2005 AMC 12B Problems/Problem 22
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.