2019 AIME II Problems/Problem 8
The polynomial has real coefficients not exceeding , and . Find the remainder when is divided by .
We have where is a primitive 6th root of unity. Then we have
We wish to find . We first look at the real parts. As and , we have . Looking at imaginary parts, we have , so . As and do not exceed 2019, we must have and . Then , so .
Denote with .
By using the quadratic formula () in reverse, we can find that is the solution to a quadratic equation of the form such that , , and . This clearly solves to , , and , so solves .
Multiplying by on both sides yields . Muliplying this by on both sides yields , or . This means that .
We can use this to simplify the equation to
As in Solution 1, we use the values and to find that and Since neither nor can exceed , they must both be equal to . Since and are equal, they cancel out in the first equation, resulting in .
Therefore, , and . ~emerald_block
Calculate the first few powers of .
We figure that the power of repeats in a cycle 6.
Since 2016 is a multiple of 6,
Using the first equation, we can get that , and using the second equation, we can get that .
Since all coefficients are less than or equal to , .
Therefore, and .
, and the remainder when it divides is
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