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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.

KKZiomek
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2 Answers2

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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 $$N\equiv 6\mod 10$$

Peter
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    But... What if I have tetration $^66$? How can I apply that to it? – KKZiomek Feb 20 '17 at 21:09
  • Yes, the proof is even easier. Again a power tower of sixes is always even and because of base $6$, we know immediately that it is congruent $1$ modulo $5$, so any power tower of sixes has end digit $6$ – Peter Feb 20 '17 at 21:42
  • The situation for a power tower of $3's$ is slightly more complicated. This time, we have an odd number. The exponent is congruent $3$ modulo $4$ (if we have at least two threes), so the power tower is congruent $3^3\equiv 2$ modulo $5$, hence the end digit is $7$. – Peter Feb 20 '17 at 21:46
  • In general, it is very difficult to find out the remainder of a large power tower modulo a given number, but usually the remainder quickly becomes stationary. This is because we can (assuming that the base is coprime to the given number) reduce the first exponent modulo $\phi(n)$, the second modulo $\phi(\phi(n))$ and so on. At some point the iterated totient function is $1$. At this point, the remainder does not change (but the stationary point often is reached earlier) – Peter Feb 20 '17 at 21:49
  • In Wikipedia , you can find the last $500$ (if I remember right) digits of Graham's number. Although it is hopeless to calculate this number (or even the first few digits) , it is possible to calculate the last digits. Graham's number is a power tower of threes, but the number of threes is completely impossible to comprehend. Google for Graham's number, it is fascinating! – Peter Feb 20 '17 at 21:53
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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.

Joffan
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