Difference between revisions of "2005 AMC 12B Problems/Problem 22"
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<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 can 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 can take values from 0 to <math>2^{2005}-1</math>, so the answer is <math>2^{2005}\Rightarrow\boxed{\mathrm{E}}</math> |
== Solution 2 == | == Solution 2 == |
Revision as of 00:36, 22 December 2020
Contents
[hide]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 1
Since , let
, where
is an argument of
.
We 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 can take values from 0 to
, so the answer is
Solution 2
Let .
Repeating through this recursive process, we can quickly see that
Thus,
. The solutions for
are
where
. Note that
for all
, so the answer is
. (Author: Patrick Yin)
Solution 3
Note that for any complex number , we have
. Therefore, the magnitude of
is always
, meaning that all of the numbers in the sequence
are of magnitude
.
Another property of complex numbers is that . For the numbers in our sequence, this means
, so
. Rewriting our recursive condition with these facts, we now have
Solving for
here, we obtain
It is seen that there are two values of
which correspond to one value of
. That means that there are two possible values of
, four possible values of
, and so on. Therefore, there are
possible values of
, giving the answer as
.
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.