Difference between revisions of "2005 AIME I Problems/Problem 8"

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== 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>
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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 ==

Revision as of 19:50, 2 January 2022

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 (ProblemsAnswer KeyResources)
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

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