When `2^(256)` is divided by 17 the remainder would be

Options

A. 1
B. 16
C. 14
D. None of these

Answer

To find the remainder when \( 2^{256} \) is divided by 17, we can use Fermat's Little Theorem, which states that if \( p \) is a prime number and \( a \) is an integer not divisible by \( p \), then:

\[ a^{p-1} \equiv 1 \mod p \]

In this case, \( a = 2 \) and \( p = 17 \). Since 2 is not divisible by 17, we can apply the theorem.

### Step 1: Apply Fermat's Little Theorem
According to Fermat's Little Theorem:

\[ 2^{16} \equiv 1 \mod 17 \]

### Step 2: Reduce the exponent modulo \( p-1 \)
Now, we need to express \( 256 \) in terms of \( 16 \):

\[ 256 \mod 16 = 0 \]

This means that \( 256 \) is a multiple of \( 16 \).

### Step 3: Substitute back into the equation
Using the result from Fermat's theorem:

\[ 2^{256} = (2^{16})^{16} \equiv 1^{16} \mod 17 \]

### Step 4: Simplify the expression
Since \( 1^{16} = 1 \):

\[ 2^{256} \equiv 1 \mod 17 \]

### Conclusion
Thus, the remainder when \( 2^{256} \) is divided by 17 is:

\[ \text{Remainder} = 1 \]