2024 AIME II Problems/Problem 13

Problem

Let $\omega\neq 1$ be a 13th root of unity. Find the remainder when $\prod_{k=0}^{12}(2-2\omega^k+\omega^{2k})$ is divided by 1000.

Solution 1

$\prod_{k=0}^{12} \left(2- 2\omega^k + \omega^{2k}\right) = \prod_{k=0}^{12} \left((1 - \omega^k)^2 + 1\right) = \prod_{k=0}^{12} \left((1 + i) - \omega^k)((1 - i) - \omega^k\right)$

Now, we consider the polynomial $x^{13} - 1$ whose roots are the 13th roots of unity. Taking our rewritten product from $0$ to $12$ , we see that both instances of $\omega^k$ cycle through each of the 13th roots. Then, our answer is:

$((1 + i)^{13} - 1)(1 - i)^{13} - 1)$

$= (-64(1 + i) - 1)(-64(1 - i) - 1)$

$= (65 + 64i)(65 - 64i)$

$= 65^2 + 64^2$

$= 8\boxed{\textbf{321}}$

~Mqnic_

Solution 2

To find $\prod_{k=0}^{12} (2 - 2w^k + w^{2k})$ , where $w\neq1$ and $w^{13}=1$ , rewrite this is as

$(r-w)(s-w)(r-w^2)(s-w^2)...(r-w^{12})(s-w^{12})$ where $r$ and $s$ are the roots of the quadratic $x^2-2x+2=0$ .

Grouping the $r$ 's and $s$ 's results in $\frac{r^{13}-1}{r-1} \cdot\frac{s^{13}-1}{s-1}$

the denomiator $(r-1)(s-1)=1$ by vietas.

the numerator $(rs)^{13} - (r^{13} + s^{13}) + 1 = 2^{13} - (-128) + 1= 8321$ by newtons sums

so the answer is $\boxed{321}$

-resources

Solution 3

Denote $r_j = e^{\frac{i 2 \pi j}{13}}$ for $j \in \left\{ 0, 1, \cdots , 12 \right\}$ .

Thus, for $\omega \neq 1$ , $\left( \omega^0, \omega^1, \cdots, \omega^{12} \right)$ is a permutation of $\left( r_0, r_1, \cdots, r_{12} \right)$ .

We have $\begin{aligned} Π_{k = 0}^{12} (2 - 2 ω^{k} + ω^{2 k}) & = Π_{k = 0}^{12} (1 + i - ω^{k}) (1 - i - ω^{k}) \\ = Π_{k = 0}^{12} (\sqrt{2} e^{i \frac{π}{4}} - ω^{k}) (\sqrt{2} e^{- i \frac{π}{4}} - ω^{k}) \\ = Π_{k = 0}^{12} (\sqrt{2} e^{i \frac{π}{4}} - r_{k}) (\sqrt{2} e^{- i \frac{π}{4}} - r_{k}) \\ = (Π_{k = 0}^{12} (\sqrt{2} e^{i \frac{π}{4}} - r_{k})) (Π_{k = 0}^{12} (\sqrt{2} e^{- i \frac{π}{4}} - r_{k})) . (1) \end{aligned}$ The third equality follows from the above permutation property.

Note that $r_0, r_1, \cdots , r_{12}$ are all zeros of the polynomial $z^{13} - 1$ . Thus, $z^{13} - 1 = \Pi_{k=0}^{12} \left( z - r_k \right) .$

Plugging this into Equation (1), we get $\begin{aligned} (1) & = ({(\sqrt{2} e^{i \frac{π}{4}})}^{13} - 1) ({(\sqrt{2} e^{- i \frac{π}{4}})}^{13} - 1) \\ = (- 2^{13 / 2} e^{i \frac{π}{4}} - 1) (- 2^{13 / 2} e^{- i \frac{π}{4}} - 1) \\ = 2^{13} + 1 + 2^{13 / 2} \cdot 2 \cos \frac{π}{4} \\ = 2^{13} + 1 + 2^{7} \\ = 8321 . \end{aligned}$

Therefore, the answer is $\boxed{\textbf{(321) }}$ .

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution

https://youtu.be/aSD8Xz0dAI8?si=PUDeOrRg-0bVXNpp

~MathProblemSolvingSkills.com

Video Solution

https://youtu.be/CtIdbP4F28Q

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

2024 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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2024 AIME II Problems/Problem 13

Contents

Problem

Solution 1

Solution 2

Solution 3

Video Solution

Video Solution

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