<commit_before>\begin{enumerate}

\item First find $27^{-1} \in \mathbb{Z}_{47}$. In order to do this we need $gcd(27,47)$:
\begin{eqnarray*}
    gcd(27,47) &=& gcd(20,27) \\
    &=& gcd(7,20) \\
    &=& gcd(6,7) \\
    &=& gcd(1,6) \\
    &=& 1
\end{eqnarray*}
Now with the Extended division algorithm we have that:
\begin{eqnarray*}
    gcd(27,47) = 1 = 7 \cdot 27 - 4 \cdot 47
\end{eqnarray*}
Therefore the inverse of $27^{-1}$ in $\mathbb{Z}_{47}$ is $7$.
Thus $27^{-1}$ mod $47 = 7$ mod $47 = 7$ 

\end{enumerate}<commit_msg>first answer for 1b<commit_after>\begin{enumerate}

\item First find $27^{-1} \in \mathbb{Z}_{47}$. In order to do this we need $gcd(27,47)$:
\begin{eqnarray*}
    gcd(27,47) &=& gcd(20,27) \\
    &=& gcd(7,20) \\
    &=& gcd(6,7) \\
    &=& gcd(1,6) \\
    &=& 1
\end{eqnarray*}
Now with the Extended division algorithm we have that:
\begin{eqnarray*}
    gcd(27,47) = 1 = 7 \cdot 27 - 4 \cdot 47
\end{eqnarray*}
Therefore the inverse of $27^{-1}$ in $\mathbb{Z}_{47}$ is $7$.
Thus $27^{-1}$ mod $47 = 7$ mod $47 = 7$ 

\item Note that $135x \equiv 15 \textrm{ mod } 235 \implies 27x \equiv 3 \textrm{ mod } 47$ (cite slide 30). \\
  Now, \begin{eqnarray*}
    27x &=& 3 + 47n \textrm{ } (\forall n \in \mathbb{Z}) \\
    \implies x &=& 27^{-1} \cdot 3 + 47n \\
    \implies x &\equiv& 21 \textrm{ } (\textrm{mod } 47) \\
  \end{eqnarray*}

So the solutions of $135x \equiv 15 \textrm{ mod } 235$ are the integers of the form $21 + 47k \textrm{, } \forall k \in \mathbb{Z}$.

\end{enumerate}
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