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> | ||
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| + | == Video Solution by OmegaLearn (Using Modular Arithmetic and Meta-solving) == | ||
| + | https://youtu.be/nnjr17q7fS0 | ||
| + | |||
| + | ~ pi_is_3.14 | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2021|ab=B|num-b=19|num-a=21}} | {{AMC12 box|year=2021|ab=B|num-b=19|num-a=21}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 21:45, 11 February 2021
Contents
Problem
Let 
and 
be the unique polynomials such that![\[z^{2021}+1=(z^2+z+1)Q(z)+R(z)\]](http://latex.artofproblemsolving.com/3/9/3/393d6f4b1ab6042556ad34f110d35ae89a84ca06.png)
and the degree of 
is less than 
What is 
Solution
Note that
![\[z^3-1\equiv 0\pmod{z^2+z+1}\]](http://latex.artofproblemsolving.com/0/0/8/0081f33cde41b2a048452f54d626c9851576eab8.png)
so if 
is the remainder when dividing by 
,
![\[F(z)\equiv R(z)\pmod{z^2+z+1}.\]](http://latex.artofproblemsolving.com/c/0/5/c052b84162cbead7ec4e63b419488cdad2ede437.png)
Now,
![\[z^{2021}+1= (z^3-1)(z^{2018} + z^{2015} + \cdots + z^2) + z^2+1\]](http://latex.artofproblemsolving.com/e/c/3/ec3b6a426c3d85eac065692b8fb8a74ec9e6b916.png)
So 
, and
![\[R(z)\equiv F(z) \equiv -z\pmod{z^2+z+1}\]](http://latex.artofproblemsolving.com/7/3/a/73a98ce4138d4061f15e4f359a7bd277f7b76bd3.png)
The answer is 
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. 
