Find the remainder when `2^(2013)` in divided by 17.

Answer

To find the remainder when \( 2^{2013} \) is divided by 17, we can use properties of modular arithmetic and the concept of exponents.

### Step-by-Step Solution:

1. **Identify the Modulus**:
We need to find \( 2^{2013} \mod 17 \).

2. **Use Fermat's Little Theorem**:
According to Fermat's Little Theorem, if \( p \) is a prime number and \( a \) is an integer not divisible by \( p \), then:
\[ a^{p-1} \equiv 1 \mod p \]
Here, \( p = 17 \) and \( a = 2 \). Since 2 is not divisible by 17, we can apply the theorem:
\[ 2^{16} \equiv 1 \mod 17 \]

3. **Reduce the Exponent**:
We need to reduce the exponent 2013 modulo 16 (since \( 2^{16} \equiv 1 \)):
\[ 2013 \div 16 = 125 \quad \text{(quotient)} \]
\[ 2013 = 16 \times 125 + 13 \quad \text{(remainder)} \]
Thus, \( 2013 \mod 16 = 13 \).

4. **Rewrite the Original Expression**:
Now we can rewrite \( 2^{2013} \) using the reduced exponent:
\[ 2^{2013} \equiv 2^{13} \mod 17 \]

5. **Calculate \( 2^{13} \mod 17 \)**:
We can calculate \( 2^{13} \) directly or break it down:
\[ 2^1 = 2 \]
\[ 2^2 = 4 \]
\[ 2^3 = 8 \]
\[ 2^4 = 16 \equiv -1 \mod 17 \]
\[ 2^5 = 2 \times 2^4 = 2 \times (-1) = -2 \equiv 15 \mod 17 \]
\[ 2^6 = 2 \times 2^5 = 2 \times 15 = 30 \equiv 13 \mod 17 \]
\[ 2^7 = 2 \times 2^6 = 2 \times 13 = 26 \equiv 9 \mod 17 \]
\[ 2^8 = 2 \times 2^7 = 2 \times 9 = 18 \equiv 1 \mod 17 \]
\[ 2^9 = 2 \times 2^8 = 2 \times 1 = 2 \mod 17 \]
\[ 2^{10} = 4 \mod 17 \]
\[ 2^{11} = 8 \mod 17 \]
\[ 2^{12} = 16 \equiv -1 \mod 17 \]
\[ 2^{13} = 2 \times 2^{12} = 2 \times (-1) = -2 \equiv 15 \mod 17 \]

6. **Final Result**:
Therefore, the remainder when \( 2^{2013} \) is divided by 17 is:
\[ \boxed{15} \]