Try this beautiful problem from the Pre-RMO, 2017 based on Geometric Progression.
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]
Geometric Progression
Greatest Integer
Real Number
But try the problem first...
Answer: is 10.
PRMO, 2017, Question 14
Elementary Algebra by Hall and Knight
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.
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]
Geometric Progression
Greatest Integer
Real Number
But try the problem first...
Answer: is 10.
PRMO, 2017, Question 14
Elementary Algebra by Hall and Knight
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.
