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

Revision as of 19:48, 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 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:

$6.25 < n < 900$

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

@@ Line 27: / Line 27: @@
 <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>\boxed{893}</math>.
+Thus <math>n \subseteq (6.25 , 900)</math>, and the number of positive integer <math>n</math> which satisfies this requirement is <math>\boxed{893}</math>.
 == See also ==

2006 AIME II (Problems • Answer Key • Resources)
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Difference between revisions of "2006 AIME II Problems/Problem 2"

Revision as of 19:48, 30 November 2019

Problem

Solution

See also