Suppose $x$ is a positive real number such that $\{x\}, [x]$ and $x$ are in a geometric progression. Find the least positive integer $n$ such that $x^n > 100$. (Here $[x]$ denotes the integer part of $x$ and $\{x\} = x - [x]$.)
1. Let \( x = a \), \(\{x\} = ar\), and \([x] = ar^2\). Since \(\{x\}\), \([x]\), and \(x\) are in geometric progression, we have:
   \[
   \{x\} \cdot x = ([x])^2
   \]
   Substituting the values, we get:
   \[
   ar \cdot a = (ar^2)^2
   \]
   Simplifying, we obtain:
   \[
   a^2 r = a^2 r^4
   \]
   Dividing both sides by \(a^2\) (assuming \(a \neq 0\)), we get:
   \[
   r = r^4
   \]
   This implies:
   \[
   r^4 - r = 0
   \]
   Factoring out \(r\), we get:
   \[
   r(r^3 - 1) = 0
   \]
   Since \(r \neq 0\), we have:
   \[
   r^3 = 1
   \]
   Therefore:
   \[
   r = 1
   \]
   However, this contradicts the assumption that \(\{x\}\), \([x]\), and \(x\) are in geometric progression. Thus, we need to re-evaluate the initial conditions.

2. Given that \(\{x\}\), \([x]\), and \(x\) are in geometric progression, we can write:
   \[
   \{x\} = a, \quad [x] = ar, \quad x = ar^2
   \]
   Since \(x = [x] + \{x\}\), we have:
   \[
   ar^2 = ar + a
   \]
   Dividing both sides by \(a\) (assuming \(a \neq 0\)), we get:
   \[
   r^2 = r + 1
   \]
   This is a quadratic equation:
   \[
   r^2 - r - 1 = 0
   \]
   Solving for \(r\) using the quadratic formula \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -1\), and \(c = -1\), we get:
   \[
   r = \frac{1 \pm \sqrt{5}}{2}
   \]
   Since \(r\) must be positive, we take:
   \[
   r = \frac{1 + \sqrt{5}}{2}
   \]

3. Now, we need to find the least positive integer \(n\) such that \(x^n > 100\). Given \(x = ar^2\) and \(a = \{x\}\), we have:
   \[
   x = \frac{1 + \sqrt{5}}{2}
   \]
   Let \(r = \frac{1 + \sqrt{5}}{2}\). Then:
   \[
   x = r
   \]
   We need to find the smallest \(n\) such that:
   \[
   r^n > 100
   \]
   Taking the natural logarithm of both sides, we get:
   \[
   \ln(r^n) > \ln(100)
   \]
   Simplifying, we obtain:
   \[
   n \ln(r) > \ln(100)
   \]
   Solving for \(n\), we get:
   \[
   n > \frac{\ln(100)}{\ln(r)}
   \]
   Substituting \(r = \frac{1 + \sqrt{5}}{2}\), we have:
   \[
   \ln(r) = \ln\left(\frac{1 + \sqrt{5}}{2}\right)
   \]
   Using the approximate value \(\ln\left(\frac{1 + \sqrt{5}}{2}\right) \approx 0.4812\), we get:
   \[
   n > \frac{\ln(100)}{0.4812} \approx \frac{4.6052}{0.4812} \approx 9.57
   \]
   Therefore, the smallest integer \(n\) is:
   \[
   n = 10
   \]

The final answer is \(\boxed{10}\).