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 $\{x\} = x - [x]$.)
Attempt: Let ${x}=a, [x]=ar, x=ar^2$. $a+ar=ar^2$ $r^2-r-1=0$ $r=\frac{1+\sqrt{5}}{2}$ Now, $0\leq {x}<1$ $0\leq ar<r$ $0\leq [x]<1.61$ So, $[x]=0,1$ So, $x=r[x]=0,\frac{1+\sqrt{5}}{2}$ $x$ is positive, so $x=\frac{1+\sqrt{5}}{2}$ $x^n>100$ Thus, $n=10$
I'm right?
$\lfloor x \rfloor =1$ as already shown (in comments). Let $x=1+y$
Let the three numbers be $\{y, 1, 1+y\}$. Then $y\times(1+y) = 1^2=1$. Solving this, we get $y = \frac{-1\pm \sqrt{1+4}}{2}$. Since $y\in[0,1]$, $y=\frac{-1\pm \sqrt{5}}{2} \simeq 0.6180 = \phi-1$, where $\phi$ is the golden ratio.
Thus the numbers in geometric progression are $\{\phi-1, 1, \phi\}$=$\{\phi', 1, \phi\}$ where $\phi'=\phi-1=\frac{1}{\phi}$ is also known as the conjugate of the golden ratio
Now, $\phi^{10}$ crosses 100 for the first time. So the required $n=10$
See here for more information on the golden ratio