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>\displaystyle (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>(x+1)^{48}</math>.
  
 
== Solution ==
 
== Solution ==
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<math>\displaystyle {} (\sqrt[2^n]{5} + 1)(\sqrt[2^n]{5} - 1) = (\sqrt[2^n]{5})^2 - 1^2 = \sqrt[2^{n-1}]{5} - 1 </math>.
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<math>{} (\sqrt[2^n]{5} + 1)(\sqrt[2^n]{5} - 1) = (\sqrt[2^n]{5})^2 - 1^2 = \sqrt[2^{n-1}]{5} - 1 </math>.
 
</center>
 
</center>
  
It now becomes apparent that if we multiply the [[numerator]] and [[denominator]] of <math> \displaystyle \frac{4}{ (\sqrt{5}+1) (\sqrt[4]{5}+1) (\sqrt[8]{5}+1) (\sqrt[16]{5}+1) } </math> by <math>\displaystyle (\sqrt[16]{5} - 1) </math>, the denominator will [[telescope]] to <math> \displaystyle \sqrt[1]{5} - 1 = 4 </math>, so  
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It now becomes apparent that if we multiply the [[numerator]] and [[denominator]] of <math>\frac{4}{ (\sqrt{5}+1) (\sqrt[4]{5}+1) (\sqrt[8]{5}+1) (\sqrt[16]{5}+1) } </math> by <math>(\sqrt[16]{5} - 1) </math>, the denominator will [[telescope]] to <math>\sqrt[1]{5} - 1 = 4 </math>, so  
  
 
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<center>
<math> \displaystyle x = \frac{4(\sqrt[16]{5} - 1)}{4} = \sqrt[16]{5} - 1</math>.
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<math>x = \frac{4(\sqrt[16]{5} - 1)}{4} = \sqrt[16]{5} - 1</math>.
 
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[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Revision as of 22:22, 4 July 2013

Problem

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

Solution

We note that in general,

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

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

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

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
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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