Suppose x is a positive real number such that {x}, [x] and x are in the geometric progression. Find the least positive integer n such that `x ^(n) gt 100.` (Here [x] denotes the integer part of x and `{x} = x- [x])`

Answer

To solve the problem, we start by interpreting the given conditions. We have three quantities: the fractional part of \( x \) denoted as \( \{x\} \), the integer part of \( x \) denoted as \( [x] \), and \( x \) itself. We know that these three quantities are in geometric progression (GP).

1. **Define the Variables:**
Let \( [x] = k \) (the integer part of \( x \)), and thus the fractional part \( \{x\} = x - k \). The three quantities in GP can be expressed as:
\[ \{x\}, [x], x \implies (x - k), k, x \]

2. **Set Up the GP Condition:**
For three numbers \( a, b, c \) to be in GP, the condition is:
\[ b^2 = ac \]
Applying this to our quantities:
\[ k^2 = (x - k) \cdot x \]

3. **Expand the Equation:**
Expanding the equation gives:
\[ k^2 = x^2 - kx \]
Rearranging this, we have:
\[ x^2 - kx - k^2 = 0 \]

4. **Solve the Quadratic Equation:**
We can use the quadratic formula to solve for \( x \):
\[ x = \frac{k \pm \sqrt{k^2 + 4k^2}}{2} = \frac{k \pm \sqrt{5k^2}}{2} = \frac{k(1 \pm \sqrt{5})}{2} \]
We take the positive root since \( x \) is a positive real number:
\[ x = \frac{k(1 + \sqrt{5})}{2} \]

5. **Find the Fractional Part:**
The fractional part \( \{x\} = x - k \):
\[ \{x\} = \frac{k(1 + \sqrt{5})}{2} - k = k \left(\frac{1 + \sqrt{5}}{2} - 1\right) = k \left(\frac{\sqrt{5} - 1}{2}\right) \]

6. **Set the Range for \( k \):**
Since \( \{x\} \) must be in the interval \( [0, 1) \):
\[ 0 < k \left(\frac{\sqrt{5} - 1}{2}\right) < 1 \]
This gives:
\[ 0 < k < \frac{2}{\sqrt{5} - 1} \]
Calculating \( \frac{2}{\sqrt{5} - 1} \) gives approximately \( 1.618 \). Thus, \( k \) can take the integer value \( 1 \).

7. **Substituting \( k = 1 \):**
If \( k = 1 \):
\[ x = \frac{1(1 + \sqrt{5})}{2} = \frac{1 + \sqrt{5}}{2} \]
This value is known as the golden ratio \( \phi \approx 1.618 \).

8. **Finding \( n \) such that \( x^n > 100 \):**
We need to find the smallest integer \( n \) such that:
\[ \left(\frac{1 + \sqrt{5}}{2}\right)^n > 100 \]
Taking logarithms:
\[ n \log\left(\frac{1 + \sqrt{5}}{2}\right) > \log(100) \]
Since \( \log(100) = 2 \):
\[ n > \frac{2}{\log\left(\frac{1 + \sqrt{5}}{2}\right)} \]

9. **Calculating \( \log\left(\frac{1 + \sqrt{5}}{2}\right) \):**
We know \( \frac{1 + \sqrt{5}}{2} \approx 1.618 \), and \( \log(1.618) \approx 0.208 \):
\[ n > \frac{2}{0.208} \approx 9.615 \]
Thus, the least integer \( n \) is \( 10 \).

### Final Answer:
The least positive integer \( n \) such that \( x^n > 100 \) is \( n = 10 \).