The remainder when `2^(2003)` is divided by 17 is:

Options

A. 1
B. 2
C. 8
D. None of these

Answer

To find the remainder when \( 2^{2003} \) 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, \( p = 17 \) and \( a = 2 \). 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 16
Now, we need to reduce the exponent 2003 modulo 16 to simplify our calculation:

\[ 2003 \mod 16 \]

To find \( 2003 \mod 16 \), we divide 2003 by 16:

\[ 2003 \div 16 = 125.1875 \quad \text{(take the integer part, which is 125)} \]
\[ 125 \times 16 = 2000 \]
\[ 2003 - 2000 = 3 \]

Thus,

\[ 2003 \mod 16 = 3 \]

### Step 3: Substitute back into the exponent
Now we can rewrite \( 2^{2003} \) using the result from the modulo operation:

\[ 2^{2003} \equiv 2^3 \mod 17 \]

### Step 4: Calculate \( 2^3 \)
Now we calculate \( 2^3 \):

\[ 2^3 = 8 \]

### Step 5: Find the remainder
Thus, we have:

\[ 2^{2003} \equiv 8 \mod 17 \]

So the remainder when \( 2^{2003} \) is divided by 17 is:

\[ \boxed{8} \]