Difference between revisions of "2005 AMC 12B Problems/Problem 22"
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<math>z_{2005}=1\Rightarrow \theta_{2005}=2k\pi</math>, where <math>k</math> is an integer. Therefore, | <math>z_{2005}=1\Rightarrow \theta_{2005}=2k\pi</math>, where <math>k</math> is an integer. Therefore, | ||
<cmath> | <cmath> | ||
− | 2^{2005}(\theta_0+\frac{\pi}{2})=(2k+\frac{1}{2})\pi | + | 2^{2005}(\theta_0+\frac{\pi}{2})=(2k+\frac{1}{2})\pi</cmath> |
− | \theta_0=\frac{k}{2^{2004}}\pi+\left(\frac{1}{2^{2006}}-\frac{1}{2}\right)\pi | + | <cmath>\theta_0=\frac{k}{2^{2004}}\pi+\left(\frac{1}{2^{2006}}-\frac{1}{2}\right)\pi |
</cmath> | </cmath> | ||
The value of <math>\theta_0</math> only matters [[modulo]] <math>2\pi</math>. Since <math>\frac{k+2^{2005}}{2^{2004}}\pi\equiv\frac{k}{2^{2004}}\pi\mod 2\pi</math>, k only needs to take values from 0 to <math>2^{2005}-1</math>, so the answer is <math>2^{2005}\Rightarrow\boxed{E}</math> | The value of <math>\theta_0</math> only matters [[modulo]] <math>2\pi</math>. Since <math>\frac{k+2^{2005}}{2^{2004}}\pi\equiv\frac{k}{2^{2004}}\pi\mod 2\pi</math>, k only needs to take values from 0 to <math>2^{2005}-1</math>, so the answer is <math>2^{2005}\Rightarrow\boxed{E}</math> |
Revision as of 11:52, 31 December 2015
Problem
A sequence of complex numbers is defined by the rule
where is the complex conjugate of and . Suppose that and . How many possible values are there for ?
Solution
Since , let , where is an argument of . I will prove by induction that , where .
Base Case: trivial
Inductive Step: Suppose the formula is correct for , then Since the formula is proven
, where is an integer. Therefore, The value of only matters modulo . Since , k only needs to take values from 0 to , so the answer is
See Also
2005 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.