Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2006 AIME I Problems/Problem 2: Difference between revisions

← Older editNewer edit →
1=2 (talk | contribs)
4,129 edits
m wik, i don't think this falls under combinatorics..
Line 1: Line 1:
== Problem ==
== Problem ==


The lengths of the sides of a triangle with positive area are <math>\log_{10} 12</math>, <math>\log_{10} 75</math>, and <math>\log_{10} n</math>, where <math>n</math> is a positive integer. Find the number of possible values for <math>n</math>.
The lengths of the sides of a [[triangle]] with positive area are <math>\log_{10} 12</math>, <math>\log_{10} 75</math>, and <math>\log_{10} n</math>, where <math>n</math> is a positive integer. Find the number of possible values for <math>n</math>.


== Solution ==
== Solution ==
By the [[Triangle Inequality]]:
By the [[Triangle Inequality]]:
 
<div style="text-align:center;">
<math>\log_{10} 12 + \log_{10} n > \log_{10} 75 </math>
<math>\log_{10} 12 + \log_{10} n > \log_{10} 75 </math>


Line 15: Line 13:


<math> n > \frac{75}{12} = \frac{25}{4} = 6.25 </math>
<math> n > \frac{75}{12} = \frac{25}{4} = 6.25 </math>
 
</div>
Also:
Also:
 
<div style="text-align:center;">
<math>\log_{10} 12 + \log_{10} 75 > \log_{10} n </math>
<math>\log_{10} 12 + \log_{10} 75 > \log_{10} n </math>


Line 23: Line 21:


<math> n < 900 </math>  
<math> n < 900 </math>  
 
</div>
Combining these two inequalities:  
Combining these two inequalities:  


<math> 6.25 < n < 900 </math>
<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>893</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>.
Line 33: Line 31:
{{AIME box|year=2006|n=I|num-b=1|num-a=3}}
{{AIME box|year=2006|n=I|num-b=1|num-a=3}}


[[Category:Intermediate Combinatorics Problems]]
[[Category:Intermediate Geometry Problems]]
[[Category:Intermediate Algebra Problems]]

Revision as of 15:56, 25 September 2007

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 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
Retrieved from "https://wiki.aopstest.com/wiki/index.php?title=2006_AIME_I_Problems/Problem_2&oldid=16912"