The unit digit of `(3+sqrt(5))^(2013)+(3-sqrt(5))^(2013)` is

Options

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

Answer

To find the unit digit of \( (3 + \sqrt{5})^{2013} + (3 - \sqrt{5})^{2013} \), we can use the properties of binomial expansion and the fact that \( (3 - \sqrt{5})^{2013} \) will be a very small number since \( 3 - \sqrt{5} \) is less than 1.

### Step-by-step Solution:

1. **Identify the terms**:
We have two terms: \( a = (3 + \sqrt{5})^{2013} \) and \( b = (3 - \sqrt{5})^{2013} \).

2. **Approximation of \( b \)**:
Since \( 3 - \sqrt{5} \) is a small positive number (approximately 0.764), raising it to the power of 2013 will make it extremely small. Thus, \( b \) can be considered negligible compared to \( a \).

3. **Use the Binomial Theorem**:
We can expand \( a \) using the Binomial Theorem:
\[ (3 + \sqrt{5})^{2013} = \sum_{k=0}^{2013} \binom{2013}{k} 3^{2013-k} (\sqrt{5})^k \]

4. **Combine the two expansions**:
The expression we want to evaluate is:
\[ (3 + \sqrt{5})^{2013} + (3 - \sqrt{5})^{2013} \]
When we expand both terms, the odd powers of \( \sqrt{5} \) will cancel out, and we will only be left with the even powers.

5. **Focus on the even powers**:
The even powers of \( \sqrt{5} \) can be expressed as \( 5^{k/2} \), where \( k \) is even. Therefore, we can rewrite our expression as:
\[ 2 \sum_{k \text{ even}} \binom{2013}{k} 3^{2013-k} 5^{k/2} \]

6. **Finding the unit digit**:
To find the unit digit of this expression, we need to focus on the unit digits of \( 3^{2013} \) and \( 5^{k/2} \) for even \( k \):
- The unit digit of \( 3^{2013} \) follows a cycle: \( 3, 9, 7, 1 \). Since \( 2013 \mod 4 = 1 \), the unit digit of \( 3^{2013} \) is \( 3 \).
- The unit digit of \( 5^{k/2} \) for any \( k \geq 0 \) is always \( 5 \).

7. **Calculate the unit digit of the sum**:
The unit digit of \( 2 \times (3 + 5) \) is:
\[ 2 \times 8 = 16 \]
Thus, the unit digit is \( 6 \).

### Final Answer:
The unit digit of \( (3 + \sqrt{5})^{2013} + (3 - \sqrt{5})^{2013} \) is **6**.