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 "2005 AIME I Problems/Problem 8"
(Undo revision 227435 by Charlestang (talk))
(Tag: Undo)
 
(10 intermediate revisions by 6 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
−
The [[equation]] <math> 2^{333x-2} + 2^{111x+2} = 2^{222x+1} + 1 </math> has three [[real]] [[root]]s. Given that their sum is <math> \frac mn </math> where <math> m </math> and <math> n </math> are [[relatively prime]] [[positive integer]]s, find <math> m+n. </math>
+
The [[equation]] <math> 2^{333x-2} + 2^{111x+2} = 2^{222x+1} + 1 </math> has three [[real]] [[root]]s. Given that their sum is <math>m/n</math> where <math> m </math> and <math> n </math> are [[relatively prime]] [[positive integer]]s, find <math> m+n. </math>
  
 
== Solution ==
 
== Solution ==
−
Let <math>y = 2^{111x}</math>.  Then our equation reads <math>\frac{1}{4}y^3 + 4y = 2y^2 + 1</math> or <math>y^3 - 8y^2 + 16y - 4 = 0</math>.  Thus, if this equation has roots <math>r_1, r_2</math> and <math>r_3</math>, we have <math>r_1\cdot r_2\cdot r_3 = 4</math>.  Let the corresponding values of <math>x</math> be <math>x_1, x_2</math> and <math>x_3</math>.  Then the previous statement says that <math>2^{111\cdot(x_1 + x_2 + x_3)} = 4</math> so that taking a [[logarithm]] gives <math>111(x_1 + x_2 + x_3) = 2</math> and <math>x_1 + x_2 + x_3 = \frac{2}{111}</math>.  Thus the answer is <math>111 + 2 = 113</math>.
+
Let <math>y = 2^{111x}</math>.  Then our equation reads <math>\frac{1}{4}y^3 + 4y = 2y^2 + 1</math> or <math>y^3 - 8y^2 + 16y - 4 = 0</math>.  Thus, if this equation has roots <math>r_1, r_2</math> and <math>r_3</math>, by [[Vieta's formulas]] we have <math>r_1\cdot r_2\cdot r_3 = 4</math>.  Let the corresponding values of <math>x</math> be <math>x_1, x_2</math> and <math>x_3</math>.  Then the previous statement says that <math>2^{111\cdot(x_1 + x_2 + x_3)} = 4</math> so taking a [[logarithm]] of that gives <math>111(x_1 + x_2 + x_3) = 2</math> and <math>x_1 + x_2 + x_3 = \frac{2}{111}</math>.  Thus the answer is <math>111 + 2 = \boxed{113}</math>.
  
 +
== See also ==
 +
{{AIME box|year=2005|n=I|num-b=7|num-a=9}}
  
−
== See also ==
+
[[Category:Intermediate Algebra Problems]]
−
* [[2005 AIME I Problems/Problem 7 | Previous problem]]
+
{{MAA Notice}}
−
* [[2005 AIME I Problems/Problem 9 | Next problem]]
 
−
* [[2005 AIME I Problems]]
 
−
* [[Exponent]]
 

Latest revision as of 16:46, 17 September 2024

Problem

The equation $2^{333x-2} + 2^{111x+2} = 2^{222x+1} + 1$ has three real roots. Given that their sum is $m/n$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$

Solution

Let $y = 2^{111x}$. Then our equation reads $\frac{1}{4}y^3 + 4y = 2y^2 + 1$ or $y^3 - 8y^2 + 16y - 4 = 0$. Thus, if this equation has roots $r_1, r_2$ and $r_3$, by Vieta's formulas we have $r_1\cdot r_2\cdot r_3 = 4$. Let the corresponding values of $x$ be $x_1, x_2$ and $x_3$. Then the previous statement says that $2^{111\cdot(x_1 + x_2 + x_3)} = 4$ so taking a logarithm of that gives $111(x_1 + x_2 + x_3) = 2$ and $x_1 + x_2 + x_3 = \frac{2}{111}$. Thus the answer is $111 + 2 = \boxed{113}$.

See also

2005 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2005_AIME_I_Problems/Problem_8&oldid=227438"