Difference between revisions of "2021 AMC 12B Problems/Problem 20"
(Created page with "==Problem 20== Let <math>Q(z)</math> and <math>R(z)</math> be the unique polynomials such that<cmath>z^{2021}+1=(z^2+z+1)Q(z)+R(z)</cmath>and the degree of <math>R</math> is l...") |
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− | ==Problem | + | ==Problem== |
Let <math>Q(z)</math> and <math>R(z)</math> be the unique polynomials such that<cmath>z^{2021}+1=(z^2+z+1)Q(z)+R(z)</cmath>and the degree of <math>R</math> is less than <math>2.</math> What is <math>R(z)?</math> | Let <math>Q(z)</math> and <math>R(z)</math> be the unique polynomials such that<cmath>z^{2021}+1=(z^2+z+1)Q(z)+R(z)</cmath>and the degree of <math>R</math> is less than <math>2.</math> What is <math>R(z)?</math> | ||
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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> | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC12 box|year=2021|ab=B|num-b=19|num-a=21}} | ||
+ | {{MAA Notice}} |
Revision as of 19:36, 11 February 2021
Problem
Let and
be the unique polynomials such that
and the degree of
is less than
What is
Solution
Note that
so if
is the remainder when dividing by
,
Now,
So
, and
The answer is
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.