If ` 5^(99) ` is divided by 13, the remainder is

Options

A. 2
B. 4
C. 6
D. 8

Explanation

To find the remainder when \( 5^{99} \) is divided by 13, 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 \( a^{p-1} \equiv 1 \mod p \).

### Step-by-step Solution:

1. **Identify the parameters**: Here, \( a = 5 \) and \( p = 13 \). Since 5 is not divisible by 13, we can apply Fermat's Little Theorem.

2. **Apply Fermat's Little Theorem**: According to the theorem, since \( p = 13 \), we have:
\[ 5^{12} \equiv 1 \mod 13 \]

3. **Reduce the exponent**: We need to express \( 5^{99} \) in terms of \( 5^{12} \):
\[ 99 \mod 12 \]
To find \( 99 \mod 12 \):
\[ 99 \div 12 = 8 \quad \text{(whole part)} \]
\[ 99 - (12 \times 8) = 99 - 96 = 3 \]
Thus, \( 99 \mod 12 = 3 \).

4. **Rewrite the exponent**: Therefore, we can rewrite \( 5^{99} \) as:
\[ 5^{99} \equiv 5^3 \mod 13 \]

5. **Calculate \( 5^3 \)**:
\[ 5^3 = 125 \]

6. **Find the remainder when dividing by 13**:
\[ 125 \div 13 = 9 \quad \text{(whole part)} \]
\[ 125 - (13 \times 9) = 125 - 117 = 8 \]
Thus, the remainder when \( 125 \) is divided by \( 13 \) is \( 8 \).

### Final Answer:
The remainder when \( 5^{99} \) is divided by \( 13 \) is \( \boxed{8} \).