Prime Factorisation

Step 1: A prime number has exactly two positive factors. Step 2: (1) has only one positive factor, so it has no prime factor. Step 3: Treating (1) as prime is a common mistake.

Step 1: An odd factor must not contain (2). Step 2: Remove (2^3) and keep (3^2 \times 5=45). Step 3: For the greatest odd factor, remove all powers of (2).

Step 1: The given number is odd because it has no factor (2). Step 2: To make the smallest even multiple, multiply by only one (2). Step 3: Do not increase exponents unnecessarily when the smallest value is asked.

Step 1: Trailing zeros are formed by pairs of (2) and (5). Step 2: The number of pairs equals the smaller exponent of (a) and (b), so (\min(a,b)=4). Step 3: For trailing zeros, count the minimum exponent, not the sum.

Step 1: Write (72=8 \times 9). Step 2: (8=2^3) and (9=3^2), so (72=2^3 \times 3^2). Step 3: Recognising simple squares and cubes speeds up factorisation.

Step 1: The total number of factors is found by multiplying ((exponent+1)). Step 2: ((2+1)(3+1)(1+1)=3 \times 4 \times 2=24). Step 3: If no exponent is written, treat it as (1).

Step 1: When taking a square root, halve the prime exponents. Step 2: (2^4) becomes (2^2) and (3^2) becomes (3), so (\sqrt{N}=2^2 \times 3). Step 3: In square roots, halve exponents, not bases.

Step 1: In a cube root, divide each prime exponent by (3). Step 2: (2^6) becomes (2^2) and (5^3) becomes (5). Step 3: The cube root is an integer only when all exponents are multiples of (3).

Step 1: The prime factors of the first number are (2,3,5). Step 2: The second number has (2,3,7), so (5) is not common. Step 3: To check common factors, compare the prime bases.

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

Step 1: (12=2^2 \times 3). Step 2: Each division by (12) uses (2^2) and (3). From (2^5), this can happen (2) times, and from (3^2), also (2) times. Step 3: For a composite divisor, check the limiting prime exponent.

Step 1: (625=25 \times 25). Step 2: Since (25=5^2), (625=5^2 \times 5^2=5^4). Step 3: Do not write (25) in prime factorisation because (25) is not prime.

Step 1: In prime factorisation, every base must be prime. Step 2: (9) is not prime because (9=3^2), so the first option is not prime factorisation. Step 3: Every base must be prime; writing powers alone is not enough.

Step 1: In a perfect square, all exponents must be even. Step 2: Make (2^3) into (2^4) and (5) into (5^2); (3^2) is already fine. Step 3: For the smallest square multiple, increase only the necessary exponents.

Step 1: In a perfect cube, exponents must be multiples of (3). Step 2: (2^2) must become (2^3), and (3^5) must become (3^6). Step 3: For the smallest cube multiple, move each exponent to the next multiple of (3).

Step 1: A factor divisible by (3) must have exponent of (3) at least (1). Step 2: The exponent of (2) has (5) choices, and exponent of (3) has (1,2,3), so (3) choices; total (5 \times 3=15). Step 3: For conditional factor counts, adjust the exponent choices.

Step 1: A factor divisible by (10) must contain at least one (2) and one (5). Step 2: Exponent choices for (2) are (1) to (5), so (5) choices; for (5), (1) to (4), so (4) choices. Total (20). Step 3: Divisibility by (10) needs both prime factors.

Step 1: The total number of factors is ((a+1)(b+1)). Step 2: With (a=4), (5(b+1)=20), so (b+1=4) and (b=3). Step 3: Make an equation when finding an exponent from factor count.

Step 1: In a square factor, every prime exponent must be even. Step 2: For each prime, the exponent can be (0) or (2), giving (2) choices. Total (2 \times 2 \times 2=8). Step 3: Count even exponent choices separately and multiply.

Step 1: The fundamental theorem of arithmetic says every integer greater than (1) has a unique prime factorisation apart from order. Step 2: So the given factorisation of (N) is based on this property. Step 3: Link uniqueness of prime factorisation with this theorem.