Difference between revisions of "2005 AIME II Problems/Problem 7"
(Format; category tag) |
m (→See Also) |
||
| Line 21: | Line 21: | ||
== See Also == | == See Also == | ||
| + | *[[2005 AIME II Problems/Problem 6| Previous problem]] | ||
| + | *[[2005 AIME II Problems/Problem 8| Next problem]] | ||
*[[2005 AIME II Problems]] | *[[2005 AIME II Problems]] | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
Revision as of 19:57, 7 September 2006
Problem
Let ![$x=\frac{4}{(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1)(\sqrt[16]{5}+1)}.$](http://latex.artofproblemsolving.com/6/f/7/6f71d4755273a86295f916ae662a281e75af55c6.png)
Find 
Solution
We note that in general,
![$\displaystyle {} (\sqrt[2^n]{5} + 1)(\sqrt[2^n]{5} - 1) = (\sqrt[2^n]{5})^2 - 1^2 = \sqrt[2^{n-1}]{5} - 1$](http://latex.artofproblemsolving.com/e/f/c/efce902fd8efc20643c59a8fde2500a88952e03b.png)
.
It now becomes apparent that if we multiply the numerator and denominator of ![$\displaystyle \frac{4}{ (\sqrt{5}+1) (\sqrt[4]{5}+1) (\sqrt[8]{5}+1) (\sqrt[16]{5}+1) }$](http://latex.artofproblemsolving.com/3/1/a/31a02128c8c0f574d51feb9d4c4128dfca1dad75.png)
by ![$\displaystyle (\sqrt[16]{5} - 1)$](http://latex.artofproblemsolving.com/3/d/8/3d899f8802dd745f978572b19b4c3362c8e9eee8.png)
, the denominator will telescope to ![$\displaystyle \sqrt[1]{5} - 1 = 4$](http://latex.artofproblemsolving.com/b/2/2/b2240d179bebdcd1785041036e57dd80c688181f.png)
, so
![$\displaystyle x = \frac{4(\sqrt[16]{5} - 1)}{4} = \sqrt[16]{5} - 1$](http://latex.artofproblemsolving.com/1/7/7/17714c9e6e5feb3e6b059b13cbf2a25f57d359ac.png)
.
It follows that ![$(x + 1)^{48} = (\sqrt[16]5)^{48} = 5^3 = 125$](http://latex.artofproblemsolving.com/7/1/2/712816585976400e539a06ff0b568daa40f4e1da.png)
