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 | ||
Line 10: | Line 10: | ||
\end{align*}</cmath> | \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 = 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{ | + | 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>. |
-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}} | ||
+ | |||
+ | [[Category:Intermediate Algebra Problems]] |
Latest revision as of 22:18, 28 December 2023
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.