The greatest integer contained in `(sqrt3+1)^(6)` is

Options

A. 208
B. 416
C. 415
D. 207

Answer

To find the greatest integer contained in \((\sqrt{3} + 1)^{6}\), we can use the binomial theorem to expand the expression and then analyze the result.

### Step-by-Step Solution:

1. **Use the Binomial Theorem**:
The binomial theorem states that \((a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r\). Here, we will expand \((\sqrt{3} + 1)^{6}\).

\[ (\sqrt{3} + 1)^{6} = \sum_{r=0}^{6} \binom{6}{r} (\sqrt{3})^{6-r} (1)^{r} \]

2. **Calculate the Terms**:
The expansion gives us the following terms:
- For \(r = 0\): \(\binom{6}{0} (\sqrt{3})^{6} = 1 \cdot 3^{3} = 27\)
- For \(r = 1\): \(\binom{6}{1} (\sqrt{3})^{5} = 6 \cdot 3^{2.5} = 6 \cdot 3 \sqrt{3} = 18\sqrt{3}\)
- For \(r = 2\): \(\binom{6}{2} (\sqrt{3})^{4} = 15 \cdot 3^{2} = 15 \cdot 9 = 135\)
- For \(r = 3\): \(\binom{6}{3} (\sqrt{3})^{3} = 20 \cdot 3^{1.5} = 20 \cdot 3 \sqrt{3} = 60\sqrt{3}\)
- For \(r = 4\): \(\binom{6}{4} (\sqrt{3})^{2} = 15 \cdot 3^{1} = 15 \cdot 3 = 45\)
- For \(r = 5\): \(\binom{6}{5} (\sqrt{3})^{1} = 6 \cdot \sqrt{3} = 6\sqrt{3}\)
- For \(r = 6\): \(\binom{6}{6} (\sqrt{3})^{0} = 1\)

3. **Combine the Terms**:
Now we can combine all the terms:
\[ (\sqrt{3} + 1)^{6} = 27 + 135 + 45 + (18\sqrt{3} + 60\sqrt{3} + 6\sqrt{3}) + 1 \]
\[ = 207 + 84\sqrt{3} \]

4. **Calculate \((\sqrt{3} - 1)^{6}\)**:
Next, we will also consider \((\sqrt{3} - 1)^{6}\) to help find the greatest integer:
\[ (\sqrt{3} - 1)^{6} = \sum_{r=0}^{6} \binom{6}{r} (\sqrt{3})^{6-r} (-1)^{r} \]
The odd terms will cancel out when we add \((\sqrt{3} + 1)^{6}\) and \((\sqrt{3} - 1)^{6}\).

5. **Analyze \((\sqrt{3} - 1)^{6}\)**:
The value of \((\sqrt{3} - 1)\) is less than 1, hence \((\sqrt{3} - 1)^{6}\) will be a small positive number.

6. **Final Calculation**:
The sum of \((\sqrt{3} + 1)^{6}\) and \((\sqrt{3} - 1)^{6}\) gives:
\[ (\sqrt{3} + 1)^{6} + (\sqrt{3} - 1)^{6} = 207 + 84\sqrt{3} + \text{(small positive number)} \]
Since \((\sqrt{3} - 1)^{6}\) is between 0 and 1, we can conclude:
\[ \lfloor (\sqrt{3} + 1)^{6} \rfloor = 416 - (\sqrt{3} - 1)^{6} \]
Therefore, the greatest integer contained in \((\sqrt{3} + 1)^{6}\) is \(415\).

### Conclusion:
The greatest integer contained in \((\sqrt{3} + 1)^{6}\) is \(\boxed{415}\).