Difference between revisions of "2021 AMC 12B Problems/Problem 20"
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<cmath>R(z)\equiv F(z) \equiv -z\pmod{z^2+z+1}</cmath> | <cmath>R(z)\equiv F(z) \equiv -z\pmod{z^2+z+1}</cmath> | ||
The answer is <math>\boxed{\textbf{(A) }-z}.</math> | The answer is <math>\boxed{\textbf{(A) }-z}.</math> | ||
+ | |||
+ | ==Solution 1b (More Thorough Version of 1)== | ||
+ | Instead of dealing with a nasty <math>z^2+z+1</math>, we can instead deal with the nice <math>z^3 - 1</math>, as <math>z^2+z+1</math> is a factor of <math>z^3-1</math>. Then, we try to see what <math>\frac{z^{2021} + 1}{z^3 - 1}</math> is. Of course, we will need a <math>z^{2018}</math>, getting <math>z^{2021} - z^{2018}</math>. Then, we've gotta get rid of the <math>z^{2018}</math> term, so we add a <math>z^{2015}</math>, to get <math>z^{2021} - z^{2015}</math>. This pattern continues, until we add a <math>z^2</math> to get rid of <math>z^5</math>, and end up with <math>z^{2021} - z^2</math>. We can't add anything more to get rid of the <math>z^2</math>, so our factor is <math>z^{2018} + z^{2015} + z^{2012} + \cdots + z^2</math>. Then, to get rid of the <math>z^2</math>, we must have a remainder of <math>+z^2</math>, and to get the <math>+1</math> we have to also have a <math>+1</math> in the remainder. So, our product is <cmath>z^{2021}+1= (z^3-1)(z^{2018} + z^{2015} + \cdots + z^2) + z^2+1.</cmath> Then, our remainder is <math>z^2+1</math>. The remainder when dividing by <math>z^3-1</math> must be the same when dividing by <math>z^2+z+1</math>, modulo <math>z^2+z+1</math>. So, we have that <math>z^2 + 1 \equiv R(z) \pmod{z^2+z+1}</math>, or <math>R(z) \equiv -z\pmod{z^2+z+1}</math>. This corresponds to answer choice <math>\boxed{\textbf{(A)} ~ -z}</math>. ~rocketsri | ||
==Solution 2 (Complex numbers)== | ==Solution 2 (Complex numbers)== |
Revision as of 16:40, 12 February 2021
Contents
Problem
Let and
be the unique polynomials such that
and the degree of
is less than
What is
Solution 1
Note that
so if
is the remainder when dividing by
,
Now,
So
, and
The answer is
Solution 1b (More Thorough Version of 1)
Instead of dealing with a nasty , we can instead deal with the nice
, as
is a factor of
. Then, we try to see what
is. Of course, we will need a
, getting
. Then, we've gotta get rid of the
term, so we add a
, to get
. This pattern continues, until we add a
to get rid of
, and end up with
. We can't add anything more to get rid of the
, so our factor is
. Then, to get rid of the
, we must have a remainder of
, and to get the
we have to also have a
in the remainder. So, our product is
Then, our remainder is
. The remainder when dividing by
must be the same when dividing by
, modulo
. So, we have that
, or
. This corresponds to answer choice
. ~rocketsri
Solution 2 (Complex numbers)
One thing to note is that takes the form of
for some constants A and B.
Note that the roots of
are part of the solutions of
They can be easily solved with roots of unity:
Obviously the right two solutions are the roots of
We substitute
into the original equation, and
becomes 0. Using De Moivre's theorem, we get:
Expanding into rectangular complex number form:
Comparing the real and imaginary parts, we get:
The answer is
. ~Jamess2022(burntTacos;-;)
Video Solution by OmegaLearn (Using Modular Arithmetic and Meta-solving)
~ pi_is_3.14
See Also
2021 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.