Difference between revisions of "2019 AIME II Problems/Problem 8"
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− | ==Problem | + | ==Problem== |
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>. | 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== | + | ==Solution 1== |
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 | 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 | ||
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-scrabbler94 | -scrabbler94 | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Denote <math>\frac{1+\sqrt{3}i}{2}</math> with <math>\omega</math>. | ||
+ | |||
+ | By using the quadratic formula (<math>\frac{-b\pm\sqrt{b^2-4ac}}{2a}</math>) in reverse, we can find that <math>\omega</math> is the solution to a quadratic equation of the form <math>ax^2+bx+c=0</math> such that <math>2a=2</math>, <math>-b=1</math>, and <math>b^2-4ac=-3</math>. This clearly solves to <math>a=1</math>, <math>b=-1</math>, and <math>c=1</math>, so <math>\omega</math> solves <math>x^2-x+1=0</math>. | ||
+ | |||
+ | Multiplying <math>x^2-x+1=0</math> by <math>x+1</math> on both sides yields <math>x^3+1=0</math>. Muliplying this by <math>x^3-1</math> on both sides yields <math>x^6-1=0</math>, or <math>x^6=1</math>. This means that <math>\omega^6=1</math>. | ||
+ | |||
+ | We can use this to simplify the equation <math>a\omega^{2018}+b\omega^{2017}+c\omega^{2016}=f(\omega)=2015+2019\sqrt{3}i</math> to <math>a\omega^2+b\omega+c=2015+2019\sqrt{3}i.</math> | ||
+ | |||
+ | As in Solution 1, we use the values <math>\omega=\frac{1}{2}+\frac{\sqrt{3}}{2}i</math> and <math>\omega^2=-\frac{1}{2}+\frac{\sqrt{3}}{2}i</math> to find that <math>-\frac{1}{2}a+\frac{1}{2}b+c=2015</math> and <math>\frac{\sqrt{3}}{2}a+\frac{\sqrt{3}}{2}b=2019\sqrt{3} \implies a+b=4038.</math> Since neither <math>a</math> nor <math>b</math> can exceed <math>2019</math>, they must both be equal to <math>2019</math>. Since <math>a</math> and <math>b</math> are equal, they cancel out in the first equation, resulting in <math>c=2015</math>. | ||
+ | |||
+ | Therefore, <math>f(1)=a+b+c=2019+2019+2015=6053</math>, and <math>6053\bmod1000=\boxed{053}</math>. ~[[User:emerald_block|emerald_block]] | ||
+ | |||
+ | |||
+ | ==Solution 3== | ||
+ | Calculate the first few powers of <math>\frac{1+\sqrt{3}i}{2}</math>. | ||
+ | |||
+ | <math>(\frac{1+\sqrt{3}i}{2})^1=\frac{1+\sqrt{3}i}{2}</math> | ||
+ | |||
+ | <math>(\frac{1+\sqrt{3}i}{2})^2=\frac{-1+\sqrt{3}i}{2}</math> | ||
+ | |||
+ | <math>(\frac{1+\sqrt{3}i}{2})^3=-1</math> | ||
+ | |||
+ | <math>(\frac{1+\sqrt{3}i}{2})^4=\frac{-1-\sqrt{3}i}{2}</math> | ||
+ | |||
+ | <math>(\frac{1+\sqrt{3}i}{2})^5=\frac{1-\sqrt{3}i}{2}</math> | ||
+ | |||
+ | <math>(\frac{1+\sqrt{3}i}{2})^6=1</math> | ||
+ | |||
+ | We figure that the power of <math>\frac{1+\sqrt{3}i}{2}</math> repeats in a cycle 6. | ||
+ | |||
+ | <math>f(\frac{1+\sqrt{3}i}{2})=(a(\frac{1+\sqrt{3}i}{2})^2+b(\frac{1+\sqrt{3}i}{2})+c)(\frac{1+\sqrt{3}i}{2})^{2016}</math> | ||
+ | |||
+ | Since 2016 is a multiple of 6, <math>(\frac{1+\sqrt{3}i}{2})^{2016}=1</math> | ||
+ | |||
+ | <math>f(\frac{1+\sqrt{3}i}{2})=a(\frac{-1+\sqrt{3}i}{2})+b(\frac{1+\sqrt{3}i}{2})+c</math> | ||
+ | |||
+ | <math>f(\frac{1+\sqrt{3}i}{2})=(-\frac{1}{2}a+\frac{1}{2}b+c)+(\frac{\sqrt{3}i}{2}a+\frac{\sqrt{3}i}{2}b)=2015+2019\sqrt{3}i</math> | ||
+ | |||
+ | Therefore, <math>-\frac{1}{2}a+\frac{1}{2}b+c=2015</math> and <math>\frac{\sqrt{3}i}{2}a+\frac{\sqrt{3}i}{2}b=2019\sqrt{3}i</math> | ||
+ | |||
+ | Using the first equation, we can get that <math>-a+b+2c=4030</math>, and using the second equation, we can get that <math>a+b=4038</math>. | ||
+ | |||
+ | Since all coefficients are less than or equal to <math>2019</math>, <math>a=b=2019</math>. | ||
+ | |||
+ | Therefore, <math>2c=4030</math> and <math>c=2015</math>. | ||
+ | |||
+ | <math>f(1)=a+b+c=2019+2019+2015=6053</math>, and the remainder when it divides <math>1000</math> is <math>\boxed{053}.</math> | ||
+ | |||
+ | ~Interstigation | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2019|n=II|num-b=7|num-a=9}} | {{AIME box|year=2019|n=II|num-b=7|num-a=9}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 01:41, 21 February 2021
Problem
The polynomial has real coefficients not exceeding , and . Find the remainder when is divided by .
Solution 1
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 .
-scrabbler94
Solution 2
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
Solution 3
Calculate the first few powers of .
We figure that the power of repeats in a cycle 6.
Since 2016 is a multiple of 6,
Therefore, and
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
~Interstigation
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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