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Olympiad workbook
Exam
Olympiad workbook
Subject
IOQM
Chapter
Sequences and Series

Olympiad workbook · IOQM · Sequences and Series

Suppose x is a positive real number such that \ x\ .[x] and x are...

Suppose x is a positive real number such that {x}.[x]\{\mathrm{x}\} .[\mathrm{x}] and x are in the geometric progression. Find the least positive integer nn such that xn>100x^n>100. (Here [x][x] denotes the integer part of xx and {x}=x−[x])\{x\}=x-[x])

Correct answer

10

Step-by-step solution

[x]2=x{x}{x}=a[x]=arx=ar2a+ar=ar2r2−r−1=0r=1±52⇒r=1+52ar=Ia=2I(1+5)a=I(5−1)2\begin{aligned} & {[x]^2=x\{x\}} \\ & \{x\}=a \\ & {[x]=a r} \\ & x=a r^2 \\ & a+a r=a r^2 \\ & r^2-r-1=0 \\ & r=\frac{1 \pm \sqrt{5}}{2} \Rightarrow r=\frac{1+\sqrt{5}}{2} \\ & a r=I \\ & a=\frac{2 I}{(1+\sqrt{5})} \\ & a=\frac{I(\sqrt{5}-1)}{2}\end{aligned} 0<a<10<I(5−1)2<10<I<25−10<I<5+12I=1\begin{aligned} & 0 < a < 1 \\ & 0 < \frac{I(\sqrt{5}-1)}{2} < 1 \\ & 0 < I < \frac{2}{\sqrt{5}-1} \\ & 0 < I < \frac{\sqrt{5}+1}{2} \\ & I=1\end{aligned} ar=1⇒a=21+5=5−12x=ar2=r=5+12(5+12)N>100⇒Nlog⁡10(5+12)>2N>9.5⇒Nmin⁡=10\begin{aligned} & a r=1 \Rightarrow a=\frac{2}{1+\sqrt{5}}=\frac{\sqrt{5}-1}{2} \\ & x=a r^2=r=\frac{\sqrt{5}+1}{2} \\ & \left(\frac{\sqrt{5}+1}{2}\right)^N>100 \Rightarrow N \log _{10}\left(\frac{\sqrt{5}+1}{2}\right)>2 \\ & N>9.5 \Rightarrow N_{\min }=10\end{aligned}
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