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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+ | https://youtu.be/aSD8Xz0dAI8?si=PUDeOrRg-0bVXNpp | ||
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+ | ~MathProblemSolvingSkills.com | ||
==Video Solution== | ==Video Solution== |
Revision as of 16:30, 9 April 2024
Contents
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 \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,
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.