Find last 3 digits of $ 2032^{2031^{2030^{\dots^{2^{1}}}}}$

Question

Find the last 3 digits of this number $$ 2032^{2031^{2030^{\dots^{2^{1}}}}} $$ So obviously we are looking for $x$ so that $$ 2032^{2031^{2030^{\dots^{2^{1}}}}} \equiv x \quad \text{mod}\hspace{0.1cm} 1000 $$ I also know that usually you use Euler' theorem here, but that only works when the numbers $a$ and $n$ are coprime, but $2032$ and $1000$ are not coprime? I can easily find $\varphi(1000)$, that is not a problem. Am I looking for wrong numbers to be coprime here or is there another way instead of Euler' theorem?

Answer 1

This can be done in a couple minutes of mental arithmetic using the $\!\bmod\!$ Distributive Law
i.e. using $\ ab\bmod ac \,=\, a(b \bmod c)\ $ to factor out $\,a=8\,$ (and later $\,a = 5)\,$ below

$\bmod 1000\!:\ 32^{\large 2031^{\LARGE 2k}}\!\!\equiv\, 8\left[\dfrac{\color{#0a0}{32^{\large 2031^{\LARGE 2k}}}}8 \bmod \color{#0a0}{125}\right]\! \equiv 8\left[\dfrac{\color{#c00}{32}}8\bmod 125\right]\! \equiv 32,\ $ by

$\ \,\begin{align} \!\bmod \color{#0a0}{125}\!:\ \color{#0a0}{32^{\large 2031^{\LARGE 2k}}}\!\! &\equiv\, 2^{\large 5\cdot 2031^{\LARGE 2k}\! \bmod 100}\ \ \ {\rm by\ \ } 100 = \phi(125)\ \ \ \rm [Euler\ totient]\\ &\equiv\,2^{\large 5(\color{#b6f}{2031}^{\LARGE \color{#d4f}2k}\! \bmod 20)}\ \ \ {\rm by\ \ \ mod\ Distributive\ Law}\\ &\equiv\,{2^{\large 5(\color{#b6f}1^{\LARGE k})}}\equiv\, \color{#c00}{32}\ \ \ \ \ \ \ \ \ \ \ {\rm by}\ \ \ \color{#b6f}{2031^{\large 2}}\!\equiv 11^{\large 2}\equiv\color{#b6f} 1\!\!\!\pmod{\!20}\\ \end{align} $

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Example 2 from here (further examples are here and here).

It's a mechanical calculation using mod order reduction to reduce $\rm\color{#0a0}{expo}\color{darkorange}{nents}$ mod Euler $\phi,\,$ and also that: $\,\ n\bmod 77 = 7\left[\dfrac{\color{#c00}n}7\bmod 11\right]\,$ if $\,7\:\!|\:\!n\ $ [mod distributive law = $\rm\color{#90f}{MDL}$ = CRT] as below

$\bmod 77\!:\ 2016^{2017^{2018}}\!\! \equiv 7\overbrace{\left[\dfrac{\bbox[5px,border:1px solid #c00]{\color{#c00}{2016^{\color{#0a0}{2017^{\color{darkorange}{2018}}}}}}}7\bmod 11\right]}^{\textstyle\bbox[5px,border:1px solid #c00]{\color{#c00}4}/7\equiv 4/(-4)\equiv -1}\!\! \equiv 7[-1]\ $ by $\rm\color{#90f}{MDL}$ & reduce $\bbox[1px,border:1px solid #c00]{\rm\color{#c00}{top}}$ as below

$\bmod 11\!:\ \bbox[5px,border:1px solid #c00]{\color{#c00}{2016^{\,\bbox[3px,border:1px solid #0a0]{\color{#0a0}{2017^{\color{darkorange}{2018}}}}}}}\equiv 3^{\,\bbox[1px,border:1px solid #0a0]{\color{#0a0}{-1}}}\equiv \dfrac{1}{3}\equiv \dfrac{12}3\equiv\bbox[5px,border:1px solid #c00]{\color{#c00}4}\, $ by reduce $\rm\bbox[1px,border:1px solid #0a0]{\color{#0a0}{expt}}$ mod $\phi(11)\!=\!10$ as below

$\bmod 10\!:\ \bbox[5px,border:1px solid #0a0]{\color{#0a0}{2017^{\,\bbox[1px,border:1px solid darkorange]{\color{darkorange}{2018}}}}}\equiv \color{#0a0}7^{\,\bbox[1px,border:1px solid darkorange]{\color{darkorange}2}}\equiv \bbox[5px,border:1px solid #0a0]{\color{#0a0}{\!\!\!-1}}\, $ by reduce $\bbox[1px,border:1px solid #c00]{\rm\color{darkorange}{expt}}$ mod $\phi(10)\!=\!4,\,$ i.e. $\,\bbox[3px,border:1px solid darkorange]{\color{darkorange}{2018}}\equiv_4 \bbox[2px,border:1px solid darkorange]{\color{darkorange}2}$

We computed the modular fractions $\frac{4}7,\ \frac{1}3\bmod 11$ via twiddling or inverse reciprocity.

