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: Understanding the Question:
This question is from Number Systems, focusing on the prime factorization of a composite number.
We need to express the integer \(2304\) as a product of powers of its prime factors (which are 2 and 3).

Step 2: Key Formula or Approach:
We perform successive division of \(2304\) by prime numbers, starting with the smallest prime, 2, until we obtain 1.

Step 3: Detailed Explanation:
Let us divide \(2304\) successively by 2:
\[ 2304 \div 2 = 1152 \]
\[ 1152 \div 2 = 576 \]
\[ 576 \div 2 = 288 \]
\[ 288 \div 2 = 144 \]
\[ 144 \div 2 = 72 \]
\[ 72 \div 2 = 36 \]
\[ 36 \div 2 = 18 \]
\[ 18 \div 2 = 9 \]
Since 9 is not divisible by 2, we divide by the next prime number, 3:
\[ 9 \div 3 = 3 \]
\[ 3 \div 3 = 1 \]
Now count the number of times each prime factor was used:
- The prime number 2 was used 8 times.
- The prime number 3 was used 2 times.
Therefore, the prime factorization is:
\[ 2304 = 2^8 \times 3^2 \]

Step 4: Final Answer:
The prime factorization of \(2304\) is \(3^2 \times 2^8\).