Difference between revisions of "2024 AIME II Problems/Problem 13"

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-resources
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==Solution 3==
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Denote <math>r_j = e^{\frac{i 2 \pi j}{13}}</math> for <math>j \in \left\{ 0, 1, \cdots , 12 \right\}</math>.
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Thus, for <math>\omega \neq 1</math>, <math>\left( \omega^0, \omega^1, \cdots, \omega^{12} \right)</math> is a permutation of <math>\left( r_0, r_1, \cdots, r_{12} \right)</math>.
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We have
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\begin{align*}\
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\Pi_{k = 0}^{12} \left( 2 - 2 \omega^k + \omega^{2k} \right)
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& = \Pi_{k=0}^{12} \left( 1 + i - \omega^k \right)
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\left( 1 - i - \omega^k \right) \
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& = \Pi_{k=0}^{12} \left( \sqrt{2} e^{i \frac{\pi}{4}} - \omega^k \right)
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\left( \sqrt{2} e^{-i \frac{\pi}{4}} - \omega^k \right) \
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& = \Pi_{k=0}^{12} \left( \sqrt{2} e^{i \frac{\pi}{4}} - r_k \right)
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\left( \sqrt{2} e^{-i \frac{\pi}{4}} - r_k \right) \
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& = \left(
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\Pi_{k=0}^{12} \left( \sqrt{2} e^{i \frac{\pi}{4}} - r_k \right)
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\right)
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\left(
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\Pi_{k=0}^{12} \left( \sqrt{2} e^{-i \frac{\pi}{4}} - r_k \right)
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\right) . \hspace{1cm} (1)
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\end{align*}
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The third equality follows from the above permutation property.
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Note that <math>r_0, r_1, \cdots , r_{12}</math> are all zeros of the polynomial <math>z^{13} - 1</math>.
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Thus,
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<cmath>
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\[
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z^{13} - 1
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= \Pi_{k=0}^{12} \left( z - r_k \right) .
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\]
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</cmath>
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Plugging this into Equation (1), we get
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\begin{align*}
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(1)
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& = \left( \left( \sqrt{2} e^{i \frac{\pi}{4}} \right)^{13} - 1 \right)
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\left( \left( \sqrt{2} e^{-i \frac{\pi}{4}} \right)^{13} - 1 \right) \
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& = \left( - 2^{13/2} e^{i \frac{\pi}{4}} - 1 \right)
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\left( - 2^{13/2} e^{-i \frac{\pi}{4}} - 1 \right) \
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& = 2^{13} + 1 + 2^{13/2} \cdot 2 \cos \frac{\pi}{4} \
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& = 2^{13} + 1 + 2^7 \
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& = 8321 .
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\end{align*}
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Therefore, the answer is \boxed{\textbf{(321) }}.
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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==Video Solution==
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 +
https://youtu.be/CtIdbP4F28Q
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 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
  
 
==See also==
 
==See also==

Revision as of 16:22, 9 February 2024

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  Πk=012(22ωk+ω2k)=Πk=012(1+iωk)(1iωk)=Πk=012(2eiπ4ωk)(2eiπ4ωk)=Πk=012(2eiπ4rk)(2eiπ4rk)=(Πk=012(2eiπ4rk))(Πk=012(2eiπ4rk)).(1) 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 (1)=((2eiπ4)131)((2eiπ4)131)=(213/2eiπ41)(213/2eiπ41)=213+1+213/22cosπ4=213+1+27=8321.

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

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

Video Solution

https://youtu.be/CtIdbP4F28Q

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

See also

2024 AIME II (ProblemsAnswer KeyResources)
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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