2005 AMC 12B Problems/Problem 22
Contents
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 thus
where
. Note that
for all
, so the answer is
. (Author: Patrick Yin)
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 |
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