The digit at the hundred's place of $33^{33}$ I would want to know how to start with the question. And if you get hung up somewhere there's the answer it's $5$.
Any help is appreciated thanks,
My approach was to look at the factors to somehow crack the nut. But still in vain. Any help or tip or approach is alright as I didn't get a clue but for that. Yeah and also do suggest a category as well if number systems isn't appropriate.
 A: We just have to compute $33^{33}\pmod{1000}$, hence $33^{33}\pmod{8}$ and $33^{33}\pmod{125}$.
$$33^{33}\equiv 33^{1}\equiv 1\pmod{8}\tag{1}$$
is easy to find, and 
$$ 33^{33}= 33\cdot 33^{32} = 33\cdot ((((33^2)^2)^2)^2)^2\tag{2}$$
gives:
$$ 33^{33} \equiv 13\pmod{125}, \tag{3}$$
hence, by the Chinese theorem:
$$ 33^{33}\equiv \color{red}{5}13\pmod{1000}.\tag{4} $$
A: Here goes, to see how quickly it can be done. Jack's extraction of a factor $8$ from the modulus is helpful.
Let's work with "the last three digits", using $\equiv$ when the thousands are dropped, but allow negative numbers if that simplifies things.
$$33^2=1089$$
$$33^4\equiv89^2=(90-1)^2=7921\equiv-79$$
$$33^8\equiv (-79)^2=(80-1)^2=6241\equiv 241$$
It is here that being able to work mod $125$ would simplify things markedly, because you would have $33^{16}\equiv (-9)^2=81$ and $33^{32}\equiv 81^2=6561\equiv 61$ (modulo $125$) but I didn't spot that. So I'd do $$33^{16}\equiv 241^2=(250-9)^2=62500-4500+81\equiv81$$ which comes to the same thing and $$33^{32}\equiv 81^2=6561\equiv 561$$
$$561\times 33 =11 \times 1683 = \dots 513$$
There isn't a lot of thinking time, but done confidently and quickly, there is nothing there to take lots of time - and it helps to know two digit squares.
