Difference between revisions of "2021 AMC 12B Problems/Problem 20"
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<math>\textbf{(A) }-z \qquad \textbf{(B) }-1 \qquad \textbf{(C) }2021\qquad \textbf{(D) }z+1 \qquad \textbf{(E) }2z+1</math> | <math>\textbf{(A) }-z \qquad \textbf{(B) }-1 \qquad \textbf{(C) }2021\qquad \textbf{(D) }z+1 \qquad \textbf{(E) }2z+1</math> | ||
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+ | ==Video Solution using long division(not brutal)== | ||
+ | https://youtu.be/kxPDeQRGLEg | ||
+ | ~hippopotamus1 | ||
==Solution 1== | ==Solution 1== |
Revision as of 12:11, 13 February 2021
Contents
Problem
Let and
be the unique polynomials such that
and the degree of
is less than
What is
Video Solution using long division(not brutal)
https://youtu.be/kxPDeQRGLEg ~hippopotamus1
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.