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

m (Solution)
Line 3: Line 3:
  
 
==Solution==
 
==Solution==
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>
+
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>
  
 
== 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 02:10, 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 $71$ 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 $n = \boxed{071.}$

See also

2012 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions
Invalid username
Login to AoPS