It is given that $ {^{2021}2021}$ is a tetration and I want to find the remainder when it is divided by $13$.
Briefly , I am looking for $ {^{2021}2021}\pmod{13}$
My work :
$\color{purple}{Firstly:}$ I found $2021\equiv 6\pmod{13}$
$\color{blue}{Secondly:}$ I thought that $6^{\color{red}{a}}\equiv 6^{\color{red}{b}}\pmod{13}$ should be satisfied.
$\color{orange}{Thirdly:}$ $a\equiv b\pmod{12}$ by Euler's phi function.Then , i said that $a\equiv 11$ and $b\equiv -1$
$\color{red}{Fourthly:}$ I said that $6^{11^m}\equiv 6^{(-1)^m}\pmod{13}$ and there left $2020$ superscript for tetration. Therefore , I said that it must be equal to $6^{(-1)^{2020}}\equiv6\pmod{13}$
$\therefore$ Answer is $6$
Is my solution correct , if not , can you share any trick or shortcut for finding remainders when we encounter tetrations..
$\color{red}{NOTE:}$ WHAT IS TETRATION $\rightarrow $ https://en.wikipedia.org/wiki/Tetration
$\color{red}{NOTE-2:}$ I would rather colorful answer for the sake of @terasalisbon 🙂
$\color{GREEN}{THANKS…}$
Best Answer
The only thing you need to know is that looking at a power $a^b$ modulo $k$ you can reduce the base $a$ modulo $k$ and the exponent $b$ modulo $\varphi(k)$.
Hence, when you consider the a tetration $$ {}^{2021}2021 = \color{purple}{2021}^{\color{blue}{2021}^{\color{orange}{2021}^{\cdots^{\color{red}{2021}}}}}, $$ you can reduce the base modulo $13$ the first exponent modulo $\varphi(13)=12$, the third exponent modulo $\varphi(12)=4$, where already $2021\equiv 1 \pmod 4$.
Hence $$ {}^{2021}2021 \equiv \color{purple}6^{\color{blue}5^{\color{orange}1^{\cdots^{\color{red}{2021}}}}} \equiv 6^5 \equiv 7776 \equiv 2 \pmod{13}. $$