<s>[INST] 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, $13, 43, 61, 101,103$.
 [/INST] In addition to $p=121-60=61$ we can use the extended Euclidean algorithm to obtain $5^{-1}=-12$, $6^{-1}=-10$ and $11^{-1}=-11$ in $\mathbb{Z}/61 \mathbb{Z}$. Hence
$$
\frac{1}{5}+\frac{1}{6}=-12-10=-22=2\cdot (-11)=\frac{2}{11}.
$$
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