AoPSWiki
Trying to get to the USAMO in 2010? Our AIME Problem Series can help you get there! Click here to enroll today!
Personal tools

2005 AIME II Problems/Problem 7

From AoPSWiki

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 (Problems • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/2005_AIME_II_Problems/Problem_7"
Looking for a challenging algebra text? Preparing for MATHCOUNTS or the AMC exams?
Check out Art of Problem Solving's Introduction to Algebra by Richard Rusczyk.
© Copyright 2008 AoPS Incorporated. All Rights Reserved. • Foundation • Privacy • Contact Us