Determine $71^{−1} \pmod {771}$.
By definition $x = 71^{-1} \pmod {771}$ is the unique solution to
$$ 71x \equiv 1 \pmod {771} $$
Here $g=\gcd(771,71)=1$ and so there is a unique solution.
The equivalent Diophantine equation is
$$ 71x + 771y = 1 $$
Let's apply the Euclidean Algorithm to 771 and 71.
$ 771 = 10(71) + 61 $
$ 71 = 1(61) + 10 $
$ 61 = 6(10) + 1 $
$ 10 = 10(1) + 0 $
And so
$ 1 = 61 - 6(10) = 61 - 6(71 - 61) = - 6(71) +7(61) $
$ 1 = 7(771 - 10(71)) - 6(71) = 771(7) + 71(-76)$
So $x=-76, y = 7$ is a solution to the Diophantine equation.
Noting that $-76 \equiv 695 \pmod {771}$, the unique solution to $ 71x \equiv 1 \pmod {771} $ is $695 \pmod {771}$.
And so the multiplicative inverse $7^{-1} \pmod{771}$ is $ 695 \pmod{771}$.