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

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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, applying the well-known logarithmic property <math>\log_{c} a \cdot \log_{c} b = \log_{c} ab</math>, we have
 
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<math>\log_{10} 12 + \log_{10} 75 > \log_{10} n </math>
 
<math>\log_{10} 12 + \log_{10} 75 > \log_{10} n </math>

Revision as of 19:45, 30 November 2019

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:

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

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

$12n > 75$

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

Also, applying the well-known logarithmic property $\log_{c} a \cdot \log_{c} b = \log_{c} ab$, we have

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

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

$n < 900$

Combining these two inequalities:

\[6.25 < n < 900\]

The number of possible integer values for $n$ is the number of integers over the interval $(6.25 , 900)$, which 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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