The greatest integer less than or equal to `(sqrt(3)+1)^(6)` is

Options

A. 208
B. 104
C. 416
D. None of these

Answer

To find the greatest integer less than or equal to \((\sqrt{3} + 1)^6\), we can use the Binomial Theorem to expand the expression and then analyze the result.

### Step-by-Step Solution:

1. **Understanding the Expression**:
We need to calculate \((\sqrt{3} + 1)^6\). According to the Binomial Theorem, this can be expanded as:
\[ (\sqrt{3} + 1)^6 = \sum_{k=0}^{6} \binom{6}{k} (\sqrt{3})^k (1)^{6-k} \]

2. **Expanding the Expression**:
The expansion gives us:
\[ = \binom{6}{0} (\sqrt{3})^6 + \binom{6}{1} (\sqrt{3})^5 + \binom{6}{2} (\sqrt{3})^4 + \binom{6}{3} (\sqrt{3})^3 + \binom{6}{4} (\sqrt{3})^2 + \binom{6}{5} (\sqrt{3})^1 + \binom{6}{6} (1)^0 \]
Which simplifies to:
\[ = 27 + 6 \cdot \sqrt{3}^5 + 15 \cdot \sqrt{3}^4 + 20 \cdot \sqrt{3}^3 + 15 \cdot \sqrt{3}^2 + 6 \cdot \sqrt{3} + 1 \]

3. **Calculating Individual Terms**:
- \((\sqrt{3})^6 = 27\)
- \((\sqrt{3})^5 = 3^{5/2} = 3 \cdot \sqrt{3} \approx 5.196\)
- \((\sqrt{3})^4 = 9\)
- \((\sqrt{3})^3 = 3 \cdot \sqrt{3} \approx 5.196\)
- \((\sqrt{3})^2 = 3\)
- \((\sqrt{3})^1 = \sqrt{3} \approx 1.732\)

4. **Putting It All Together**:
Now substituting these values back into the expansion:
\[ = 27 + 6 \cdot (3 \cdot \sqrt{3}) + 15 \cdot 9 + 20 \cdot (3 \cdot \sqrt{3}) + 15 \cdot 3 + 6 \cdot \sqrt{3} + 1 \]
Simplifying this:
\[ = 27 + 18\sqrt{3} + 135 + 60\sqrt{3} + 45 + 6\sqrt{3} + 1 \]
Combining like terms:
\[ = 203 + (18 + 60 + 6)\sqrt{3} = 203 + 84\sqrt{3} \]

5. **Estimating \(\sqrt{3}\)**:
Using \(\sqrt{3} \approx 1.732\):
\[ 84\sqrt{3} \approx 84 \cdot 1.732 \approx 145.488 \]
Thus,
\[ (\sqrt{3} + 1)^6 \approx 203 + 145.488 \approx 348.488 \]

6. **Finding the Greatest Integer**:
The greatest integer less than or equal to \(348.488\) is \(348\).

### Final Answer:
The greatest integer less than or equal to \((\sqrt{3} + 1)^6\) is **348**.