how to compute the last 2 digits of $3^{3^{3^{3}}}$ to n times? Input $n$, output the last $2$ digits of the result.
n=1 03 3=3  
n=2 27 3^3=27  
n=3 87 3^27=7625597484987
n=4 ?? 3^7625597484987=??

Sorry guys, the formula given is $T_n=3^{T_{n-1}}$, I have updated the example.
I was asked this question during an interview, but I didn't have any clue. (The interviewer kind of gave me a hint that for $n>10$, the last $10$ digits of all results would be the same?)
 A: Notice $$3^{100} = 515377520732011331036461129765621272702107522001 \equiv 1 \pmod{100}$$
If we define $p_n$ such that $p_1 = 3$ and $p_n = 3^{p_{n-1}}$ recursively and
split $p_n$ as $100 q_n + r_n$ where $q_n, r_n \in \mathbb{Z}$, $0 \le r_n < 100$, we have
$$p_n = 3^{p_{n-1}} = 3^{100 q_{n-1} + r_{n-1}} \equiv 1^{q_{n-1}}3^{r_n} \equiv 3^{r_{n-1}} \pmod{100} \\ \implies r_n \equiv 3^{r_{n-1}} \pmod{100}$$
This means to obtain $r_n$, the last two digit of $p_n$, we just need to start with
$r_1 = 3$ and repeat iterate it. We find 
$$\begin{align}
r_1 &= 3\\
r_2 &\equiv 3^3 = 27 \pmod{100}\\
r_3 &\equiv 3^{3^3} = 7625597484987 \equiv 87 \pmod{100}
\end{align}$$
Notice 
$$3^{87} = 323257909929174534292273980721360271853387 \equiv 87 \pmod{100}$$
So after the third (not fourth) iteration, we have $r_n = 87$ for all $n \ge 3$. 
A: These are the last 2 digits of $3^m$ for $m\in\{0..19\}$
01,03,09,27,81,43,29,87,61,83,49,47,41,23,69,07,21,63,89,67
Let's call this A: A[0]=01, A[1]=03, A[19]=67...

These numbers are repeated for every 20 numbers. i.e. $3^{20+x}$ mod 100=$3^x$ mod 100
Your sequence of numbers simplified is: $3^{3^{n-1}}$. 
For example for n=4: $a_4=((3^3)^3)^3=(3^3)^{3*3}=3^{3*3*3}$ using that $(x^a)^b=x^{ab}$
So, you just have to find the exponent modulo 20. i.e $3^{n-1}$ mod 20
These are the remainder (modulo 20) of $3^m$ for $m\in\{0..3\}$
1,3,9,7

These numbers are repeated for every 4 numbers. i.e. $3^{4+x}$ mod 20=$3^x$ mod 20
So what you need is: 


*

*A[1]=03 (=B[1])

*A[3]=27 (=B[2])

*A[9]=83 (=B[3])

*A[7]=87 (=B[0])


These numbers are repeated. So in order to find an answer for a given n, you will have to compute n mod 4 and then get B[n mod 4].
