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
Problem
Let ![\[P(x) = x^{2022} + x^{1011} + 1.\]](http://latex.artofproblemsolving.com/c/6/6/c66458c6088164c71caaad852c39637d9cfb6740.png)
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 ![\[x^{1011} = \frac{-1\pm i\sqrt{3}}{2}\]](http://latex.artofproblemsolving.com/b/a/a/baa714e9ca788c6c9a1fb3bb0b4f231a8be038eb.png)
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, ![\[x^3= \frac{-1\pm i\sqrt{3}}{2}\]](http://latex.artofproblemsolving.com/5/e/1/5e1a03efb85bc4bfddd2819ccfca5e7e00335974.png)
for the solution we are testing. Now notice that we can rewrite the roots of as ![\[x^3 = cis{\frac{2\pi}{3}}, cis{\frac{4\pi}{3}}\]](http://latex.artofproblemsolving.com/b/b/0/bb0d330d1386ce4b1e457931d4d90fcb3b2ccfc0.png)
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 ![\[x^6+x^3+1 | x^{2022}+x^{1011}+1\]](http://latex.artofproblemsolving.com/c/f/4/cf4a152021171080b28ca990bd705c93d856fe41.png)
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. 
