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 $\log_{c} a \cdot \log_{c} b = \log_{c} ab$ , 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:

$6.25 < n < 900$

The number of possible integer values for $n$ is the number of integers over the interval $(6.25 , 900)$ , which is $\boxed{893}$ .

2006 AIME II (Problems • Answer Key • Resources)
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

Page

Toolbox

Search

2006 AIME II Problems/Problem 2

Problem

Solution

See also