Prime Factorisation

Step 1: Write (75=3 \times 25). Step 2: Since (25=5^2), (75=3 \times 5^2). Step 3: Do not leave (25) in the final answer because it is not prime.

Step 1: (10=2 \times 5). Step 2: The exponent of (2) is (4) and of (5) is (1), so only (1) pair of (10) can be formed. Step 3: The maximum number of divisions by (10) is decided by the smaller exponent.

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

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

Step 1: The prime factors are the base numbers. Step 2: Here (2,3,5) are prime factors, and the greatest is (5). Step 3: When the greatest prime factor is asked, do not choose a composite number.

Step 1: A factor divisible by (3) must have exponent of (3) at least (1). Step 2: The exponent of (2) has (3) choices (0,1,2), and exponent of (3) has (3) choices (1,2,3). Total (3 \times 3=9). Step 3: For conditional factors, adjust exponent choices carefully.

Step 1: A factor divisible by (10) must contain both (2) and (5). Step 2: The exponent of (2) can be (1) to (4), giving (4) choices, and the exponent of (5) can be (1) to (2), giving (2) choices. Total (4 \times 2=8). Step 3: Divisibility by (10) needs both prime factors.

Step 1: In a square factor, every prime exponent must be even. Step 2: For (2), choices are (0,2), so (2) choices; for (3), choices are (0,2), so (2) choices; for (5), only (0), so (1) choice. Total (2 \times 2 \times 1=4). Step 3: Count even exponent choices separately.

Step 1: Every integer greater than (1) has a unique prime factorisation apart from the order. Step 2: This comes from the fundamental theorem of arithmetic. Step 3: Remember uniqueness of prime factorisation with this theorem.

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: Do not make the mistake of treating (1) as prime.

Step 1: Write (210) as (21 \times 10). Step 2: (21=3 \times 7) and (10=2 \times 5), so (210=2 \times 3 \times 5 \times 7). Step 3: After factorisation, check that every base is prime.

Step 1: Write (168) as (8 \times 21). Step 2: (8=2^3) and (21=3 \times 7), so (168=2^3 \times 3 \times 7). Step 3: To find the required exponent, you do not need to calculate any extra value.

Step 1: Evaluate the powers first. Step 2: (2^2=4) and (3^3=27), so (4 \times 27 \times 5=540). Step 3: Solving powers before multiplication is a safe method.

Step 1: Write (196) as (14 \times 14). Step 2: Since (14=2 \times 7), (196=2^2 \times 7^2). Step 3: Do not leave composite bases like (4) or (14) in the final answer.

Step 1: Write (252) as (4 \times 63). Step 2: (4=2^2) and (63=3^2 \times 7), so (a=2) and (b=2). Hence (a+b=4). Step 3: Match prime bases to identify unknown exponents.

Step 1: To count total factors, add (1) to each exponent. Step 2: ((4+1)(2+1)=5 \times 3=15). Step 3: While counting factors, focus on the prime exponents.

Step 1: Total frequency is found by adding the exponents. Step 2: The exponents are (3,1,2), so the total is (3+1+2=6). Step 3: If no exponent is written, treat it as (1).

Step 1: Write (320) as (32 \times 10). Step 2: (32=2^5) and (10=2 \times 5), so (320=2^6 \times 5). Step 3: Do not keep (10) in the final form because it is not prime.

Step 1: Write (432) as (16 \times 27). Step 2: (16=2^4) and (27=3^3), so (432=2^4 \times 3^3). Step 3: Recognising squares and cubes makes factorisation faster.

Step 1: (2^2=4). Step 2: (4 \times 3 \times 11=12 \times 11=132). Step 3: Forming small products first reduces mistakes.