What is $\operatorname{ord}_{22}(5^6)$? 
Find $\operatorname{ord}_{22}(5^6)$.

So, basically we want to find:$$\operatorname*{arg\,min}_k 5^{6k} \equiv 1\pmod {22}$$
I found that $5^5 \equiv 1\pmod {22}$ so I know that for $k=5$ we have $5^{6k} \equiv 1 \pmod {22}$. Therefore, $\text{ord}(5^6) \le 5$.
I guess that I could proceed with the somewhat tedious checking:


*

*$k=4$: $(5^6)^4 \equiv (5^4)^6 \equiv 5^4 \equiv 9 \pmod {22}$

*$k=3$  $(5^6)^3 \equiv (5^3)^6 \equiv 5^3 \equiv 15 \pmod {22}$

*$k=2$  $(5^6)^2 \equiv (5^2)^6 \equiv 5^2 \equiv 3 \pmod {22}$

*$k=1$  $(5^6)^1 \equiv (5^1)^6 \equiv 5 \pmod {22}$


So $\text{ord}_{22}(5^6) = 5$.
Questions:


*

*Is that what I'm expected to do? Or is there a simpler way? (I had to use a calculator or to tediously calculate it myself)

*Why can we get rid of the exponent when we do modular arithmetic? (I actually used it along the proof) i.e. $$x \equiv a \pmod m \iff x^b \equiv a \pmod m$$ 


By the Chinese remainder theorem, it is enough to find the order of $5^6$ $\!\!\pmod{2}$ (which is obviously $1$) and the order of $5^6\equiv5$ $\!\!\pmod{11}$. Since $5$ is a quadratic residue $\!\!\pmod{11}$, due to
$$\left(\frac{5}{11}\right)=\left(\frac{11}{5}\right)=\left(\frac{1}{5}\right)=1$$
the order of $5\pmod{11}$ is five and that is also the order of $5^6\pmod{22}$.
