The Exponentorial of $2018$

Relevant wiki: Euler's Theorem

Let $N=2018¡$ . We need to find $N \bmod 100$ . Since 2018 and 100 are not coprime integers, we have to consider the factors 4 and 25 of 100 separately using Chinese remainder theorem .

Consider factor 4 : $N \equiv 0 \text{ (mod 4)}$ .

Consider factor 25 :

$\large \begin{aligned} N & \equiv 2018^{\color{#3D99F6}2017^a \bmod \phi (25)} \text{ (mod 25)} & \small \color{#3D99F6} \text{Since }\gcd (2018, 25) = 1 \text{, Euler's theorem applies.} \\ & \equiv 2018^{\color{#3D99F6}2017^a \bmod 20} \text{ (mod 25)} & \small \color{#3D99F6} \text{Euler's totient function }\phi (25) = 20 \\ & \equiv 2018^{2017^{\color{#3D99F6} 2016^b \bmod \phi(20)} \bmod 20} \text{ (mod 25)} & \small \color{#3D99F6} \text{Again }\gcd (2017, 20) = 1 \text{, Euler's theorem applies.} \\ & \equiv 2018^{2017^{\color{#3D99F6} 2016^b \bmod 8} \bmod 20} \text{ (mod 25)} & \small \color{#3D99F6} \text{Euler's totient function }\phi (20) = 8 \\ & \equiv 2018^{2017^{\color{#3D99F6} 0} \bmod 20} \text{ (mod 25)} \\ & \equiv 2018^1 \text{ (mod 25)} \\ & \equiv 18 \text{ (mod 25)} \end{aligned}$

Therefore, we have:

$\begin{aligned} \implies N & \equiv 25n + 18 & \small \color{#3D99F6} \text{where }n \text{ is an integer.} \\ 25n + 18 & \equiv 0 \text{ (mod 4)} \\ n + 2 & \equiv 0 \text{ (mod 4)} \\ \implies n & \equiv 2 \text{ (mod 4)} \\ \implies N & \equiv 25(2) + 18 \text{ (mod 100)} \\ & \equiv \boxed{68} \text{ (mod 100)} \end{aligned}$

Last 2 digits of powers of 2018 will the same as for 18 and there will be a cycle (hopefully it's short.)

$18^{1}$ ends in 18

$18^{2}$ ends in 24

$18^{3}$ ends in 32

$18^{4}$ ends in 76

$18^{5}$ ends in 68

$18^{6}$ ends in 24 (same as $18^{2}$ so the cycle is length 4.)

So where do powers of 2017 get us mod 4?

2017 is 1 mod 4. Nice! 1*1=1 so all powers of 2017 are 1 mod 4.

We just need the right spot on the list. It can't be the 18 because that one is unique. Advance 4 spaces to 68.

We want to find a $x$ such that $x\equiv2018¡\pmod{100}$ .

This is equivalent to finding the unique $0≤x<100$ such that $x\equiv2018¡\pmod{4}$ $x\equiv2018¡\pmod{25}$

The first equivalence is easy to solve, as clearly, $2018¡$ is divisible by $4$ , so $x\equiv0\pmod{4}$ .

Consider the second equivalence: $x\equiv2018¡\equiv{2018}^{{2017}^{{2016}^{.^{.^{.^{2^1}}}}}}\equiv{18}^{{2017}^{{2016}^{.^{.^{.^{2^1}}}}}}\pmod{25}$

Note that $2017\equiv1\pmod{4}$ and ${18}^4\equiv{(-7)}^4\equiv{(-1)}^2\equiv1\pmod{25}$ . Thus, for a certain integer $k$ , $x\equiv{18}^{4k+1}\equiv{18}^{4k}\cdot{18}^{1}\equiv18\pmod{25}$

Thus, we're finding $x\equiv0\pmod{4}$ $x\equiv18\pmod{25}$

By the Chinese Remainder Theorem, such a $0≤x<100$ exists and is unique. By trial-and-error, we find that $x=68$ .

Reduce!

Take note that $10^{n} \equiv 0\ (\text{mod} 100)$ if $n$ is an integer greater than 2.

Similar to the fact that $k! \equiv 0\ (\text{mod} 100)$ if $k$ is greater than or equal to 10, the entire expression, when working from the top to the bottom, hits reset when the expression hits every multiple of ten.

For example.

$10^{9^{\text{etc}}} \equiv 100^{99^{\text{etc}}} \equiv 1000^{999^{\text{etc}}} \equiv 0 (\text{mod} 100)$

This is true because $11 ^ n \equiv 0 (\text{mod} 100)$ for every n that is multiple by then, therefore the reset works every time and does not get broken.

Thus,

$2018^{2017^{\text{etc}}} \equiv 18^{17^{\text{etc}}} (\text{mod} 100)$ .

Now, it should be easy to solve this.

$18^1 \equiv 18 (\text{mod} 100)$

$18^2 \equiv 24 (\text{mod} 100)$

$18^3 \equiv 32 (\text{mod} 100)$

$18^4 \equiv 76 (\text{mod} 100)$

$18^5 \equiv 68 (\text{mod} 100)$

$18^6 \equiv 24 (\text{mod} 100)$

Take note that the cycle length is four but the starting point of the cycle is 24.

Now let us figure out what is $17^{16^{\text{etc}}} (\text{mod} 4)$ . Luckily, $17 \equiv 1 (\text{mod} 4)$ , thus, $17^{16^{\text{etc}}} \equiv 1 (\text{mod} 4)$

Since the starting point of the cycle is the second number, we subtract from 2 from 1 and we get $1 - 2 = -1 \equiv 3 (\text{mod} 4)$ .

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Therefore the remainder is the 3rd position from the starting point 24.

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And we now have our answer. $\boxed{2018^{2017^{\text{etc}}} \equiv 68 \ (\text{mod} 100). }$

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P.S. Forgive me for the incomplete nested exponents for I do not know how to stack more than two exponents. But you get the idea, right? Give me advice in the comment section if you have a solution to my naivete.)