<s>[INST] What is the tens digit of $3^{100}$? Is there a general formula to calculate the n-th digit of any big number?
 [/INST] We would like to calculate $3^{100}\pmod{100}$.  This will tell us the last two digits of $3^{100}$, which includes the tens digit.
We proceed as follows: $3^4 = 81$, so:
$$\begin{align}
3^8 & \equiv (81)^2 \equiv 61 \pmod{100} \\
3^{16} & \equiv (61)^2 \equiv 21\pmod{100} \\
3^{24} & \equiv 61\cdot 21 \equiv 81\pmod{100}\\
3^{48} & \equiv (81)^2 \equiv 61 \pmod{100} \\
3^{50} & \equiv 9\cdot 61 \equiv 49\pmod{100} \\
3^{100} & \equiv (49)^2\equiv 01 \pmod {100}
\end{align}$$
So the last two digits are 01.
In general it's very quick to calculate $a^b\pmod n$.  You calculate $a^{\lfloor b/2 \rfloor}\pmod n$ (using this method recursively if necessary) and square it, again mod $n$; if $b$ is odd you multiply the result by $a$ and you are done.
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