Difference between revisions of "2024 AIME II Problems/Problem 13"
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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
[hide]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
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.