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Difference between revisions of "2005 AMC 10A Problems/Problem 13"

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How many positive integers <math>n</math> satisfy the following condition:
 
How many positive integers <math>n</math> satisfy the following condition:
  
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<math> (130n)^{50} > n^{100} > 2^{200} </math>?
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<cmath>\left(130n\right)^{50} > n^{100} > 2^{200} \ \text{?}</cmath>
  
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<math> \mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 7\qquad \mathrm{(C) \ } 12\qquad \mathrm{(D) \ } 65\qquad \mathrm{(E) \ } 125 </math>
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<math>
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\textbf{(A) } 0\qquad \textbf{(B) } 7\qquad \textbf{(C) } 12\qquad \textbf{(D) } 65\qquad \textbf{(E) } 125
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</math>
  
 
==Solution==
 
==Solution==
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We're given <math> (130n)^{50} > n^{100} > 2^{200} </math>, so
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Since <math>n > 0</math>, all <math>3</math> terms of the inequality are positive, so we may take the <math>50</math>th root, yielding
  
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<math> \sqrt[50]{(130n)^{50}} > \sqrt[50]{n^{100}} > \sqrt[50]{2^{200}} </math> (because all terms are positive) and thus
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<cmath>\begin{align*}&\sqrt[50]{(130n)^{50}} > \sqrt[50]{n^{100}} > \sqrt[50]{2^{200}} \\ \iff &130n > n^2 > 2^4 \\ \iff &130n > n^2 > 16.\end{align*}</cmath>
  
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<math> 130n > n^2 > 2^4 </math>  
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Solving each part separately, while noting that <math>n > 0</math>, therefore gives <math>n^2 > 16 \iff n > 4</math> and <math>130n > n^2 \iff n < 130</math>.
  
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<math> 130n > n^2 > 16 </math>
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Hence the solution is <math>4 < n < 130</math>, and therefore the answer is the number of positive integers in the open interval <math>(4,130)</math>, which is <math>129-5+1 = \boxed{\textbf{(E) } 125}</math>.
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Solving each part separately:
 
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<math> n^2 > 16 \Longrightarrow n > 4 </math>
 
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<math> 130n > n^2 \Longrightarrow 130 > n </math>
 
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So <math> 4 < n < 130 </math>.
 
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Therefore the answer is the number of [[positive integer]]s over the interval <math> (4,130) </math> which is <math> 125 \Longrightarrow \mathrm{(E)} </math>.
 
  
 
==See also==
 
==See also==
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{{AMC10 box|year=2005|ab=A|num-b=12|num-a=13}}
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{{AMC10 box|year=2005|ab=A|num-b=12|num-a=14}}
  
 
[[Category:Introductory Number Theory Problems]]
 
[[Category:Introductory Number Theory Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 17:42, 1 July 2025

Problem

How many positive integers $n$ satisfy the following condition:

\[\left(130n\right)^{50} > n^{100} > 2^{200} \ \text{?}\]

$\textbf{(A) } 0\qquad \textbf{(B) } 7\qquad \textbf{(C) } 12\qquad \textbf{(D) } 65\qquad \textbf{(E) } 125$

Solution

Since $n > 0$, all $3$ terms of the inequality are positive, so we may take the $50$th root, yielding

\begin{align*}&\sqrt[50]{(130n)^{50}} > \sqrt[50]{n^{100}} > \sqrt[50]{2^{200}} \\ \iff &130n > n^2 > 2^4 \\ \iff &130n > n^2 > 16.\end{align*}

Solving each part separately, while noting that $n > 0$, therefore gives $n^2 > 16 \iff n > 4$ and $130n > n^2 \iff n < 130$.

Hence the solution is $4 < n < 130$, and therefore the answer is the number of positive integers in the open interval $(4,130)$, which is $129-5+1 = \boxed{\textbf{(E) } 125}$.

See also

2005 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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