Difference between revisions of "2012 AIME I Problems/Problem 6"

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==Problem 6==
 
==Problem 6==
Let <math>z</math> and <math>w</math> be complex numbers such that <math>z^{13} = w</math> and <math>w^{11} = z</math>. If the imaginary part of <math>z</math> can be written as <math> \sin {\frac{m\pi}{n}}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers, find <math>n</math>.
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The complex numbers <math>z</math> and <math>w</math> satisfy <math>z^{13} = w,</math> <math>w^{11} = z,</math> and the imaginary part of <math>z</math> is <math>\sin{\frac{m\pi}{n}}</math>, for relatively prime positive integers <math>m</math> and <math>n</math> with <math>m<n.</math> Find <math>n.</math>
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==Solution==
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Substituting the first equation into the second, we find that <math>(z^{13})^{11} = z</math> and thus <math>z^{142} = 1.</math> So <math>z</math> must be a <math>142</math>nd root of unity, and thus the imaginary part of <math>z</math> will be <math>\sin{\frac{2m\pi}{142}} = \sin{\frac{m\pi}{71}}</math> for some <math>m</math> with <math>0 \le m < 142.</math> But note that since <math>71</math> is prime and <math>m<71</math> by the conditions of the problem, the denominator in the argument of this value will always be <math>71</math> and thus <math>n = \boxed{071.}</math>
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2012|n=I|num-b=5|num-a=7}}
 
{{AIME box|year=2012|n=I|num-b=5|num-a=7}}

Revision as of 01:11, 17 March 2012

Problem 6

The complex numbers $z$ and $w$ satisfy $z^{13} = w,$ $w^{11} = z,$ and the imaginary part of $z$ is $\sin{\frac{m\pi}{n}}$, for relatively prime positive integers $m$ and $n$ with $m<n.$ Find $n.$

Solution

Substituting the first equation into the second, we find that $(z^{13})^{11} = z$ and thus $z^{142} = 1.$ So $z$ must be a $142$nd root of unity, and thus the imaginary part of $z$ will be $\sin{\frac{2m\pi}{142}} = \sin{\frac{m\pi}{71}}$ for some $m$ with $0 \le m < 142.$ But note that since $71$ is prime and $m<71$ by the conditions of the problem, the denominator in the argument of this value will always be $71$ and thus $n = \boxed{071.}$

See also

2012 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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All AIME Problems and Solutions
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