Difference between revisions of "2024 AIME II Problems/Problem 13"
Line 38: | Line 38: | ||
-resources | -resources | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | Denote <math>r_j = e^{\frac{i 2 \pi j}{13}}</math> for <math>j \in \left\{ 0, 1, \cdots , 12 \right\}</math>. | ||
+ | |||
+ | Thus, for <math>\omega \neq 1</math>, <math>\left( \omega^0, \omega^1, \cdots, \omega^{12} \right)</math> is a permutation of <math>\left( r_0, r_1, \cdots, r_{12} \right)</math>. | ||
+ | |||
+ | We have | ||
+ | \begin{align*}\ | ||
+ | \Pi_{k = 0}^{12} \left( 2 - 2 \omega^k + \omega^{2k} \right) | ||
+ | & = \Pi_{k=0}^{12} \left( 1 + i - \omega^k \right) | ||
+ | \left( 1 - i - \omega^k \right) \ | ||
+ | & = \Pi_{k=0}^{12} \left( \sqrt{2} e^{i \frac{\pi}{4}} - \omega^k \right) | ||
+ | \left( \sqrt{2} e^{-i \frac{\pi}{4}} - \omega^k \right) \ | ||
+ | & = \Pi_{k=0}^{12} \left( \sqrt{2} e^{i \frac{\pi}{4}} - r_k \right) | ||
+ | \left( \sqrt{2} e^{-i \frac{\pi}{4}} - r_k \right) \ | ||
+ | & = \left( | ||
+ | \Pi_{k=0}^{12} \left( \sqrt{2} e^{i \frac{\pi}{4}} - r_k \right) | ||
+ | \right) | ||
+ | \left( | ||
+ | \Pi_{k=0}^{12} \left( \sqrt{2} e^{-i \frac{\pi}{4}} - r_k \right) | ||
+ | \right) . \hspace{1cm} (1) | ||
+ | \end{align*} | ||
+ | The third equality follows from the above permutation property. | ||
+ | |||
+ | Note that <math>r_0, r_1, \cdots , r_{12}</math> are all zeros of the polynomial <math>z^{13} - 1</math>. | ||
+ | Thus, | ||
+ | <cmath> | ||
+ | \[ | ||
+ | z^{13} - 1 | ||
+ | = \Pi_{k=0}^{12} \left( z - r_k \right) . | ||
+ | \] | ||
+ | </cmath> | ||
+ | |||
+ | Plugging this into Equation (1), we get | ||
+ | \begin{align*} | ||
+ | (1) | ||
+ | & = \left( \left( \sqrt{2} e^{i \frac{\pi}{4}} \right)^{13} - 1 \right) | ||
+ | \left( \left( \sqrt{2} e^{-i \frac{\pi}{4}} \right)^{13} - 1 \right) \ | ||
+ | & = \left( - 2^{13/2} e^{i \frac{\pi}{4}} - 1 \right) | ||
+ | \left( - 2^{13/2} e^{-i \frac{\pi}{4}} - 1 \right) \ | ||
+ | & = 2^{13} + 1 + 2^{13/2} \cdot 2 \cos \frac{\pi}{4} \ | ||
+ | & = 2^{13} + 1 + 2^7 \ | ||
+ | & = 8321 . | ||
+ | \end{align*} | ||
+ | |||
+ | Therefore, the answer is \boxed{\textbf{(321) }}. | ||
+ | |||
+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://youtu.be/CtIdbP4F28Q | ||
+ | |||
+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
==See also== | ==See also== |
Revision as of 16:22, 9 February 2024
Problem
Let be a 13th root of unity. Find the remainder when is divided by 1000.
Solution 1
Now, we consider the polynomial whose roots are the 13th roots of unity. Taking our rewritten product from to , we see that both instances of cycle through each of the 13th roots. Then, our answer is:
~Mqnic_
Solution 2
To find , where and , rewrite this is as
where and are the roots of the quadratic .
Grouping the 's and 's results in
the denomiator by vietas.
the numerator by newtons sums
so the answer is
-resources
Solution 3
Denote for .
Thus, for , is a permutation of .
We have
Note that are all zeros of the polynomial . Thus,
Plugging this into Equation (1), we get
Therefore, the answer is \boxed{\textbf{(321) }}.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See also
2024 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.