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Algebra Arithmetic Math Olympiad PRMO

Problem on Geometric Progression | PRMO 2017 | Question 14

Try this beautiful problem from the Pre-RMO, 2017 based on Geometric Progression. You may use sequential hints to solve the problem.

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

Check the Answer


But try the problem first…

Answer: is 10.

Source
Suggested Reading

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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