Inverse of $ 3\cdot\frac 57 \mod 7 $ Stuck with the following question: $ 3\cdot\frac 57 \mod 7 $
This can be broken down to $ 3\cdot 5\cdot \frac 17 \mod 7 $ however not sure about the module inverse of $7x = 1 \mod 7$
As per my knowledge we cannot have a inverse, as anything multiplied with $7$ will yield a $0$ when we do a $\mod 7$.

If we want to find the inverse of $3\cdot 5\cdot \dfrac 1 7 \bmod 7$, we are looking for an $x$ such that $3\cdot 5\cdot \frac 17\cdot x \equiv 1 \pmod 7$. So, we solve as follows
$$
\begin{align*}
3\cdot 5\cdot \frac 17\cdot x &\equiv 1 \bmod 7\\
3\cdot 5\cdot x &\equiv 7 \bmod 7\\
15x &\equiv 0 \bmod 7
\end{align*}
$$
which means $x=0$, so clearly $3\cdot 5\cdot \dfrac 1 7$ does not have an inverse $\bmod 7$.
