2006 AIME I Problems/Problem 2: Difference between revisions
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== Solution == | == Solution == | ||
By the [[Triangle Inequality]]: | |||
<math>\log_{10} 12 + \log_{10} n > \log_{10} 75 </math> | |||
<math>\log_{10} 12n > \log_{10} 75 </math> | |||
<math> 12n > 75 </math> | |||
<math> n > \frac{75}{12} = \frac{25}{4} = 6.25 </math> | |||
Also: | |||
<math>\log_{10} 12 + \log_{10} 75 > \log_{10} n </math> | |||
<math>\log_{10} 12\cdot75 > \log_{10} n </math> | |||
<math> n < 900 </math> | |||
Combining these two inequalities: | |||
<math> 6.25 < n < 900 </math> | |||
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>893</math>. | |||
== See also == | == See also == | ||
Revision as of 12:56, 25 September 2007
Problem
The lengths of the sides of a triangle with positive area are 
, 
, and 
, where 
is a positive integer. Find the number of possible values for .
Solution
By the Triangle Inequality:
Also:
Combining these two inequalities:
The number of possible integer values for is the number of integers over the interval 
, which is 
.
See also
| 2006 AIME I (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 | ||
