Difference between revisions of "2022 AMC 12A Problems/Problem 21"
(→Solution 2) |
(→Solution) |
||
| Line 4: | Line 4: | ||
<math>\textbf{(A)} \, x^2 -x + 1 \qquad\textbf{(B)} \, x^2 + x + 1 \qquad\textbf{(C)} \, x^4 + 1 \qquad\textbf{(D)} \, x^6 - x^3 + 1 \qquad\textbf{(E)} \, x^6 + x^3 + 1 </math> | <math>\textbf{(A)} \, x^2 -x + 1 \qquad\textbf{(B)} \, x^2 + x + 1 \qquad\textbf{(C)} \, x^4 + 1 \qquad\textbf{(D)} \, x^6 - x^3 + 1 \qquad\textbf{(E)} \, x^6 + x^3 + 1 </math> | ||
| − | ==Solution== | + | ==Solution 1== |
<math>P(x) = x^{2022} + x^{1011} + 1</math> is equal to <math>\frac{x^{3033}-1}{x^{1011}-1}</math> by difference of powers. | <math>P(x) = x^{2022} + x^{1011} + 1</math> is equal to <math>\frac{x^{3033}-1}{x^{1011}-1}</math> by difference of powers. | ||
Revision as of 20:57, 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
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. 
