Difference between revisions of "2014 AIME I Problems/Problem 14"

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== Solution ==
 
== Solution ==
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The first step is to notice that the 4 on the right hand side can simplify the terms on the left hand side. If we distribute 1 to <math>3/x-3</math>, then the fraction becomes of the form <math>(x/x - 3)</math>. A similar cancellation happens with the other four terms.
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== See also ==
 
== See also ==
 
{{AIME box|year=2014|n=I|num-b=13|num-a=15}}
 
{{AIME box|year=2014|n=I|num-b=13|num-a=15}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 19:32, 14 March 2014

Problem 14

Let $m$ be the largest real solution to the equation

$\frac{3}{x-3}+\frac{5}{x-5}+\frac{17}{x-17}+\frac{19}{x-19}=x^2-11x-4$

There are positive integers $a$, $b$, and $c$ such that $m=a+\sqrt{b+\sqrt{c}}$. Find $a+b+c$.

Solution

The first step is to notice that the 4 on the right hand side can simplify the terms on the left hand side. If we distribute 1 to $3/x-3$, then the fraction becomes of the form $(x/x - 3)$. A similar cancellation happens with the other four terms.


See also

2014 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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