2024 AIME II Problems/Problem 13
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)
Solution 4
Since is a root of unity, and is a prime, we have by the Fundamental Theorem of Algebra. Next, observe that the quadratic factors as Take the product of the above identity over to get the product of interest \begin{align*} P &:= \prod_{k = 0}^{12}(2 - 2\omega^k + \omega^{2k}) \\ &= \prod_{k = 0}^{12}(1 - i - \omega^k) \cdot \prod_{k = 0}^{12}(1 + i - \omega^k) \\ &= f(1-i) \cdot f(1+i) \\ &= \overline{f(1+i)} \cdot f(1+i) \\ P &= \big| f(1+i) \big|^2. \end{align*} (Here, we use the fact that whenever is a polynomial of real coefficients.) Next, notice that which means . So And we are done. Alternatively, to add some geometric flavor, we can also compute by law of cosines.
-- VensL.
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 |
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