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Problem on Geometric Progression | PRMO 2017 | Question 14

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

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.

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

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.

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