8 is Nice

Denote by $^n x$ the tetration operation, that is, $^n x=\underbrace{x^{x^{x^{.^{.}}}}}_{n\textrm{ times}}$ .

Lemma 1: $^x8\equiv 0\pmod{4}~,~x\geq 1$ .

Proof: Since $8$ is a multiple of $4$ , any arbitrary positive exponentiation of $8$ will also be divisible by $4$ according to the rules of modular arithmetic. Hence, the lemma is established and proved.

Then, our problem becomes:

$\textrm{Find the value of the following: }~~^88\bmod 100.$

Note that $\gcd(8,100)\neq 1$ . So, we can't use Euler's Theorem directly. Instead, we first find the residues of $^88$ modulo $25$ and $4$ .

Using Lemma 1 , we have $^88\bmod 4=0$ . Now, for modulo $25$ , note that $\gcd(25,8)=1$ . So, we use Euler's theorem with $\phi(25)=20$ to get,

$^88\equiv \begin{cases}0\pmod{4}\\ \left(8^{^78~\bmod~20}\right)\pmod{25}\end{cases}$

Now, again, note that $\gcd(20,8)\neq 1$ . So, we find the residues of $^78$ modulo $4$ and $5$ . As before, by Lemma 1 , we have $^78\bmod 4=0$ . Now, for modulo $5$ , we use Fermat's Little Theorem and Lemma 1 to get,

$^78\equiv\begin{cases}0\pmod{4}\\ \left(8^{^68~\bmod~4}\right)\equiv 8^0\equiv 1\pmod{5}\end{cases}\implies ^78\equiv\begin{cases}0\pmod{4}\\1\pmod{5}\end{cases}$

Combine these results using Chinese Remainder theorem to get $^78\bmod 20=16$ . Now, using this value, our first congruence system becomes,

$^88\equiv \begin{cases}0\pmod{4}\\ 8^{16}\equiv (64)^8\equiv 14^8\equiv (-4)^4\equiv 6\pmod{25}\end{cases}\implies ^88\equiv \begin{cases}0\pmod{4}\\ 6\pmod{25}\end{cases}$

Again, combine the results obtained using Chinese Remainder Theorem to get,

$\boxed{^88\equiv 56\pmod{100}}$

We will study numbers of the form $8^{8^{8n}}$ , where $n$ is a positive integer. This format includes $^m8$ for $m\geq3$ .

Now $8^{8n}=16^{6n}\equiv16\pmod{20}$ since they are congruent both mod 5 and mod 4. Then $8^{8^{8n}}\equiv8^{16}=2^{48}\equiv2^8 \pmod{25}$ because $20=\phi(25)$ . Since all terms are divisible by 4, we have in effect $8^{8^{8n}}\equiv2^8=256\equiv56 \pmod{100}$ .

In fact, $^3 8,^4 8,^5 8\ldots$ all give the same last two digits. We'll prove it for $^n 8$ , $n\ge 3$ .

$100=4\cdot 25$ . Obviously $^n 8=4k$ for some $k\in\mathbb Z$ .

By Euler's Theorem, $^n 8\equiv 8^{^{n-1} 8\pmod {20}}\equiv 2^{3(^{n-1}8)\pmod{20}}\pmod {25}$

$^{n-1} 8\equiv 16\pmod {20}$ , since $((8^2)^2)^2\equiv ((-4)^2)^2$ $\equiv 16^2\equiv 16\pmod {20}$ and this pattern of squaring $8$ modulo $20$ continues (remember - $^{n-2} 8$ is a power of $2$ ).

$4k\equiv 2^{3(^{n-1} 8)\pmod {20}}\equiv 2^{48\pmod{20}}\equiv 2^{8}$

$\stackrel{: 4}\Rightarrow\, k\equiv 2^6\equiv 64\equiv 14\pmod {25}$

$^8 8=4(25l+14)=100l+56$ for some $l\in\mathbb Z$ .

I got lucky hahaha I used a calculator and found that the last numbers are 2, 4, 6, 8 And the last second numbers are almost any number I tried 78 then 56 OMG ITS CORRECT

. Obviously for some . By Euler's Theorem, , since and this pattern of squaring modulo continues (remember - is a power of ). for som

very nice solution,