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

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

(→‎Problem)
(→‎Solution)
Line 4: Line 4:
  
 
== Solution ==
 
== Solution ==
−
The smallest <math>S</math> is <math>1+2+ \ldots +90 = 91 \cdot 45 = 4095</math>. The largest <math>S</math> is <math>11+12+ \ldots +100=111\cdot 45=4995</math>. All numbers between <math>4095</math> and <math>4995</math> are possible values of S, so the number of possible values of S is <math>4995-4095+1=901</math>.
+
 
 +
 
 +
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 $\log_{10} 12$, $\log_{10} 75$, 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://artofproblemsolving.com/wiki/index.php?title=2006_AIME_I_Problems/Problem_2&oldid=16797"