Difference between revisions of "2022 AMC 12A Problems/Problem 21"
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Let <math>x^{1011} = u</math>, now we can rewrite our polynomial as <math>u^2+u+1</math>. Using the quadratic formula to solve for the roots of this polynomial, we have <cmath>x^{1011} = \frac{-1\pm i\sqrt{3}}{2}</cmath> Looking at our answer choices, we want to find a polynomial whose roots satisfy this expression. Since the expression <math>x^6+x^3+1</math> is in a similar form to our original polynomial, except with <math>x^3</math> in place of <math>x^{1011}</math>, this would be a good place to start. Solving for the roots of <math>x^3</math> in a similar fashion, <cmath>x^3= \frac{-1\pm i\sqrt{3}}{2}</cmath> for the solution we are testing. Now notice that we can rewrite the roots of <math>x^3</math> as <cmath>x^3 = cis{\frac{2\pi}{3}}, cis{\frac{4\pi}{3}}</cmath> Both of which are third roots of unity. We want to now check if this value of <math>x^3</math> satisfies <math>x^{1011} = \frac{-1\pm i\sqrt{3}}{2}</math>. Notice that <math>x^{1011} = (x^{3})^{112\cdot3}\cdot x^3</math>, and since both values of <math>x^3</math> are roots of unity, we can simplify the expression we want satisfiedto the expression to <math>x^{1011}=x^3</math>. Since both values of <math>x^3</math> are also values of <math>x^{1011}</math>, the roots for our <math>x^6+x^3+1</math> are also roots of <math>x^{2022}+x^{1011}+1</math>, meaning that <cmath>x^6+x^3+1 | x^{2022}+x^{1011}+1</cmath> so Therefore, the answer is <math>\boxed{E}</math>. | Let <math>x^{1011} = u</math>, now we can rewrite our polynomial as <math>u^2+u+1</math>. Using the quadratic formula to solve for the roots of this polynomial, we have <cmath>x^{1011} = \frac{-1\pm i\sqrt{3}}{2}</cmath> Looking at our answer choices, we want to find a polynomial whose roots satisfy this expression. Since the expression <math>x^6+x^3+1</math> is in a similar form to our original polynomial, except with <math>x^3</math> in place of <math>x^{1011}</math>, this would be a good place to start. Solving for the roots of <math>x^3</math> in a similar fashion, <cmath>x^3= \frac{-1\pm i\sqrt{3}}{2}</cmath> for the solution we are testing. Now notice that we can rewrite the roots of <math>x^3</math> as <cmath>x^3 = cis{\frac{2\pi}{3}}, cis{\frac{4\pi}{3}}</cmath> Both of which are third roots of unity. We want to now check if this value of <math>x^3</math> satisfies <math>x^{1011} = \frac{-1\pm i\sqrt{3}}{2}</math>. Notice that <math>x^{1011} = (x^{3})^{112\cdot3}\cdot x^3</math>, and since both values of <math>x^3</math> are roots of unity, we can simplify the expression we want satisfiedto the expression to <math>x^{1011}=x^3</math>. Since both values of <math>x^3</math> are also values of <math>x^{1011}</math>, the roots for our <math>x^6+x^3+1</math> are also roots of <math>x^{2022}+x^{1011}+1</math>, meaning that <cmath>x^6+x^3+1 | x^{2022}+x^{1011}+1</cmath> so Therefore, the answer is <math>\boxed{E}</math>. |
Revision as of 22:58, 13 November 2022
Contents
[hide]Problem
Let Which of the following polynomials is a factor of ?
Solution 1
is equal to by difference of powers.
Therefore, the answer is a polynomial that divides but not .
Note that any polynomial divides if and only if is a factor of .
The prime factorizations of and are and , respectively.
Hence, is a divisor of but not .
By difference of powers, . Therefore, the answer is .
Solution 2
We simply test roots for each, as are multiples of three, we need to make sure the roots are in the form of , so we only have to look at .
If we look at choice , which works perfectly, the answer is just
~bluesoul
Solution 3
Let , now we can rewrite our polynomial as . Using the quadratic formula to solve for the roots of this polynomial, we have Looking at our answer choices, we want to find a polynomial whose roots satisfy this expression. Since the expression is in a similar form to our original polynomial, except with in place of , this would be a good place to start. Solving for the roots of in a similar fashion, for the solution we are testing. Now notice that we can rewrite the roots of as Both of which are third roots of unity. We want to now check if this value of satisfies . Notice that , and since both values of are roots of unity, we can simplify the expression we want satisfiedto the expression to . Since both values of are also values of , the roots for our are also roots of , meaning that so Therefore, the answer is .
- DavidHovey
Video Solution by ThePuzzlr
~ MathIsChess
See Also
2022 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 20 |
Followed by Problem 22 |
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.