Difference between revisions of "2012 AIME I Problems/Problem 6"
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==Problem 6== | ==Problem 6== | ||
− | + | 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== | ||
+ | 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 and satisfy and the imaginary part of is , for relatively prime positive integers and with Find
Solution
Substituting the first equation into the second, we find that and thus So must be a nd root of unity, and thus the imaginary part of will be for some with But note that since is prime and by the conditions of the problem, the denominator in the argument of this value will always be and thus
See also
2012 AIME I (Problems • Answer Key • Resources) | ||
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 |