How I can determine a generator of $\mathbb{Z}^{*}_{242}$. Is the first time that I studied this.
The first I do, was write $242=11^2*2$, then $\mathbb{Z}^{*}_{242}\simeq (\mathbb{Z}^{*}_{2})\times(\mathbb{Z}^{*}_{11^2})$ and now an element is generator if and only if has order $\phi(242)=242\left(1-\dfrac{1}{11}\right)\left(1-\dfrac{1}{2}\right)=110$, i.e., $a^{110}=1$ and $a^n\neq 1$ if $n<20$. Thanks!
you need to find a generator $\bmod 2$ and $\bmod 11$.
The generator $\bmod 2$ is cleary $1$.
To find a generator $\bmod 121$ we find one $\bmod 11$ first by trial and error.
$2,4,8,5,-1,-2,-4-8,-5,1$. Bingo! $2$ is a generator.
we now use this lemma to make it easier:
if $a$ is a generator $\bmod p$ then one of $a$ and $a+p$ has order $p-1$ and the other is a generator.
since $2^{10}-1=1027$ is not a multiple of $11$ we are done. $2$ is a generator $\bmod 11^2$.
So now we just need to solve the congruence equation $x\equiv 1 \bmod 2,x\equiv 2 \bmod 121$ which is clearly solved by $123\bmod 242$. We are done.
Most cases are not as easy as this one, but the ideas in the solution are useful most of the times.(something that we didn't need but always helps is that if $x$ is a generator $\bmod p^2$ then $x$ is a generator $\bmod p^k$ for all $k$, (when $p$ is an odd prime))