arithmetic - Number in tens place

A number in tens place in result of $4^{2015} \cdot 9^{2016}$ is?

Obviously without using calculator, though I doubt it could count with those high numbers.

By tens place I mean, for example if you have number $2451$, the number in tens place here is $5$.

I know the answer, but I don't know how to get it, so if you got any ideas, please share. :)

Answer

$$4^{2015}\cdot9^{2016} = 9\cdot36^{2015}$$

So look at $9\cdot36^n$ for $n=1,2,3,\ldots$:
$$\begin{align} 9\cdot36^1 & = 324 \\ & = \cdots24 \\[6pt] 9\cdot36^2 & = \cdots64 \\ 9\cdot36^3 & = \cdots04 \\ 9\cdot36^4 & = \cdots44 \\ 9\cdot36^5 & = \cdots84 \\ 9\cdot36^6 & = \cdots24 \longleftarrow\text{Now we're back to where we were when }n=1, \\ & \phantom{=\cdots24 \longleftarrow\text{a}}\text{so it starts over again.} \end{align}$$
After five steps we return to where we started; thus at $n=1,6,11,16,21,26,\ldots$ we get $\cdots24$.

Of course, in order for this to make sense, you have to realize that when you multiply two numbers, the last two digits of the answer are determined by the last two digits of each of the numbers you're multiplying, and are not affected by any earlier digits. That is clear if you think about the algorithm for multiplication that you learned in elementary school, and you can also show it via some simple algebra.

So we get $24$ whenever the exponent is congruent mod $5$ to $1$, i.e. the last digit is $1$ or $6$. So $9\cdot36^{2016}=24$ and $9\cdot{2015}$ is one step before that, thus $84$.