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 polar form of complex numbers. Let 
. We have two cases to consider. Either 
, or 
and 
z^{6!} = z^{5!}
e^{6! \theta i} = e^{5! \theta i}
720 \theta \equiv 120 \theta \mod 2\pi
600\theta \equiv 0 \mod 2\pi
e^{6! \theta i} = e^{(\pi - 5!\theta)i}
840 \theta \equiv \pi mod 2\pi
\boxed{1440}$.
See also
| 2018 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
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