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

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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>.
 
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==
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==See also==
*[[2005 AMC 10A Problems]]
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{{AMC10 box|year=2005|ab=A|num-b=12|num-a=13}}
  
*[[2005 AMC 10A Problems/Problem 12|Previous Problem]]
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[[Category:Introductory Number Theory Problems]]
 
 
*[[2005 AMC 10A Problems/Problem 14|Next Problem]]
 
 
 
[[Category:Introductory Algebra Problems]]
 
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 14:34, 13 August 2019

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

2005 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 13
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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