# 2006 AIME II Problems/Problem 2

## 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 (Problems • Answer Key • Resources) Preceded byProblem 1 Followed byProblem 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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