Difference between revisions of "2019 AIME II Problems/Problem 8"
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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> | <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}} |
Revision as of 02:41, 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.