Difference between revisions of "1984 AHSME Problems/Problem 5"

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==Solution==
 
==Solution==
Since both sides are positive, we can take the <math> 100th </math> root of both sides to find the largest integer <math> n </math> such that <math> n^2<5^3 </math>. Fortunately, this is simple to evaluate: <math> 5^3=125 </math>, and the largest square less than <math> 125 </math> is <math> 11^2=121 </math>, so the largest <math> n </math> is <math> 11, \boxed{\text{D}} </math>.
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Since both sides are positive, we can take the <math> 100th </math> root of both sides to find the largest integer <math> n </math> such that <math> n^2<5^3 </math>. Fortunately, this is simple to evaluate: <math> 5^3=125 </math>, and the largest [[Perfect square|square]] less than <math> 125 </math> is <math> 11^2=121 </math>, so the largest <math> n </math> is <math> 11, \boxed{\text{D}} </math>.
  
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1984|num-b=4|num-a=6}}
 
{{AHSME box|year=1984|num-b=4|num-a=6}}

Revision as of 20:59, 16 June 2011

Problem 5

The largest integer $n$ for which $n^{200}<5^{300}$ is

$\mathrm{(A) \ }8 \qquad \mathrm{(B) \ }9 \qquad \mathrm{(C) \ } 10 \qquad \mathrm{(D) \ }11 \qquad \mathrm{(E) \ } 12$

Solution

Since both sides are positive, we can take the $100th$ root of both sides to find the largest integer $n$ such that $n^2<5^3$. Fortunately, this is simple to evaluate: $5^3=125$, and the largest square less than $125$ is $11^2=121$, so the largest $n$ is $11, \boxed{\text{D}}$.

See Also

1984 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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All AHSME Problems and Solutions