Difference between revisions of "2005 AIME II Problems/Problem 7"
m (→Solution) |
|||
| Line 17: | Line 17: | ||
</center> | </center> | ||
| − | It follows that <math>(x + 1)^{48} = (\sqrt[16]5)^{48} = 5^3 = 125</math>. | + | It follows that <math>(x + 1)^{48} = (\sqrt[16]5)^{48} = 5^3 = \boxed{125}</math>. |
== See Also == | == See Also == | ||
Revision as of 21:44, 22 July 2017
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,
![${} (\sqrt[2^n]{5} + 1)(\sqrt[2^n]{5} - 1) = (\sqrt[2^n]{5})^2 - 1^2 = \sqrt[2^{n-1}]{5} - 1$](http://latex.artofproblemsolving.com/9/3/7/937cf890b84b97c40d1f2d4b7da71a843b154b17.png)
.
It now becomes apparent that if we multiply the numerator and denominator of ![$\frac{4}{ (\sqrt{5}+1) (\sqrt[4]{5}+1) (\sqrt[8]{5}+1) (\sqrt[16]{5}+1) }$](http://latex.artofproblemsolving.com/2/2/d/22d5c7311cbb44d266e83b88a340bac91137c07d.png)
by ![$(\sqrt[16]{5} - 1)$](http://latex.artofproblemsolving.com/8/4/a/84a092dd3852e33ad6405250f9cc6b2aa98b38db.png)
, the denominator will telescope to ![$\sqrt[1]{5} - 1 = 4$](http://latex.artofproblemsolving.com/4/3/6/436de5ac5ee3053ac8b6cbfd9a2eecf0e0d6d718.png)
, so
![$x = \frac{4(\sqrt[16]{5} - 1)}{4} = \sqrt[16]{5} - 1$](http://latex.artofproblemsolving.com/7/7/c/77c13ab43c7c8bc2202efc5ac6500904d9f9ce6e.png)
.
It follows that ![$(x + 1)^{48} = (\sqrt[16]5)^{48} = 5^3 = \boxed{125}$](http://latex.artofproblemsolving.com/6/a/c/6ac848f1207fe4ad7d8d8e26c47265fe0d4fb5e7.png)
.
See Also
| 2005 AIME II (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
