Difference between revisions of "2021 AMC 12B Problems/Problem 20"
Jamess2022 (talk | contribs) (→Solution 2 (Somewhat of a long method)) |
Jamess2022 (talk | contribs) (→Solution 2 (Somewhat of a long method)) |
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<cmath>z^3 = 1</cmath> | <cmath>z^3 = 1</cmath> | ||
<cmath>z^3 = e^{i 0}</cmath> | <cmath>z^3 = e^{i 0}</cmath> | ||
− | <cmath>z = e^{i 0}, e^{i \frac{2\pi}{3}}, e^{i -\frac{2\pi}{3}}</cmath> | + | <cmath>z = e^{i 0}, e^{i \frac{2\pi}{3}}, e^{i -\frac{2\pi}{3}}</cmath> <cmath>\newline</cmath> |
Obviously the right two solutions are the roots of <math>z^2 + z + 1 = 0</math> | Obviously the right two solutions are the roots of <math>z^2 + z + 1 = 0</math> | ||
We substitute <math>e^{i \frac{2\pi}{3}}</math> into the original equation, and <math>z^2 + z + 1</math> becomes 0. Using De Moivre's theorem, we get: | We substitute <math>e^{i \frac{2\pi}{3}}</math> into the original equation, and <math>z^2 + z + 1</math> becomes 0. Using De Moivre's theorem, we get: |
Revision as of 11:36, 12 February 2021
Contents
Problem
Let and be the unique polynomials such thatand 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 2 (Somewhat of a long method)
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.