Prime Factorisation

Step 1: The condition requires (2^4 \times 5^2). Step 2: (2^4 \times 5^2=16 \times 25=400), and it has no factor (3). Step 3: Check the exponent of every prime before choosing.

Step 1: Write (840=84 \times 10). Step 2: Since (84=2^2 \times 3 \times 7) and (10=2 \times 5), (840=2^3 \times 3 \times 5 \times 7). The exponent of (7) is (1). Step 3: A prime appearing once has exponent (1).

Step 1: A divisor must not require prime exponents higher than those available. Step 2: (45=3^2 \times 5), which is fully contained in (3^2 \times 5^3). Step 3: Compare exponents to test divisibility.

Step 1: For a square factor, every prime exponent must be even. Step 2: For (2^3), even choices are (0,2); for (3^2), (0,2); for (5^2), (0,2). Total (2 \times 2 \times 2=8). Step 3: Count only even exponent choices for square factors.

Step 1: Write (540=54 \times 10). Step 2: (54=2 \times 3^3) and (10=2 \times 5), so (540=2^2 \times 3^3 \times 5). Step 3: Splitting a number into two easy parts saves time.

Step 1: First calculate (2^2=4). Step 2: (4 \times 3 \times 13=12 \times 13=156). Step 3: Break multiplication into smaller steps.

Step 1: In a perfect cube, every prime exponent must be a multiple of (3). Step 2: (2^6) is fine, but (3^2) needs one more (3) to become (3^3). Step 3: For cubes, make exponents multiples of (3).

Step 1: In a perfect square, all prime exponents are even. Step 2: (2^5) has an odd exponent, so multiplying by (2) makes it (2^6); (7^2) is already even. Step 3: For squares, fix only the odd exponents.

Step 1: One pair of (2) and (5) makes (10), giving a trailing zero. Step 2: Remove one (2) with the (5), leaving (2^3 \times 3^2=72), so the last non-zero digit is (2). Step 3: After removing trailing-zero pairs, check the last digit of the remaining product.

Step 1: A trailing zero comes from a pair (10=2 \times 5). Step 2: The exponent of (2) is (3) and of (5) is (1), so there is (1) pair. Step 3: For trailing zeros, take the smaller exponent of (2) and (5).

Step 1: Write (1260=126 \times 10). Step 2: (126=2 \times 3^2 \times 7) and (10=2 \times 5), so (1260=2^2 \times 3^2 \times 5 \times 7). It does not include (11). Step 3: Match the options with the prime factorisation.

Step 1: Write (900=9 \times 100). Step 2: (9=3^2) and (100=2^2 \times 5^2), so (900=2^2 \times 3^2 \times 5^2). Step 3: Using square numbers makes factorisation easier.

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

Step 1: In a perfect cube, all prime exponents are multiples of (3). Step 2: Each exponent is (3), so (N) is a perfect cube. Step 3: To identify a cube, check divisibility of exponents by (3).

Step 1: (144=12^2). Step 2: Since (12=2^2 \times 3), (144=2^4 \times 3^2); the sum is (4+2=6). Step 3: When squaring a number, its prime exponents double.

Step 1: A number is even if its prime factorisation contains (2). Step 2: Only the first option contains (2), so it represents an even number. Step 3: To check evenness, look only for the factor (2).

Step 1: An odd number has no factor (2) in its prime factorisation. Step 2: The first option contains only (3,5,7), so it is odd. Step 3: The presence of (2) makes a number even.

Step 1: Total frequency means the sum of exponents. Step 2: The exponents are (3) and (4), so the total frequency is (3+4=7). Step 3: Do not confuse distinct prime factors with total prime-factor frequency.

Step 1: Write (1080=108 \times 10). Step 2: (108=2^2 \times 3^3) and (10=2 \times 5), so (1080=2^3 \times 3^3 \times 5). The exponent of (2) is (3). Step 3: Repeated division by (2) is also useful for finding this exponent.

Step 1: To count factors, add (1) to each exponent. Step 2: For (p^2q^3), the number of factors is ((2+1)(3+1)=12). Step 3: When prime bases are distinct, factor count comes directly from exponents.