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

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<math> 130n > n^2 > 16 </math>
 
<math> 130n > n^2 > 16 </math>
  
Solving each part seperatly:  
+
Solving each part separately:  
  
 
<math> n^2 > 16 \Longrightarrow n > 4 </math>  
 
<math> n^2 > 16 \Longrightarrow n > 4 </math>  

Revision as of 12:20, 8 November 2015

Problem

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

$(130n)^{50} > n^{100} > 2^{200}$?

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

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 $125 \Longrightarrow \mathrm{(E)}$.

See Also

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