Difference between revisions of "2005 AIME II Problems/Problem 7"

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== Problem ==
 
== Problem ==
  
Let <math> x=\frac{4}{(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1)(\sqrt[16]{5}+1)}. </math> Find <math> (x+1)^{48}. </math>
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Let <math> x=\frac{4}{(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1)(\sqrt[16]{5}+1)}. </math> Find <math>\displaystyle (x+1)^{48}</math>.
  
 
== Solution ==
 
== Solution ==
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</center>
 
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It follows that <math>(x + 1)^{48} = (\sqrt[16]5)^{48} = 5^3 = 125</math>
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It follows that <math>(x + 1)^{48} = (\sqrt[16]5)^{48} = 5^3 = 125</math>.
  
 
== See Also ==
 
== See Also ==
 
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{{AIME box|year=2005|n=II|num-b=6|num-a=8}}
*[[2005 AIME II Problems/Problem 6| Previous problem]]
 
*[[2005 AIME II Problems/Problem 8| Next problem]]
 
*[[2005 AIME II Problems]]
 
 
 
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Revision as of 18:55, 21 March 2007

Problem

Let $x=\frac{4}{(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1)(\sqrt[16]{5}+1)}.$ Find $\displaystyle (x+1)^{48}$.

Solution

We note that in general,

$\displaystyle {} (\sqrt[2^n]{5} + 1)(\sqrt[2^n]{5} - 1) = (\sqrt[2^n]{5})^2 - 1^2 = \sqrt[2^{n-1}]{5} - 1$.

It now becomes apparent that if we multiply the numerator and denominator of $\displaystyle \frac{4}{ (\sqrt{5}+1) (\sqrt[4]{5}+1) (\sqrt[8]{5}+1) (\sqrt[16]{5}+1) }$ by $\displaystyle (\sqrt[16]{5} - 1)$, the denominator will telescope to $\displaystyle \sqrt[1]{5} - 1 = 4$, so

$\displaystyle x = \frac{4(\sqrt[16]{5} - 1)}{4} = \sqrt[16]{5} - 1$.

It follows that $(x + 1)^{48} = (\sqrt[16]5)^{48} = 5^3 = 125$.

See Also

2005 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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All AIME Problems and Solutions
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