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

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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>
 
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>
  
==Solution==
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==Solutions==
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===Solution 1===
 
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 <math>71</math> is prime and <math>m<71</math> by the conditions of the problem, so the denominator in the argument of this value will always be <math>71</math> and thus <math>n = \boxed{071.}</math>
 
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 <math>71</math> is prime and <math>m<71</math> by the conditions of the problem, so the denominator in the argument of this value will always be <math>71</math> and thus <math>n = \boxed{071.}</math>
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===Solution 2===
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Note that <math>w^{143}=w</math> and similar for <math>z</math>, and they are not equal to <math>0</math> because the question implies the imaginary part is positive.  Thus <math>w^{142}=z^{142}=1</math>, so each is of the form <math>sin(2*pi*k/142)</math> where <math>k</math> is a positive integer between <math>0</math> and <math>141</math> inclusive.  This simplifies to <math>sin(pi*k/71)</math>, and <math>071</math> is prime, so it is the only possible denominator, and thus is the answer.
  
 
== 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 13:41, 24 August 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 $n.$

Solutions

Solution 1

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 $71$ is prime and $m<71$ by the conditions of the problem, so the denominator in the argument of this value will always be $71$ and thus $n = \boxed{071.}$

Solution 2

Note that $w^{143}=w$ and similar for $z$, and they are not equal to $0$ because the question implies the imaginary part is positive. Thus $w^{142}=z^{142}=1$, so each is of the form $sin(2*pi*k/142)$ where $k$ is a positive integer between $0$ and $141$ inclusive. This simplifies to $sin(pi*k/71)$, and $071$ is prime, so it is the only possible denominator, and thus is the answer.

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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