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Difference between revisions of "2019 AIME II Problems/Problem 8"

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==Problem==
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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>.
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==Solution 1==
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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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<cmath>\begin{align*}
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f(\omega) &= a\omega^{2018} + b\omega^{2017} + c\omega^{2016}\\
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&= a\omega^2 + b\omega + c
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\end{align*}</cmath>
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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 = 2019</math> and <math>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>.
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-scrabbler94
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==Solution 2==
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Denote <math>\frac{1+\sqrt{3}i}{2}</math> with <math>\omega</math>.
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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>.
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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>.
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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>
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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>.
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Therefore, <math>f(1)=a+b+c=2019+2019+2015=6053</math>, and <math>6053\bmod1000=\boxed{053}</math>. ~[[User:emerald_block|emerald_block]]
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==Solution 3==
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Calculate the first few powers of <math>\frac{1+\sqrt{3}i}{2}</math>.
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<math>(\frac{1+\sqrt{3}i}{2})^1=\frac{1+\sqrt{3}i}{2}</math>
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<math>(\frac{1+\sqrt{3}i}{2})^2=\frac{-1+\sqrt{3}i}{2}</math>
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<math>(\frac{1+\sqrt{3}i}{2})^3=-1</math>
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<math>(\frac{1+\sqrt{3}i}{2})^4=\frac{-1-\sqrt{3}i}{2}</math>
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<math>(\frac{1+\sqrt{3}i}{2})^5=\frac{1-\sqrt{3}i}{2}</math>
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<math>(\frac{1+\sqrt{3}i}{2})^6=1</math>
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We figure that the power of <math>\frac{1+\sqrt{3}i}{2}</math> repeats in a cycle 6.
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<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>
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Since 2016 is a multiple of 6, <math>(\frac{1+\sqrt{3}i}{2})^{2016}=1</math>
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<math>f(\frac{1+\sqrt{3}i}{2})=a(\frac{-1+\sqrt{3}i}{2})+b(\frac{1+\sqrt{3}i}{2})+c</math>
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<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>
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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>
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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>.
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Since all coefficients are less than or equal to <math>2019</math>, <math>a=b=2019</math>.
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Therefore, <math>2c=4030</math> and <math>c=2015</math>.
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<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>
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~Interstigation
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==See Also==
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{{AIME box|year=2019|n=II|num-b=7|num-a=9}}
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{{MAA Notice}}
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[[Category:Intermediate Algebra Problems]]

Latest revision as of 21:18, 28 December 2023

Problem

The polynomial $f(z)=az^{2018}+bz^{2017}+cz^{2016}$ has real coefficients not exceeding $2019$, and $f\left(\tfrac{1+\sqrt{3}i}{2}\right)=2015+2019\sqrt{3}i$. Find the remainder when $f(1)$ is divided by $1000$.

Solution 1

We have $\frac{1+\sqrt{3}i}{2} = \omega$ where $\omega = e^{\frac{i\pi}{3}}$ is a primitive 6th root of unity. Then we have

\begin{align*} f(\omega) &= a\omega^{2018} + b\omega^{2017} + c\omega^{2016}\\ &= a\omega^2 + b\omega + c \end{align*}

We wish to find $f(1) = a+b+c$. We first look at the real parts. As $\text{Re}(\omega^2) = -\frac{1}{2}$ and $\text{Re}(\omega) = \frac{1}{2}$, we have $-\frac{1}{2}a + \frac{1}{2}b + c = 2015 \implies -a+b + 2c = 4030$. Looking at imaginary parts, we have $\text{Im}(\omega^2) = \text{Im}(\omega) = \frac{\sqrt{3}}{2}$, so $\frac{\sqrt{3}}{2}(a+b) = 2019\sqrt{3} \implies a+b = 4038$. As $a$ and $b$ do not exceed 2019, we must have $a = 2019$ and $b = 2019$. Then $c = \frac{4030}{2} = 2015$, so $f(1) = 4038 + 2015 = 6053 \implies f(1) \pmod{1000} = \boxed{053}$.