Generally $\,\ {{\color{#c00}{\underbrace{(14\!+\!77i)}_{\!\!\!\!\!\!\!\!\!\!\!\!2016\ \equiv\ 14\pmod{\!77}}}^{\!\color{#0a0}{\large (\pm7+10j)^{\LARGE\color{darkorange}{2+4k}}}}}}\!\equiv -7\pmod{\!77}\ $ exactly as above.

Above we have $\ \color{#c00}{\frac{77}7\!=\! 11}\overset{\phi}\to \color{#0a0}{10} \overset{\phi}\to \color{darkorange}4\ $ are the mod expt reductions recursively applied in the tower.

Answer 2

It's a lot simpler than it looks. I shall call the number $N$.

You will know the residue modulo $10^3$, thus the last three digits, if you first get the residues modulo $2^3=8$ and modulo $5^3=125$.

$N$ is obviously a multiple of $8$, thus $N\equiv 0\bmod 8$. Which leaves $\bmod 125$.

The base $2032\equiv 32$. When this is raised to a power, the residue of this power depends only on the residue of the exponent $\bmod 100$ where $100$ is the Euler totient of $125$. But the exponent on $2032$ has the form

$2031^{10k}=(2030+1)^{10k}=(\text{binomial expansion})=100m+1$

So $N\equiv 32^1\equiv 32\bmod 125$. The only multiple of $8$ between $0$ and $999$ satisfying this result is $32$ itself so ... $N\equiv 32\bmod 1000$. Meaning the last three digits were there all along, the $\color{blue}{032}$ in the base $2032$!

Answer 3

Don't be scared. If it turns into a monster and eats you, run away after you throw stones at it. Don't run away before throwing stones just because it looks like a monster.

$2032^{monster}$ and $1000$ are relatively prime so we can't use Euler theorem but we can break it down with Chinese remainder theorem.

$2032^{monster} = 0 \pmod 8$ and so we just need to solve $2032^{monster} \pmod {125}$ and for that we can use Euler Theorem.

$\phi(125=5^3) = (5-1)*5^{3-1} = 100$.

So $2032^{monster} \equiv 32^{monster \% 100}$.

And $monster = 2031^{littlemonster}\equiv 31^{littlemonster}\pmod {100}$

$31$ and $100$ are relatively prime and $\phi(100)= 40$ so

$31^{littlemonster} \equiv 31^{littlemonster \% 40} \pmod {100}$.

$littlemonster = 2030^{smallmonster}$ but $5|2030$ and as $smallmonster > 2$ we know $8|2^{smallmonster}$ and $2^{smallmonster}|2030^{smallmonster}$.

So $littlemonster \equiv 0 \pmod {40}$.

$2031^{littlemonster} \equiv 31^0 \equiv 1 \pmod {100}$

So $2032^{monster} \equiv 32 \pmod {125}$

So $2032^{monster} \equiv 0 \pmod 8$ and $2032^{monster} \equiv 32 \pmod {125}$.

As $8|32$ we are done. $2032^{monster} \equiv 32 \pmod {1000}$.

and the last three digits are $032$.

Answer 4

By the Chinese Remainder Theorem, if you want to find what remainder a given number has when divided by $1000$, you can split that into 2 problems: Find the remainder$\mod 8$ and$\mod 125$. Obviously

$$z_0:=2032^{2031^{2030^{\dots^{2^{1}}}}} \equiv 0 \pmod 8$$

What remains to be found is $x_0 \in [0,124]$ in

$$z_0 \equiv x_0 \pmod {125}.$$

As $z_0$ is now coprime to $125$, you can apply Euler's theorem now. With

$$z_1:=2031^{2030^{\dots^{2^{1}}}}$$

and

$$\phi(125)=100$$ the new problem becomes to find $x_1 \in [0,99]$ in

$$z_1 \equiv x_1 \pmod {100}$$

and then use

$$32^{x_1} \equiv x_0 \pmod {125}$$

to find $x_0$.

So this reduced the original problem$\mod 1000$ to a smaller problem$\mod 100$.

Applying this reduction procedure a few more times (using the Chinese Remainder Theorem if appropriate), should result in congruences with smaller and smaller module that can in the end be solved (e.g $\mod 2$).

Then to solve the original problem you need to back-substitute the calculated $x_i$ to get $x_{i-1}$, just as outlined for $x_1,x_0$ above.