The order of $33\pmod{83}$ 
Find $\gamma_{83}(33)$

My attempt:
stupid approach:
$33^1\equiv 33 \pmod{83}$
$33^2\equiv 10 \pmod{83}$
$33^3\equiv 81 \pmod{83}$
$33^4\equiv 17 \pmod{83}$
$33^5\equiv 63 \pmod{83}$
$\vdots$
$33^{41}\equiv 1 \pmod{83}$
I am sure that there is a better approach
 A: As $\phi(83)=82=2\cdot41,$ and the order must divide $82,$
it must be one of $1,2,41,82$
So, as ord$_{83}33>2,$ show that $33$ is a Quadratic Residue $\pmod{83}$ 
A: In practical terms, your "stupid approach" is not so stupid if you are calculating in a spreadsheet and can use a simple formula that builds from row to row, taking the modulus each time. For numbers much larger than $83$ you could improve the technique, especially with a prime number $p$ where $p-1$ is divisible by a larger power of $2$.
The binary method of exponentiation that André mentions in his comment reduces the number of calculations, with the knowledge from Fermat's little theorem that $33^{82} \equiv 1 \bmod 83$ and thus that $\gamma_{83}(33)$ must divide $82$, giving $\gamma_{83}(33) \in \{2,41,82\}$.
Then, calculating $\bmod 83$,  you have:
$$
\begin{align} 33^{1} &&\equiv 33 \\
33^{2} &= 1089 &\equiv 10 &\quad \text{note: }\neq 1\\
33^{4} &\equiv 10^2 = 100 &\equiv 17 \\
33^{8} &\equiv 17^2 = 289 &\equiv 40 \\
33^{16} &\equiv 40^2 = 1600 &\equiv 23 \\
33^{32} &\equiv 23^2 = 529 &\equiv 31 \\
\\
41 &=32+8+1\\
33^{41} &\equiv 31\times 40 \times 33 = 1023 \times 40 \\
&\equiv 27\times 40 = 1080 &\equiv 1 \\
\end{align}
$$
giving your result. 
Until we calculated that last number, of course, we didn't know whether $\gamma_{83}(33)=41$ or $\gamma_{83}(33)=82$. However our only two options were $33^{41} \equiv 1 \bmod 83$ or $33^{41} \equiv 82 \bmod 83$, in view of Fermat's little theorem.
