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:

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

In this case, \( a = 5 \) and \( p = 13 \). Since 5 is not divisible by 13, we can apply the theorem.

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

\[ 5^{12} \equiv 1 \ (\text{mod} \ 13) \]

### Step 2: Find \( 99 \mod 12 \)
Now, we need to reduce the exponent 99 modulo 12 to use the theorem effectively.

\[ 99 \div 12 = 8 \quad \text{(quotient)} \]
\[ 99 - (12 \times 8) = 3 \quad \text{(remainder)} \]

So, \( 99 \mod 12 = 3 \).

### Step 3: Rewrite \( 5^{99} \)
Using the result from the modulo operation, we can rewrite \( 5^{99} \):

\[ 5^{99} \equiv 5^3 \ (\text{mod} \ 13) \]

### Step 4: Calculate \( 5^3 \)
Now, we calculate \( 5^3 \):

\[ 5^3 = 125 \]

### Step 5: Find \( 125 \mod 13 \)
Next, we find the remainder of 125 when divided by 13:

\[ 125 \div 13 = 9 \quad \text{(quotient)} \]
\[ 125 - (13 \times 9) = 125 - 117 = 8 \quad \text{(remainder)} \]

Thus,

\[ 125 \equiv 8 \ (\text{mod} \ 13) \]

### Conclusion
Therefore, the remainder when \( 5^{99} \) is divided by 13 is:

\[ \boxed{8} \]