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

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~Mqnic_
 
~Mqnic_
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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 10:33, 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

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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