Difference between revisions of "2019 AIME II Problems/Problem 8"
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− | + | ==Problem 8== | |
+ | The polynomial <math>f(z)=az^{2018}+bz^{2017}+cz^{2016}</math> has real coefficients not exceeding <math>2019</math>, and <math>f\left(\tfrac{1+\sqrt{3}i}{2}\right)=2015+2019\sqrt{3}i</math>. Find the remainder when <math>f(1)</math> is divided by <math>1000</math>. | ||
+ | |||
+ | ==Solution== | ||
+ | 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 | ||
+ | |||
+ | <cmath>\begin{align*} | ||
+ | f(\omega) &= a\omega^{2018} + b\omega^{2017} + c\omega^{2016}\\ | ||
+ | &= a\omega^2 + b\omega + c | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | We wish to find <math>f(1) = a+b+c</math>. We first look at the real parts. As <math>\text{Re}(\omega^2) = -\frac{1}{2}</math> and <math>\text{Re}(\omega) = \frac{1}{2}</math>, we have <math>-\frac{1}{2}a + \frac{1}{2}b + c = 2015 \implies -a+b + 2c = 4030</math>. Looking at imaginary parts, we have <math>\text{Im}(\omega^2) = \text{Im}(\omega) = \frac{\sqrt{3}}{2}</math>, so <math>\frac{\sqrt{3}}{2}(a+b) = 2019\sqrt{3} \implies a+b = 4038</math>. As <math>a</math> and <math>b</math> do not exceed 2019, we must have <math>a+b = 2019</math>. Then <math>c = \frac{4030}{2} = 2015</math>, so <math>f(1) = 4038 + 2015 = 6053 \implies f(1) \pmod{1000} = \boxed{053}</math>. | ||
+ | |||
+ | -scrabbler94 | ||
+ | |||
+ | ==See Also== | ||
+ | {{AIME box|year=2019|n=II|num-b=7|num-a=9}} | ||
+ | {{MAA Notice}} |
Revision as of 16:43, 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.