How to compute $\,3^{3^{3^{\:\!\phantom{}^{.^{.^.}}}}}\!\!\!\!\bmod 46,$ for power tower height $2020$?

Question

What is the remainder of $\,3^{3^{3^{\:\!\phantom{}^{.^{.^.}}}}}\!\!\!$ divided by $46$? The level of powers is $2020$.

First there is no parenthesis so it means 3 power of 3 which is also power 3 and so on 2020 times

Second I think that we can use Fermat's Little Theorem but I don't know how, and maybe there is a better way.

Note that $a^{(b^c)}\neq (a^b)^c$ in general, so your expression is ambiguous. — Arturo Magidin, Feb 24 '20 at 06:55
No, Fermat's little theorem isn't going to work, at least not by itself, because $46$ is not a prime. There are ways to fix this, however. — Arthur, Feb 24 '20 at 07:03
@SakethMalyala: Don’t you think you should let the OP answer? — Arturo Magidin, Feb 24 '20 at 07:04
You can describe this number with the notation for teration $$3\uparrow \uparrow 2020$$ which is a power tower consisting of $2020$ $3's$. — Peter, Feb 24 '20 at 16:43
@Peter That argument is not correct. See my answer for the correct result. — Bill Dubuque, Feb 25 '20 at 00:34
@ArturoMagidin That's not really a matter of opinion, $a^{b^c}=a^{(b^c)}$ is simply the most common convention, especially since $(a^b)^c=a^{bc}$ can already be written more simply as such. — Simply Beautiful Art, Feb 25 '20 at 02:09
@SimplyBeautifulArt It is not so rigorously however as , for example , the convention for the square-root. That said, in the context of power towers the purpose is usually to produce huge numbers, which only works if we assume that they are calculated from above. — Peter, Feb 25 '20 at 14:18
@Peter I've no idea what you mean by rigorous here, but I've never seen $a^{b^c}$ mean $(a^b)^c$ in any context, as having $a^{b\widehat~c}=a^{b\times c}$ is simply confusing at best. — Simply Beautiful Art, Feb 25 '20 at 17:04

score 1 · Answer 1 · answered Feb 25 '20 at 02:22

We can easily apply Euler's totient theorem here several times

$\varphi(46)=\varphi(2)\varphi(23)=22$

$\varphi(22)=\varphi(2)\varphi(11)=10$

$\varphi(10)=\varphi(2)\varphi(5)=4$

$\varphi(4)=2$

$\varphi(2)=1$

since they are all coprime to $3$. This gives the result:

\begin{align}R&=3\widehat~(3\widehat~(3\widehat~(3\widehat~(3\widehat~n))))\bmod46\\&=3\widehat~(3\widehat~(3\widehat~(3\widehat~(3\widehat~(n\bmod1)\bmod2)\bmod4)\bmod10)\bmod22)\bmod46\\&=3\widehat~(3\widehat~(3\widehat~(3\widehat~(3\widehat~1\bmod2)\bmod4)\bmod10)\bmod22)\bmod46\\&=3\widehat~(3\widehat~(3\widehat~(3\widehat~1\bmod4)\bmod10)\bmod22)\bmod46\\&=3\widehat~(3\widehat~(3\widehat~3\bmod10)\bmod22)\bmod46\\&=3\widehat~(3\widehat~7\bmod22)\bmod46\\&=3\widehat~9\bmod46\\&=41\end{align}

for any value of $n$.

Bill Dubuque · Answer 2 · 2020-05-15T18:33:55.663

Power tower mods are all resolvable by iteratively applying MOR = modular order reduction.

$\!\!\bmod 23\!:\,\ 3^{\large 11}\!\equiv 1\,\Rightarrow\, 3^{\Large 3^{\Large 3^{\Large 3^n}}}\!\!\!\!\!\! \equiv 3^{\Large\color{#c00}{3^{\Large 3^{\Large 3^n}}\!\!\!\!\bmod 11}}\!\equiv 3^{\large\color{#c00}9}\equiv 18\ $ by MOR, by

$\!\!\color{#c00}{\bmod 11}\!:\,\ \ 3^{\large 5}\equiv 1\,\Rightarrow\,\color{#c00}{3^{\Large 3^{\Large 3^n}}}\!\!\equiv 3^{\Large\color{#0a0}{3^{\Large 3^n}\!\bmod 5}}\equiv 3^{\large\color{#0a0}2}\equiv\color{#c00}9\, $ by ditto, by

$\!\!\color{#0a0}{\bmod 5}\!:\ \ \ \ \ 3^{\large 4}\equiv 1\,\Rightarrow\ \color{#0a0}{3^{\Large 3^n}}\!\equiv\ 3^{\Large 3^n\bmod 4}\equiv 3^{\large 3}\equiv\color{#0a0}2,\ \ {\rm by}\ \bmod 4\!:\! \smash{\overbrace{3^n \equiv 3}^{\!\!\large (-1)^{\Large n}\, \equiv\ -1}}\!\!\!,\ {\rm by}\ n\ \rm odd$

Hence $\,n\,$ odd $\,\Rightarrow\, x:= 3^\smash{\Large 3^{\Large 3^{{\Large 3^n}}}}\!\!\!\!\!\equiv 18\pmod{\!23},\,$ so $\,x\,$ odd $\Rightarrow\, x\equiv 18\!+\!23\equiv 41\pmod{\!46}$

For $\,n\,$ even we get $\, \color{#0a0}{3\pmod{\! 5}},\,$ so $\,3^{\large \color{#0a0}3}\equiv\color{#c00}{5\pmod{\! 11}},\:$ so $\,\ x\equiv 3^{\large\color{#c00}5}\equiv 13\pmod{\!23\ \&\ 46}$

Hence, more generally $\,\ \bbox[1px,border:1px solid #0a0]{\bbox[8px,border:1px solid #050]{\!\!\bmod 46\!:\,\ 3^{\Large 3^{\Large 3^{{\Large 3^n}}}}\!\!\!\equiv\, \left\{\begin{align} &41,\ \,n\,\ \rm odd\\ &13,\ \,n\,\ \rm even\end{align}\right.}}$

We used $ \bmod 23!:, \left[3\equiv 7^{\large 2}\right]^{\large 11}!!\Rightarrow, 3^{\large 11}\equiv 7^{\large 22}\equiv 1.,$ Similarly $!\color{#c00}{\bmod 11},$ using $,3\equiv 6^2\ \ $ — Bill Dubuque, Feb 25 '20 at 02:43
Note $ $ The same method works for any height power towers, but we may need to additionally apply the mod distributive law (or CRT) when the base is not coprime to the modulus, e.g. see here and here — Bill Dubuque, May 19 '20 at 22:10

How to compute $\,3^{3^{3^{\:\!\phantom{}^{.^{.^.}}}}}\!\!\!\!\bmod 46,$ for power tower height $2020$?

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