Which are the two last digits of $a_n$? We have the sequence $(a_n)$ with $a_1=3$ and the recursive formula:
$$a_{n+1}=3^{a_n} , \forall n$$
Which are the two last digits of $a_n$ ?
How can I find them? I have to find $a_n \pmod {100}$ ,right?So,do I have to find the formula of $a_n$ ?
Could I use the Eulers'Theorem?
We would have:
$$(3,100)=1$$
$$3^{\phi(100)} \equiv 1 \pmod{100} \Rightarrow 3^{40} \equiv 1 \pmod {100}$$
But...how can I use this?
 A: You noticed that $\varphi(100)=40$ hence $3^{40}=1\pmod{100}$. It happens that $3^{20}=1\pmod{100}$ hence, for every integers $(k,\color{blue}{j})$,
$$
3^{20\cdot k+\color{blue}{j}}=3^\color{blue}{j}\pmod{100}.
$$
This remark allows to compute recursively the congruence of $a_n$ modulo $100$ without computing $a_n$ itself.
For example, $a_1=3$ hence $a_2=27$ hence the remark above yields $$a_3=3^{27}=3^7=2187=\color{red}{87}\pmod{100}.$$ Using the same remark once again, one sees that $3^{\color{red}{87}}=3^7=\color{red}{87}\pmod{100}$ hence, for every $n\geqslant3$, $$a_n=\color{red}{87}\pmod{100}.$$
A: As for integer $\displaystyle n>0, a_n\equiv-1\pmod4\equiv3,a_n-1\equiv2\pmod4\implies\frac{a_n-1}2\equiv1\pmod2$
$$a_{n+1}=3^{a_n}=3\left(10-1\right)^{\frac{a_n-1}2}\equiv3\left(-1+\binom{\frac{a_n-1}2}110\right)\pmod{100}$$
$$\implies a_{n+1}\equiv15a_n-18\pmod{100}$$
Now let us find the period of $\displaystyle15\pmod{100}$ 
which will be same as that of $\displaystyle\frac{15}5\pmod{\frac{100}5}$ i.e., the period of $\displaystyle3\pmod{20}$
$\displaystyle3^1\equiv3,3^2\equiv9,3^3=27\equiv7,3^4=81\equiv1\pmod{20}$ or usingCarmichael Function $\displaystyle\lambda(20)=4$
$\displaystyle\implies a_n$ will have a period $4\pmod{100}$ or a divisor of $4$
Finally,
$$a_1=3$$
$a_2=3^3=27$ also $\equiv15a_1-18\pmod{100}\equiv15\cdot3-18\equiv27$
$$a_3\equiv15a_2-18\pmod{100}\equiv?$$
$$a_4\equiv15a_3-18\pmod{100}\equiv?$$
