Difference between revisions of "2019 AIME II Problems/Problem 8"
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− | We have <math>\frac{1+\sqrt{3}i}{2} = \omega</math> where <math>\omega = e^{\frac{\pi}{3}}</math> is a primitive 6th root of unity. Then we have | + | We have <math>\frac{1+\sqrt{3}i}{2} = \omega</math> where <math>\omega = e^{\frac{i\pi}{3}}</math> is a primitive 6th root of unity. Then we have |
<cmath>\begin{align*} | <cmath>\begin{align*} |
Revision as of 18:00, 22 March 2019
Problem 8
The polynomial has real coefficients not exceeding , and . Find the remainder when is divided by .
Solution
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 . Then , so .
-scrabbler94
See Also
2019 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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