2018 AIME I Problems/Problem 6
Problem
Let be the number of complex numbers
with the properties that
and
is a real number. Find the remainder when
is divided by
.
Solution 1
Let . This simplifies the problem constraint to
. This is true if
. Let
be the angle
makes with the positive x-axis. Note that there is exactly one
for each angle
. This must be true for
values of
(it may help to picture the reference angle making one orbit from and to the positive x-axis; note every time
). For each of these solutions for
, there are necessarily
solutions for
. Thus, there are
solutions for
, yielding an answer of
.
Solution 2
The constraint mentioned in the problem is equivalent to the requirement that the imaginary part is equal to . Since
, let
, then we can write the imaginary part of
. Using the sum-to-product formula, we get
or
. The former yields
solutions, and the latter yields
solutions, giving a total of
solution, so our answer is
.
Solution 3
As mentioned in solution one, for the difference of two complex numbers to be real, their imaginary parts must be equal. We use exponential form of complex numbers. Let . We have two cases to consider. Either
, or
and
are reflections across the imaginary axis.
If
, then
. Thus,
or
, giving us 600 solutions.
For the second case,
. This means
, giving us 840 solutions.
Our total count is thus
, yielding a final answer of
.
Solution 4
Because we know that
Hence
Because
is real, it is equal to its complex conjugate. Hence
Substituting the expression we that we derived earlier, we get
This leaves us with a polynomial whose leading term is
Hence our answer is
.
Solution 5
Since , let
. For
to be real, the imaginary parts of
and
must be equal, so
. We need to find all solutions for
in the interval
. This can be done by graphing
and
and finding their intersections. Since the period of
is
and the period of
is
, the common period of both graphs is
. Therefore, we only graph the functions in the domain
. We can clearly see that there are twelve points of intersection. However, since we only graphed
of the interval
, we need to multiply our answer by
. The answer is
See also
2018 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.