<s>[INST] Solve $x^{5} \equiv 2$ mod $221\ $ [Taking modular $k$'th roots if unique] We know that $221 = 17*13$. So we can check if the system has roots to both of those equations separately, which it does:
$x^{5} \equiv 2$ mod $13$ has the solution $6 + 13n$ and $x^{5} \equiv 2$ mod $17$ has the solution $15 + 17n$. 
I got these numbers from wolfram, I have no idea how to solve this problem WITHOUT a calculator. And even after finding these numbers. How would one obtain a solution modulo $221$? I was thinking the Chinese Remainder Theorem but I am under the assumption that CRT only applies to problems with powers of $x$ which are $1$.
Thanks.
 [/INST] We work all the time modulo $221$ from now on tacitly. Since $2$ is relatively prime to $221$, a/the solution $x$ of  $x^5=2$ (modulo $221$ - i write equivalences modulo $221$ as equalities from now on) is also relatively prime to $221$. The Euler indicator function of $221$ is $\varphi(221)=\varphi(17\cdot13)=\varphi(17)\cdot\varphi(13)= 16\cdot 12=192$.
The inverse of $5$ modulo $192$ is $77$, explicitly
$5\cdot 77=385=2\cdot192+1$. From $x^5=2$ we get then (implications in one direction)
$$
\begin{aligned}
x &= (x^{192})^2\cdot x
&&\text{ Euler, since $x^{192}=x^{\varphi(221)}=1$ modulo $221$}
\\
&=x^{385}=(x^5)^{77}=2^{77}=(2^8)^9\cdot 2^5
&&\text{ ($2^8$ is "close" to $221$)}
\\
&=256^9\cdot 32 = 35^9\cdot 32=(35^3)^3\cdot 32
\\
&=42875^3\cdot 32
=1^3\cdot 32=\color{blue}{\boxed{32}}\ .
\end{aligned}
$$
Check (for the other direction):
$$
32^5
=
(2^5)^5=2^{25}
=2^{24}\cdot 2
=(2^8)^3\cdot 2=42875^3\cdot 2=1^3\cdot2=2\ .
$$
</s>