Difference between revisions of "2021 AMC 12B Problems/Problem 20"
Pi is 3.14 (talk | contribs) (→Solution) |
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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
[hide]Problem
Let and be the unique polynomials such thatand 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
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.