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

Options

A. 8
B. 4
C. 2
D. none

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, \(a = 2\) 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 16

Next, we need to reduce the exponent 2003 modulo 16:

\[ 2003 \div 16 = 125 \quad \text{(quotient)} \]
\[ 2003 - (125 \times 16) = 2003 - 2000 = 3 \]

So, \(2003 \equiv 3 \mod 16\).

### Step 3: Substitute back into the equation

Now we can substitute back into our original expression:

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

### Step 4: Calculate \(2^3\)

Now, we calculate \(2^3\):

\[ 2^3 = 8 \]

### Step 5: Find the remainder

Thus, we find:

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

### Conclusion

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

\[ \boxed{8} \]

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