What is $3^{99} \pmod{100}$? I saw this post on how to solve $3^{123}\pmod{100}$ using Euler's Totient Theorem.
How about for $3^{99}\pmod{100}$? It seems more complicated because applying Euler's Totient Theorem gets us $3^{40}\equiv 1\pmod{100}$. This means $3^{80}\equiv 1 \pmod{100}$, which isn't enough, because we still need to find $3^{19}\pmod{100}$.
Now, when the terms are listed, a pattern does emerge.
$$\begin{array}{|c|c|c|c|}
\hline
3 & 9 & 27 & 81 \\ \hline
43 & 29 & 87 & 61 \\ \hline
83 & 49 & 47 & 41\\ \hline
23 & 69 & 07 & 21 \\ \hline
03 & 89 & 67 & 01 \\ \hline
\end{array}$$
And $3^{19}$ ends in $67$.
But how can I find this in other methods, besides bashing? Is there some sort of theorem that I can use?

You can do this using the Chinese remainder theorem and by calculating some powers.
$3^{99} = (3^3)^{33} \equiv 2^{33} \pmod{25} = (2^7)^4 \times 2^5 \equiv 3^4 \times 2^5 = 3^3 \times 3 \times 2^5 \equiv 3 \times 2^6 \equiv 17$.
$3^{99} \equiv (-1)^{99} \equiv 3 \pmod{4}$.
So we need the unique number satisfying those constraints in $[0, 100)$, which is $67$.

Your way comes out much the same: $3^{19}$ mod $4$ is $3$ again, and $3^{19}$ mod $25$ is $(3^3)^6 \times 3$ which is $2^6 \times 3$ which is again $17$.
