Difference between revisions of "2005 AIME II Problems/Problem 7"
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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>. | 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 1== |
We note that in general, | We note that in general, | ||
| Line 19: | Line 19: | ||
It follows that <math>(x + 1)^{48} = (\sqrt[16]5)^{48} = 5^3 = \boxed{125}</math>. | It follows that <math>(x + 1)^{48} = (\sqrt[16]5)^{48} = 5^3 = \boxed{125}</math>. | ||
| + | == Solution 2 (Bashing) == | ||
| + | Let <math>y=\sqrt[16]{5}</math>, then eypanding the denominator results in: | ||
| + | <cmath>(y^8+1)(y^4+1) =(y^{12}+y^8+y^4+1)</cmath> | ||
| + | <cmath>(y^{12}+y^8+y^4+1)(y^2+1)=(y^{14}+y^{12}+y^{10}+y^8+y^6+y^4+y^2+1)</cmath> | ||
| + | <cmath>(y^{14}+y^{12}+y^{10}+y^8+y^6+y^4+y^2+1)(y+1) = (y^{15}+y^{14}+y^{13}+y^{12}+y^{11}+y^{10}+y^9+y^8+y^7+y^6+y^5+y^4+y^3+y^2+y+1)=\frac{y^{16}-1}{y-1}</cmath> | ||
| + | |||
| + | Therefore: | ||
| + | <cmath>\frac{4}{y^{16}-1/y-1} = \frac{4(y-1)}{y^{16}-1}=\frac{4(y-1)}{4} = y-1 </cmath> | ||
| + | |||
| + | It is evident that $ x+1 = (y-1)+1 = \sqrt[16]5 as Solution 1 states. | ||
== See Also == | == See Also == | ||
{{AIME box|year=2005|n=II|num-b=6|num-a=8}} | {{AIME box|year=2005|n=II|num-b=6|num-a=8}} | ||
Revision as of 21:50, 16 August 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 1
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)
.
Solution 2 (Bashing)
Let ![$y=\sqrt[16]{5}$](http://latex.artofproblemsolving.com/b/5/6/b56201dcc7984ba91688214b8f2d667c34b7b589.png)
, then eypanding the denominator results in:
![\[(y^8+1)(y^4+1) =(y^{12}+y^8+y^4+1)\]](http://latex.artofproblemsolving.com/7/5/b/75bb9952564c9050cb01b5c6d1e8326de19533b9.png)
![\[(y^{12}+y^8+y^4+1)(y^2+1)=(y^{14}+y^{12}+y^{10}+y^8+y^6+y^4+y^2+1)\]](http://latex.artofproblemsolving.com/f/1/d/f1d45e7276aa0c2356f5f8c9baa8c79a6c914849.png)
![\[(y^{14}+y^{12}+y^{10}+y^8+y^6+y^4+y^2+1)(y+1) = (y^{15}+y^{14}+y^{13}+y^{12}+y^{11}+y^{10}+y^9+y^8+y^7+y^6+y^5+y^4+y^3+y^2+y+1)=\frac{y^{16}-1}{y-1}\]](http://latex.artofproblemsolving.com/e/6/b/e6bdf699db84346cf2a4172d6ce33c617e352a21.png)
Therefore:
![\[\frac{4}{y^{16}-1/y-1} = \frac{4(y-1)}{y^{16}-1}=\frac{4(y-1)}{4} = y-1\]](http://latex.artofproblemsolving.com/6/7/9/679afde6a91a3d12a45a35031da0c81a6caf977c.png)
It is evident that $ x+1 = (y-1)+1 = \sqrt[16]5 as Solution 1 states.
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. 
