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
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 and 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.
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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