2018 AIME I Problems/Problem 6
Contents
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
Solution 6(really easy 30 second solution)
We need to find all such that
is real number.
We can now factor out
to get
Because the
doesn't change whether it is real or not we can just get rid of it. Now we need to find all the solutions where
is
or
. Now we have the 2 equations
and
There are 720 solutions for each equation making a total of 1440 solutions. This means the answer is 440.
-harsha12345
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 |
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