`2^16 -1 ` is divisible by

Options

A. 11
B. 13
C. 17
D. 19

Answer

To determine the divisibility of \(2^{16} - 1\), we can use the difference of squares factorization. Here’s a step-by-step solution:

### Step 1: Factor \(2^{16} - 1\)
We can express \(2^{16} - 1\) as a difference of squares:
\[ 2^{16} - 1 = (2^8)^2 - 1^2 = (2^8 - 1)(2^8 + 1) \]

### Step 2: Calculate \(2^8 - 1\) and \(2^8 + 1\)
Now, we calculate \(2^8 - 1\) and \(2^8 + 1\):
\[ 2^8 = 256 \implies 2^8 - 1 = 256 - 1 = 255 \]
\[ 2^8 + 1 = 256 + 1 = 257 \]

### Step 3: Factor \(255\)
Next, we factor \(255\):
\[ 255 = 3 \times 5 \times 17 \]

### Step 4: Check the divisibility of \(257\)
Now we check if \(257\) is a prime number. It is not divisible by any prime numbers less than \(\sqrt{257}\) (which are 2, 3, 5, 7, 11, 13, 17). Therefore, \(257\) is a prime number.

### Step 5: Combine the factors
Thus, we can conclude that:
\[ 2^{16} - 1 = (2^8 - 1)(2^8 + 1) = 255 \times 257 \]
This means \(2^{16} - 1\) is divisible by \(3\), \(5\), \(17\), and \(257\).

### Step 6: Identify the correct option
From the options provided, we find that \(2^{16} - 1\) is divisible by \(17\).

### Final Answer
Thus, the answer is that \(2^{16} - 1\) is divisible by \(17\).

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