 Try this beautiful problem from the Pre-RMO, 2017 based on Geometric Progression.

## Problem on Geometric Progression – PRMO 2017

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

• is 107
• is 10
• is 840
• cannot be determined from the given information

### Key Concepts

Geometric Progression

Greatest Integer

Real Number

But try the problem first…

Source

PRMO, 2017, Question 14

Elementary Algebra by Hall and Knight

## Try with Hints

First hint

here we have $[x]^{2}$=x{x}

$\Rightarrow$ {x}=a, [x]=ar, $x=ar^{2}$

$\Rightarrow a+ar=ar^{2}$

$\Rightarrow r^{2}-r-1=0$

$\Rightarrow r=\frac{1+\sqrt{5}}{2}$

Second Hint

Let ar=I

$\Rightarrow a=\frac{2I}{1+\sqrt{5}}=\frac{I(\sqrt{5}-1)}{2}$

for 0 $\lt$ a $\lt$ 1 $\Rightarrow 0 \lt \frac{I(\sqrt{5}-1)}{2} \lt 1$

$\Rightarrow 0 \lt I \lt \frac{(\sqrt{5}+1)}{2}$

Final Step

$\Rightarrow$ I=1

$\Rightarrow$ ar=1

$\Rightarrow a=\frac{2}{\sqrt{5}+1}=\frac{\sqrt{5}-1}{2}$

$x=ar^{2}=r=\frac{\sqrt{5}+1}{2}$

$\Rightarrow (\frac{\sqrt{5}+1}{2})^{n} \gt 100$

$\Rightarrow Nlog_{10}(\frac{\sqrt{5}+1}{2}) \gt 2$

$\Rightarrow N \gt 9.5$

$\Rightarrow N_{min}$=10.