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Let $\omega$ be a primitive cube root of unity. Let $x = {\omega}^{{2009}^{{2009}^{{2009}^{\cdots 2009}}}}$ (up to $2009$ times). Simplify the value of $x$.

My attempt: Let $m = {{2009}^{{2009}^{{2009}^{\cdots 2009}}}}$ (up to $2007$ times). Then since $2009$ is odd so $2009^m$ is also odd. Let $k = 2009^m$. Now since $k$ is an odd integer so $2^k \equiv 2\ (\text {mod}\ 3)$. Also $2009 \equiv 2\ (\text {mod}\ 3)$. Therefore, $2009^k \equiv 2\ (\text {mod}\ 3)$. Let $n = 2009^k$. Then $n = 3k' + 2$ for some $k' \in \Bbb N$. Therefore $$x = {\omega}^n = {\omega}^2$$

Am I right? Please verify it.

Thanks in advance for reading.

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    $\begingroup$ You did not say what $\omega$ is $\endgroup$ – DIdier_ May 20 at 7:10
  • $\begingroup$ @Didlier $\omega$ is a primitive cube root of unity. $\endgroup$ – math maniac. May 20 at 7:11
  • $\begingroup$ Yep, you're right $\endgroup$ – Shashwat1337 May 20 at 7:15
  • $\begingroup$ As which primitive root $\omega$ is is not specified and the answer seems to be unambiguous, then it must be $1$. $\endgroup$ – Yves Daoust May 20 at 7:17
  • $\begingroup$ @Yves Daoust what must be $1$? If $\omega$ is a primitive cube root of unity then how can $\omega^2 = 1$? Usually $\frac {-1 + \sqrt 3 i} {2}$ is considered as $\omega.$ But then $\omega^2 = \frac {-1 - \sqrt 3 i} {2} \neq 1.$ $\endgroup$ – math maniac. May 20 at 7:20
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The answer is $\omega^2$ and your method is right and check my method $2009^{odd} $can be written as ${(2010-1)}^{odd}$. That is $2010 \times m - 1$($m$ is some integer), as $2010$ is divisible by $3$ it is $\frac1{\omega} = \omega^2$

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  • $\begingroup$ Very nice answer. $\endgroup$ – math maniac. May 20 at 7:37
  • $\begingroup$ Thank u bro.... $\endgroup$ – Namburu Karthik May 20 at 7:41
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$2009$ can be expressed as $6m-1$ where $m$ is any integer

Now observe that $(6m-1)^{6m-1}\equiv-1\pmod6,$ so again of the form $6m'-1$

If $w$ is a root of unity,

$$w^{6n-1}=(w^3)^{2n-1}w^2=?$$

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