Prove that if $a=\,^xx$, for $x>2$, where $2\,|\,x$ and $10\nmid x$, the last digit of $a$ is $6$.

Question

First, explanation of some notations. $^wv$ is called tetration, which is a higher order of exponentiation. Useful link -> https://en.wikipedia.org/wiki/Tetration. $v\,|\,w$ means "$v$ divides $w$", which can be also written as $w\,(\text{mod}\,\,\, v)=0$.

Now we have:

$$a=\,^xx$$

I want to prove that for every even $x$ that is greater than $2$, but not divisible by $10$, the last digit of $a$ is $6$.

I checked $4,6,8,12,14,16,18$, and so far I couldn't see any contradictions. But how can I prove that this statement is true (or false)?

I don't even know how to start or how to approach this problem.

Answer 1

We can consider the expression $x^{\large (2k)^{n+1}} (k,n\in \Bbb Z^+)$. For $x$ as constrained, any number of this form will have a final digit of $6$. It's obvious that the given ${}^xx$ will fit this format.

We will have $x\equiv \{2,4,6,8\} \bmod 10$. Then certainly an expression of the form $ (2k)^{n+1}$ will be divisible by $4$, say $4m$. Since $\phi(5) = 4$ (and $5\nmid x$), $x^{4m}\equiv 1 \bmod 5$ and since $x \equiv 0 \bmod 2$ we have $x^{4m}\equiv 6 \bmod 10$ as required.

Answer 2

If the power tower $N:=2\uparrow 2\uparrow 2\uparrow...\uparrow2$ contains at least three twos, we have $$N\equiv 6\mod 10$$

Proof :

It is clear that $N$ is even. Because of $\phi(5)=4$, we have $N\equiv 2^0=1\mod 5$.

This implies $$N\equiv 6\mod 10$$