Difference between revisions of "2024 AIME II Problems/Problem 13"
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~Mqnic_ | ~Mqnic_ | ||
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| + | ==Solution 2== | ||
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| + | To find <math>\prod_{k=0}^{12} (2 - 2w^k + w^{2k})</math>, where <math>w\neq1</math> and <math>w^{13}=1</math>, rewrite this is as | ||
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| + | <math>(r-w)(s-w)(r-w^2)(s-w^2)...(r-w^{12})(s-w^{12})</math> where <math>r</math> and <math>s</math> are the roots of the quadratic <math>x^2-2x+2=0</math>. | ||
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| + | Grouping the <math>r</math>'s and <math>s</math>'s results in <math>\frac{r^{13}-1}{r-1} \cdot\frac{s^{13}-1}{s-1}</math> | ||
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| + | the denomiator <math>(r-1)(s-1)=1</math> by vietas. | ||
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| + | the numerator <math>(rs)^{13} - (r^{13} + s^{13}) + 1 = 2^{13} - (-128) + 1= 8321</math> by newtons sums | ||
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| + | so the answer is <math>\boxed{321}</math> | ||
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| + | -resources | ||
==See also== | ==See also== | ||
Revision as of 09:33, 9 February 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
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. 
