2024 AIME II Problems/Problem 13
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
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. 
