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

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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 $\sin{\frac{2k\pi}{142}} = \sin{\frac{k\pi}{71}}$ for some $k$ with $0 \le k < 142.$ But note that $71$ is prime and $k<71$ 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 the imaginary part of each is of the form $\sin\left(\frac{2 \pi k}{142}\right)$ where $k$ is a positive integer between $1$ and $141$ inclusive. This simplifies to $\sin\left(\frac{\pi k}{71}\right)$ . Therefore, the imaginary part of $z$ is $\sin\left(\frac{\pi m}{71}\right)$ , where $0 \le m < 71$ . Since $071$ is prime, it is the only possible denominator, so $\boxed{n = 71}$ .

@@ Line 5: / Line 5: @@
 ===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{2k\pi}{142}} = \sin{\frac{k\pi}{71}}</math> for some <math>k</math> with <math>0 \le k < 142.</math> But note that <math>71</math> is prime and <math>k<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>
 ===Solution 2===

2012 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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