Difference between revisions of "2021 AMC 12B Problems/Problem 20"
Pi is 3.14 (talk | contribs) (→Solution) |
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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> | ||
− | ==Solution== | + | ==Solution 1== |
Note that | Note that | ||
<cmath>z^3-1\equiv 0\pmod{z^2+z+1}</cmath> | <cmath>z^3-1\equiv 0\pmod{z^2+z+1}</cmath> | ||
Line 15: | Line 15: | ||
The answer is <math>\boxed{\textbf{(A) }-z}.</math> | The answer is <math>\boxed{\textbf{(A) }-z}.</math> | ||
+ | ==Solution 2 (Somewhat of a long method)== | ||
+ | One thing to note is that <math>R(z)</math> takes the form of <math>Az + B</math> for some constants A and B. | ||
+ | Note that the roots of <math>z^2 + z + 1</math> are part of the solutions of <math>z^3 -1 = 0</math> | ||
+ | They can be easily solved with roots of unity: | ||
+ | <cmath>z^3 = 1</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> | ||
+ | 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: | ||
+ | <cmath>e^{i\frac{4042\pi}{3}} + 1 = A \cdot e^{i \frac{2\pi}{3}} + B</cmath> | ||
+ | <cmath>e^{i\frac{4\pi}{3}} + 1 = A \cdot e^{i \frac{2\pi}{3}} + B</cmath> | ||
+ | Expanding into rectangular complex number form: | ||
+ | <cmath>\frac{1}{2} - \frac{\sqrt{3}}{2} i = (-\frac{1}{2}A + B) + \frac{\sqrt{3}}{2} i A</cmath> | ||
+ | Comparing the real and imaginary parts, we get: | ||
+ | <cmath>A = -1, B = 0</cmath> | ||
+ | The answer is <math>\boxed{\textbf{(A) }-z}</math>. | ||
== Video Solution by OmegaLearn (Using Modular Arithmetic and Meta-solving) == | == Video Solution by OmegaLearn (Using Modular Arithmetic and Meta-solving) == |
Revision as of 11:34, 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 .
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.