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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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<math> (130n)^{50} > n^{100} > 2^{200}\ ?</math>
  
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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> \textbf{(A) } 0\qquad \textbf{(B) } 7\qquad \textbf{(C) } 12\qquad \textbf{(D) } 65\qquad \textbf{(E) } 125 </math>
  
 
==Solution==
 
==Solution==
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<math> (130n)^{50} > n^{100} > 2^{200} </math>
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We're given <math> (130n)^{50} > n^{100} > 2^{200} </math>, so
  
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<math> \sqrt[50]{(130n)^{50}} > \sqrt[50]{n^{100}} > \sqrt[50]{2^{200}} </math>
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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
  
 
<math> 130n > n^2 > 2^4 </math>  
 
<math> 130n > n^2 > 2^4 </math>  
  
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<math> 130n < n^2 < 16 </math>
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<math> 130n > n^2 > 16 </math>
  
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Solving each part seperatly:  
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Solving each part separately:  
  
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<math> n^2 > 16 </math>
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<math> n^2 > 16 \Longrightarrow n > 4 </math>  
  
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<math> n < 4 </math>  
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<math> 130n > n^2 \Longrightarrow 130 > n </math>  
  
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<math> 130n > n^2 </math>  
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So <math> 4 < n < 130 </math>.
  
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<math> 130 < n </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>129-5+1 = \boxed{\textbf{(E) }125}</math>
  
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So <math> 4 < n < 130 </math>.
 
  
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Therefore the answer is the number of positive integers over the interval <math> (4,130) </math> which is <math> 125 \Rightarrow E </math>.
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==Solution 2==
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We're given <math>\left(130n\right)^{50}>n^{100}>2^{200}</math>.
 +
 
 +
Alternatively to solution 1, first deal with the first half: <math>\left(130n\right)^{50}>\left(n^{2}\right)^{50}</math>. Because the exponents are equal, we can ignore them and solve for <math>n</math>: <math>130n>n^{2}</math>, or <math>n<130</math>.
 +
 
 +
The second half: <math>n^{100}>2^{200}</math>, or <math>n^{100}>4^{100}</math>, which means <math>n>4</math>.
 +
 
 +
Therefore <math>4<n<130</math> and <math>n=\left\{5,\ 6,\ 7,\ \cdots ,\ 128,\ 129\right\}</math> which contains the same number of elements as <math>\left\{1,\ 2,\ 3,\ \cdots ,\ 124,\ 125\right\}</math> which clearly contains <math>125</math> values or choice <math>\boxed{\textbf{(E) } 125}</math>.
  
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==See Also==
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~JH. L
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*[[2005 AMC 10A Problems]]
 
  
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*[[2005 AMC 10A Problems/Problem |Previous Problem]]
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==See also==
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{{AMC10 box|year=2005|ab=A|num-b=12|num-a=14}}
  
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*[[2005 AMC 10A Problems/Problem |Next Problem]]
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[[Category:Introductory Number Theory Problems]]
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{{MAA Notice}}

Latest revision as of 06:18, 16 July 2022

Problem

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

$(130n)^{50} > n^{100} > 2^{200}\ ?$

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

Solution

We're given $(130n)^{50} > n^{100} > 2^{200}$, so

$\sqrt[50]{(130n)^{50}} > \sqrt[50]{n^{100}} > \sqrt[50]{2^{200}}$ (because all terms are positive) and thus

$130n > n^2 > 2^4$

$130n > n^2 > 16$

Solving each part separately:

$n^2 > 16 \Longrightarrow n > 4$

$130n > n^2 \Longrightarrow 130 > n$

So $4 < n < 130$.

Therefore the answer is the number of positive integers over the interval $(4,130)$ which is $129-5+1 = \boxed{\textbf{(E) }125}$


Solution 2

We're given $\left(130n\right)^{50}>n^{100}>2^{200}$.

Alternatively to solution 1, first deal with the first half: $\left(130n\right)^{50}>\left(n^{2}\right)^{50}$. Because the exponents are equal, we can ignore them and solve for $n$: $130n>n^{2}$, or $n<130$.

The second half: $n^{100}>2^{200}$, or $n^{100}>4^{100}$, which means $n>4$.

Therefore $4<n<130$ and $n=\left\{5,\ 6,\ 7,\ \cdots ,\ 128,\ 129\right\}$ which contains the same number of elements as $\left\{1,\ 2,\ 3,\ \cdots ,\ 124,\ 125\right\}$ which clearly contains $125$ values or choice $\boxed{\textbf{(E) } 125}$.

~JH. L

See also

2005 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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