Question

The Prime factorization of the number 2304 is:

• A. \(3^2 \times 2^8\)
• B. \(3^1 \times 2^8\)
• C. \(3^2 \times 2^7\)
• D. \(3^1 \times 2^7\)

Hint:

We can write \(2304\) as a product of known squares:
\[ 2304 = 48^2 \]
Factor \(48\):
\[ 48 = 16 \times 3 = 2^4 \times 3^1 \]
Therefore, its square is:
\[ 48^2 = (2^4 \times 3^1)^2 = 2^8 \times 3^2 \]
This method is much faster than doing ten successive divisions.

Answer 1

The Correct Option is A

Step 1: Notice that 2304 looks like it could be a perfect square.
Checking, $48^2 = 2304$, so we can factor 2304 through 48 instead of dividing by 2 eight separate times.
Step 2: Factor 48 into primes. \[ 48 = 16 \times 3 = 2^4 \times 3 \]
Step 3: Square this factorisation to get back to 2304. \[ 2304 = 48^2 = (2^4 \times 3)^2 = 2^8 \times 3^2 \]
Step 4: Conclude. \[ \boxed{3^2 \times 2^8} \]