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2006 AIME II Problems/Problem 2

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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:

\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 893.

See also

2006 AIME II (ProblemsResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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