2018 AIME I Problems/Problem 6
Let be the number of complex numbers with the properties that and is a real number. Find the remainder when is divided by .
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 . We are given . Note that and . We can use these facts to create two types of solutions:
which implies that and reduces to . There are 7 solutions for this.
which implies that and reduces to . There are 5 solutions for this, totaling 12 values of .
For each of these solutions for , there are necessarily solutions for . Thus, there are solutions for , yielding an answer of .
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 .
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. (Equalities are taken modulo ) For the second case, . This means , giving us 840 solutions. Our total count is thus , yielding a final answer of .
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 .
Note: This is actually not rigorous, because how to we know that all of the roots of such a polynomial are distinct? One can proceed as follows. Factoring gives us that so this implies that OR To show no satisfies both of these conditions, notice that if and for some complex number , then which implies that which implies that This is a contradiction since then would also have to equal Therefore the total number of solutions is
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 (Official MAA)
If satisfies the given conditions, there is a such that and is real. This difference is real if and only if either the two numbers and represent the same angle or the two numbers represent supplementary angles. In the first case there is an integer such that which implies that is a multiple of In the second case there is an integer such that which implies that is plus a multiple of In the interval there are values of that are multiples of there are values that are plus a multiple of and there are no values of that satisfy both of these conditions. Therefore there must be complex numbers satisfying the given conditions. The requested remainder is
firstly we consider that obviously there are different roots
Then, we consider that or which leads to so there are roots, in all, we are done
A number is real if and only if . Also, we know holds true for all complex numbers, so a number is real if and only if . If is real, then . The magnitude of this polynomial is , meaning it has roots. The desired result is .
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