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 II Problems/Problem 7"

(Solution)
(Format; category tag)
Line 1: Line 1:
 
== 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>
 
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 ==
−
The expression for <math>x</math> looks very suspicious.  We multiply top and bottom by <math>(\sqrt[16]{5} -1)</math>.
 
  
−
<math>(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1)(\sqrt[16]{5}+1)(\sqrt[16]{5} -1) = (\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1)((\sqrt[16]{5})^2 - 1) = (\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1)(\sqrt[8]{5} - 1) = </math>
+
We note that in general,
−
<math>= (\sqrt{5}+1)(\sqrt[4]{5}+1)((\sqrt[8]{5})^2-1) = (\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[4]{5}-1) = (\sqrt{5}+1)((\sqrt[4]{5})^2-1) = (\sqrt5 + 1)(\sqrt5 - 1) = 4</math>.
+
 
−
(This is an example of a [[telescoping]] expression.  An alternative way to recognize the telescoping nature would be to write roots as [[fractional exponent]]s, to write everything in terms of <math>\sqrt[16]5</math>, or to expand the denominator out entirely.)
+
<center>
 +
<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>.
 +
</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
 +
 
 +
<center>
 +
<math> \displaystyle x = \frac{4(\sqrt[16]{5} - 1)}{4} = \sqrt[16]{5} - 1</math>.
 +
</center>
 +
 
 +
It follows that <math>(x + 1)^{48} = (\sqrt[16]5)^{48} = 5^3 = 125</math>
  
−
Thus, we have immediately that <math>x= \frac{4(\sqrt[16]5 - 1)}{4} = \sqrt[16]5 - 1</math> so <math>(x + 1)^{48} = (\sqrt[16]5)^{48} = 5^3 = 125</math>
 
 
== See Also ==
 
== See Also ==
 +
 
*[[2005 AIME II Problems]]
 
*[[2005 AIME II Problems]]
 +
 +
 +
[[Category:Introductory Algebra Problems]]

Revision as of 17:49, 7 September 2006

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,

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2005_AIME_II_Problems/Problem_7&oldid=10044"