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$\begingroup$

My thoughts: this can be rewritten as $256^{256^{256^{...}}}$

As number of 4 is odd, this number eventually will be $(256^{256^{256^{...}}})^4$

Also, among all reminders which can leave $56^{k} \mod3$ only 16 fits, as $4^k \mod 3 \equiv 1$.

So it is last to digits of $16^4$ = 36

However, I don't feel confidence here.

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    Why would it be rewriteable as $256^{256^{\dots}}$? – 2017-01-24
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    Note that $a^{b^{c^d}} = a^{(b^{(c^d)})}$, note further that $4^{4^{4^4}}$ has approximately $8\cdot 10^{153}$ **digits** whereas $(4^4)^{(4^4)}\approx 3\cdot 10^{616}$ has only $617$ digits. Power towers are always evaluated top to bottom and the exponentiation operation ^ is not associative. – 2017-01-24
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    @JMoravitz, yes, I guess then I don't have any ideas. – 2017-01-24
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    You can notice that $4^k$ is periodic mod $100$ and that the period is 10,then you can look with 2012 4's $4^{4^{4^\cdots}}$ mod 10. – 2017-01-24
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    And so to continue, to find what a tower of $2012$ 4's is mod 10, notice that $4^k$ is periodic mod $10$ and the period is ____, so you can then look at a tower of $2011$ 4's mod ___ – 2017-01-24
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    @kingW3, I don't understand how to use this – 2017-01-24
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    Related: [Last Digits of a Tetration](http://math.stackexchange.com/questions/1215461/last-digits-of-a-tetration) ; [Last few digits of $n^{n^{n^{\cdot^{\cdot^{\cdot^n}}}}}$](http://math.stackexchange.com/questions/166083/last-few-digits-of-nnn-cdot-cdot-cdotn) – 2017-01-24

2 Answers 2

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$4^{4^{4^{\cdots}}} \bmod 100$ can be found by examining the successive steps up the tower.

$4^k \bmod 100$ will have a cycle length that divides $10$, since the Carmichael function $\lambda(100)=20$ guarantees that the cycle length of $2$ will divide $20$.

Considering $k= 4^m \bmod 10$, then, this will have a cycle length dividing $\lambda(10)=4$ (actually, $2$). And since we already know that $4 \mid m$, we can calculate that immediately as $k \equiv 6 \bmod 10$.

Then $4^{4^{4^{\cdots}}} \equiv 4^6 \equiv 56\cdot 16 \equiv \fbox{96} \bmod 100$

2

$4^k\equiv 0\pmod{4}$ for any $k>0$.

So you really want to figure out $4^k\pmod{25}$.

Now, $4^{10}\equiv 1 \pmod{25}$.

Now if $k=4^l$ for some even $l>0$, the $k\equiv 0\pmod{2}$ and $k\equiv (-1)^l=1\pmod{5}$, so $k\equiv 6\pmod{10}$ and thus $4^{k}\equiv 4^{6}\equiv 21\pmod{25}$.

Then $4^{4^{l}}\equiv 96\pmod{100}$ for all even $l>0$.