Determine $p$ for one to have $\displaystyle\frac{1}{5}+\frac{1}{6}=\frac{1+1}{5+6}=\frac{2}{11}$, in $\mathbb{Z_p}$. Determine $p$ for one to have $\displaystyle\frac{1}{5}+\frac{1}{6}=\frac{1+1}{5+6}=\frac{2}{11}$, in $\mathbb{Z_p}$.
I know that $p$ must, $13, 43, 61, 101,103$.

Just making sure what has to be done here, as mentioned above: we will multiply in the following way:
$$(11 \cdot 5 \cdot 6)\left(\frac{1}{5}+\frac{1}{6}\right)=\left((11 \cdot 5 \cdot 6)\frac{2}{11}\right)$$, in $\mathbb{Z_p}$
then we get $11\cdot6 + 11\cdot 5= 121 = 5 \cdot 6\cdot 2 = 60$, but this is only possible inside the $\mathbb{Z_{61}}$, observe that in this ring (because $p$ is prime), we have that $61 = 0 \in \mathbb{Z_{61}} $ and that $122 = 0$, so $121 = 60$.