-scrabbler94

Solution 2

Denote $\frac{1+\sqrt{3}i}{2}$ with $\omega$.

By using the quadratic formula ($\frac{-b\pm\sqrt{b^2-4ac}}{2a}$) in reverse, we can find that $\omega$ is the solution to a quadratic equation of the form $ax^2+bx+c=0$ such that $2a=2$, $-b=1$, and $b^2-4ac=-3$. This clearly solves to $a=1$, $b=-1$, and $c=1$, so $\omega$ solves $x^2-x+1=0$.

Multiplying $x^2-x+1=0$ by $x+1$ on both sides yields $x^3+1=0$. Muliplying this by $x^3-1$ on both sides yields $x^6-1=0$, or $x^6=1$. This means that $\omega^6=1$.

We can use this to simplify the equation $a\omega^{2018}+b\omega^{2017}+c\omega^{2016}=f(\omega)=2015+2019\sqrt{3}i$ to $a\omega^2+b\omega+c=2015+2019\sqrt{3}i.$

As in Solution 1, we use the values $\omega=\frac{1}{2}+\frac{\sqrt{3}}{2}i$ and $\omega^2=-\frac{1}{2}+\frac{\sqrt{3}}{2}i$ to find that $-\frac{1}{2}a+\frac{1}{2}b+c=2015$ and $\frac{\sqrt{3}}{2}a+\frac{\sqrt{3}}{2}b=2019\sqrt{3} \implies a+b=4038.$ Since neither $a$ nor $b$ can exceed $2019$, they must both be equal to $2019$. Since $a$ and $b$ are equal, they cancel out in the first equation, resulting in $c=2015$.

Therefore, $f(1)=a+b+c=2019+2019+2015=6053$, and $6053\bmod1000=\boxed{053}$. ~emerald_block


Solution 3

Calculate the first few powers of $\frac{1+\sqrt{3}i}{2}$.

$(\frac{1+\sqrt{3}i}{2})^1=\frac{1+\sqrt{3}i}{2}$

$(\frac{1+\sqrt{3}i}{2})^2=\frac{-1+\sqrt{3}i}{2}$

$(\frac{1+\sqrt{3}i}{2})^3=-1$

$(\frac{1+\sqrt{3}i}{2})^4=\frac{-1-\sqrt{3}i}{2}$

$(\frac{1+\sqrt{3}i}{2})^5=\frac{1-\sqrt{3}i}{2}$

$(\frac{1+\sqrt{3}i}{2})^6=1$

We figure that the power of $\frac{1+\sqrt{3}i}{2}$ repeats in a cycle 6.

$f(\frac{1+\sqrt{3}i}{2})=(a(\frac{1+\sqrt{3}i}{2})^2+b(\frac{1+\sqrt{3}i}{2})+c)(\frac{1+\sqrt{3}i}{2})^{2016}$

Since 2016 is a multiple of 6, $(\frac{1+\sqrt{3}i}{2})^{2016}=1$

$f(\frac{1+\sqrt{3}i}{2})=a(\frac{-1+\sqrt{3}i}{2})+b(\frac{1+\sqrt{3}i}{2})+c$

$f(\frac{1+\sqrt{3}i}{2})=(-\frac{1}{2}a+\frac{1}{2}b+c)+(\frac{\sqrt{3}i}{2}a+\frac{\sqrt{3}i}{2}b)=2015+2019\sqrt{3}i$

Therefore, $-\frac{1}{2}a+\frac{1}{2}b+c=2015$ and $\frac{\sqrt{3}i}{2}a+\frac{\sqrt{3}i}{2}b=2019\sqrt{3}i$

Using the first equation, we can get that $-a+b+2c=4030$, and using the second equation, we can get that $a+b=4038$.

Since all coefficients are less than or equal to $2019$, $a=b=2019$.

Therefore, $2c=4030$ and $c=2015$.

$f(1)=a+b+c=2019+2019+2015=6053$, and the remainder when it divides $1000$ is $\boxed{053}.$

~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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