# find the remainder for $\color{purple}{TETRATION}$ such that $( {^{2021}2021}(\mod13))$

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 :)

$$\color{GREEN}{THANKS...}$$

• How did you get $11$ as the first exponent in the tower? Reducing $2021$ modulo $12$ gives $5$, which is not congruent to $11$ (or $-1$) modulo $12$. Jan 28 at 18:32
• @Christoph I thought that i should refrain from dealing with $2021's$ , so i made up two numbers which are related to $\phi 13$ Jan 28 at 18:34
• Please request for a nice colourful answer! +1 Jan 28 at 18:37
• @TeresaLisbon :))) Jan 28 at 18:38
• @TeresaLisbon is it okey now :D Jan 28 at 18:41

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 $${}^{2021}2021 \equiv \color{purple}6^{\color{blue}5^{\color{orange}1^{\cdots^{\color{red}{2021}}}}} \equiv 6^5 \equiv 7776 \equiv 2 \pmod{13}.$$

• @thanks for this elegant and colorful answer :) Jan 28 at 18:42
• $\color{green}{\text{You're welcome!}}$ Jan 28 at 18:43
• Ah, I see the rainbow is out in full force! +1 , by the way today's lucky colour is platinum, so make sure that if you are writing answers there's plenty of platinum in there. Jan 28 at 18:52

Tetration is a repeated raising to a given power, a "tower" of exponents. Effectively this is worked right to left (or "top down" looking at the tower analogy).

So this is a massively tall tower of exponents, but oddly enough that doesn't matter; we'll only be interested in the bottom few elements to find the answer.

You start well to find that $$2021\equiv 6 \bmod 13$$.

Then from Fermat's little theorem we know that $$2021^{12}\equiv 1 \bmod 13$$ so $$2021^a$$ will cycle round values on a pattern of $$12$$ length (at most).

Checking that, we see $$(6^i){\large|_{i=1}^{12}}\equiv (6,10,8,9,2,12,7,3,5,4,11,1) \bmod 13$$ in that order.

So now stepping up the tower of exponents, consider $$2021^a$$ where $$a=2021^b$$. We know that we only need to consider the values of $$a\bmod 12$$ from the above cycle, so that implies that we need to look at $$a = 2021^b = 5^b\bmod 12$$. So how does $$5$$ cycle under an exponent $$\bmod 12$$? It's easy to find by calucaltion that $$5^2 = 25\equiv 1\bmod 12$$, so the cycle now is only length two - effectively, we only need to know whether $$b$$ is odd or even.

In the tower of exponents though we know that $$b = 2021^c$$, which is odd. So we have our result:

$${}^{2021}2021 \equiv 6^{5^{\text{odd}}} \equiv 6^5\equiv 2 \bmod 13$$

$$^{2021}2021 = 2021^{^{2020}2021}\pmod{13}\equiv 2021^{^{2020}2021\pmod {12}}\pmod {13}\equiv 2021^{2021^{^{2019}2021}\pmod{12}}$$

$$6^{5^{^{2019}2021}\pmod{12}}\pmod{13}$$

Now the order of $$5^2 \equiv 1 \pmod 12$$ so $$5^{odd} \equiv 5 \pmod{12}$$ and $$5^{even}\equiv 1\pmod {12}$$ and $$^{2019}2021$$ is odd.

So $$6^{5^{^{2019}2021}\pmod{12}}\pmod {13}\equiv 6^5 \equiv 6^2*6^2*6\equiv (-3)^2*6\equiv 45\equiv 2 \pmod {13}$$.