Difference between revisions of "2006 AIME II Problems/Problem 2"

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== Solution ==
 
== Solution ==
By the [[Triangle Inequality]]:
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By the [[Triangle Inequality]] and applying the well-known logarithmic property <math>\log_{c} a + \log_{c} b = \log_{c} ab</math>, we have that
 
<div style="text-align:center;">
 
<div style="text-align:center;">
 
<math>\log_{10} 12 + \log_{10} n > \log_{10} 75 </math>
 
<math>\log_{10} 12 + \log_{10} n > \log_{10} 75 </math>
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<math> n > \frac{75}{12} = \frac{25}{4} = 6.25 </math>
 
<math> n > \frac{75}{12} = \frac{25}{4} = 6.25 </math>
 
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Also:
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Also,
 
<div style="text-align:center;">
 
<div style="text-align:center;">
 
<math>\log_{10} 12 + \log_{10} 75 > \log_{10} n </math>
 
<math>\log_{10} 12 + \log_{10} 75 > \log_{10} n </math>
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<cmath> 6.25 < n < 900 </cmath>
 
<cmath> 6.25 < n < 900 </cmath>
  
The number of possible integer values for <math>n</math> is the number of integers over the interval <math>(6.25 , 900)</math>, which is <math>893</math>.
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Thus <math>n</math> is in the set <math>(6.25 , 900)</math>; the number of positive integer <math>n</math> which satisfies this requirement is <math>\boxed{893}</math>.
  
 
== See also ==
 
== See also ==
{{AIME box|year=2006|n=I|num-b=1|num-a=3}}
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{{AIME box|year=2006|n=II|num-b=1|num-a=3}}
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 13:58, 14 April 2020

Problem

The lengths of the sides of a triangle with positive area are $\log_{10} 12$, $\log_{10} 75$, and $\log_{10} n$, where $n$ is a positive integer. Find the number of possible values for $n$.

Solution

By the Triangle Inequality and applying the well-known logarithmic property $\log_{c} a + \log_{c} b = \log_{c} ab$, we have that

$\log_{10} 12 + \log_{10} n > \log_{10} 75$

$\log_{10} 12n > \log_{10} 75$

$12n > 75$

$n > \frac{75}{12} = \frac{25}{4} = 6.25$

Also,

$\log_{10} 12 + \log_{10} 75 > \log_{10} n$

$\log_{10} 12\cdot75 > \log_{10} n$

$n < 900$

Combining these two inequalities:

\[6.25 < n < 900\]

Thus $n$ is in the set $(6.25 , 900)$; the number of positive integer $n$ which satisfies this requirement is $\boxed{893}$.

See also

2006 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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