponents by (3). Step 2: (2^6) becomes (2^2) and (3^3) becomes (3). Step 3: In cube roots, bases remain the same and only exponents change.

Step 1: When taking a square root, halve all prime exponents. Step 2: (2^4) becomes (2^2) and (3^2) becomes (3). Step 3: In square roots, halve the exponent, not the base.

Step 1: (12=2^2 \times 3). Step 2: From (2^3), (2^2) can be taken only (1) full time, while (3^2) can supply (3) twice. The limiting exponent is for (2), so the answer is (1). Step 3: For a composite divisor, check each required prime separately.

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

Step 1: For a divisor, its prime exponents must not exceed the available exponents. Step 2: (72=2^3 \times 3^2), which is fully present in the given number. Step 3: To test divisibility, match each prime exponent separately.

Step 1: (1000=10^3). Step 2: Since (10=2 \times 5), (10^3=(2 \times 5)^3=2^3 \times 5^3). Step 3: (10^3) is not final prime factorisation because (10) is not prime.

Step 1: Look at the exponent of (7) in each option. Step 2: The exponents are (1,2,3,2), and the greatest is (3). Step 3: For comparison, check only the required exponent instead of calculating each value.

Step 1: Compare the exponents. Step 2: The exponent of (2) is (2), of (3) is (1), and of (5) is (2). The greatest exponent is (2), shared by (2) and (5). Step 3: If exponents are equal, more than one prime base may be correct.

Step 1: (3^2=9) and (5^2=25). Step 2: (9 \times 25=225), so (N=225). Step 3: Remembering squares makes such calculations faster.

Step 1: Write (512) as (64 \times 8). Step 2: (64=2^6) and (8=2^3), so (512=2^9). Step 3: (8^3) may give the value, but prime factorisation must use base (2).

Step 1: The bases of the first number are (2,3,5). Step 2: The bases of the second number are (2,3,7), so the common primes are (2) and (3). Step 3: For common factors, choose only bases present in both numbers.

Step 1: In HCF, take the smaller exponent of a common prime. Step 2: The exponents of (2) are (4) and (2), so the smaller exponent is (2). Step 3: For HCF, remember the minimum exponent rule.

Step 1: In LCM, take the larger exponent of each prime. Step 2: The exponents of (3) are (2) and (3), so the larger exponent is (3). Step 3: For LCM, choose the maximum exponent.

Step 1: (81=9 \times 9). Step 2: Since (9=3^2), (81=3^2 \times 3^2=3^4). Step 3: (9^2) is not final prime factorisation because (9) is not prime.

Step 1: An even factor must have exponent of (2) at least (1). Step 2: The exponent of (2) has (2) choices (1,2), while (3) has (2) choices and (5) has (2) choices. Total (2 \times 2 \times 2=8). Step 3: While counting even factors, do not take exponent (0) for (2).

Step 1: In an odd factor, the exponent of (2) must be (0). Step 2: The exponent of (3) can be (0,1,2), giving (3) odd factors. Step 3: While counting odd factors, remove (2) completely.

Step 1: Trailing zeros are formed by pairs of (2) and (5). Step 2: The number of pairs equals the smaller of (a) and (b). So (\min(a,b)=2). Step 3: For trailing zeros, use the smaller exponent, not the sum.

Step 1: A prime number has exactly two positive factors. Step 2: (1) has only one positive factor, so it is not prime and has no prime factor. Step 3: Avoid the common mistake of treating (1) as prime.

Step 1: Every integer greater than (1) can be written as a product of prime factors. Step 2: This form is unique apart from order, and this comes from the fundamental theorem of arithmetic. Step 3: Remember uniqueness of prime factorisation with this theorem.

Step 1: Write (216) as (8 \times 27). Step 2: (8=2^3) and (27=3^3), so (216=2^3 \times 3^3). Hence (a=3, b=3). Step 3: For unknown exponents, split the number into familiar powers.