Difference between revisions of "2024 AIME II Problems/Problem 13"
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
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| + | ==Video Solution== | ||
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| + | https://youtu.be/aSD8Xz0dAI8?si=PUDeOrRg-0bVXNpp | ||
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| + | ~MathProblemSolvingSkills.com | ||
==Video Solution== | ==Video Solution== | ||
Revision as of 15:30, 9 April 2024
Contents
Problem
Let 
be a 13th root of unity. Find the remainder when
![\[\prod_{k=0}^{12}(2-2\omega^k+\omega^{2k})\]](http://latex.artofproblemsolving.com/c/c/d/ccd8d560907cc70b8b51b8c189e2bbd61d7a08a8.png)
is divided by 1000.
Solution 1
![\[\prod_{k=0}^{12} \left(2- 2\omega^k + \omega^{2k}\right) = \prod_{k=0}^{12} \left((1 - \omega^k)^2 + 1\right) = \prod_{k=0}^{12} \left((1 + i) - \omega^k)((1 - i) - \omega^k\right)\]](http://latex.artofproblemsolving.com/6/a/2/6a2122af68a748bd88cd323030bbf6ef8aa004a9.png)
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:
![\[((1 + i)^{13} - 1)(1 - i)^{13} - 1)\]](http://latex.artofproblemsolving.com/1/8/9/1894c0d63192f76c8212267393feeac4f74ce787.png)
![\[= (-64(1 + i) - 1)(-64(1 - i) - 1)\]](http://latex.artofproblemsolving.com/c/c/e/cceec0105244281350ffc2c5d982fc9675396b0e.png)
![\[= (65 + 64i)(65 - 64i)\]](http://latex.artofproblemsolving.com/2/e/9/2e91ea1b9c16412555755bb21a8527eaf64b35fc.png)
![\[= 65^2 + 64^2\]](http://latex.artofproblemsolving.com/c/6/4/c64e46ac56e097cda1ea56085e2a5607fe4ce032.png)
![\[= 8\boxed{\textbf{321}}\]](http://latex.artofproblemsolving.com/e/4/6/e46d6c4674f7ff5ad1e186447149e6b4c7160db5.png)
~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 \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 
are all zeros of the polynomial 
.
Thus,
![\[ z^{13} - 1 = \Pi_{k=0}^{12} \left( z - r_k \right) . \]](http://latex.artofproblemsolving.com/9/7/3/9739415c799cb143170b420f49c8a7b0979002f3.png)
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 
.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution
https://youtu.be/aSD8Xz0dAI8?si=PUDeOrRg-0bVXNpp
~MathProblemSolvingSkills.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. 
