$*^{-1}(mod 77) using Euler theorem.$

i. let $x = 8^{-1} (mod \ 77)$

Multiply both side by 8

$8x = 1(mod \ 77)$

$ 77 = 7 \times 11 $

$ \therefore $ It is a prime number.

Then by using Euler function,

$ \begin{align} \phi(P) &= P-1 \\ \phi (7) &= 7-1 = 6 \\ \phi (11) &= 11-1 = 10 \end{align} $

if m and n are co-prime integer

$\begin{align} \phi(mn) &= \phi (m) \phi (n) \\ \phi(77) &= \phi (7) \phi (11) \\ &= 6 \times 10 \\ &= 60 \end{align}$

using Eulers theorem, solution is given by

$ x = a^{\phi_({m)}-1} b (mod \ m) $

here a= 8, b=1 and m=77

$ x = 8^{60-1}\times 1 (mod \ 77) $

$ x = 8^59 \times (mod \ 77) - - - - - 1 $

Now,

$ \begin{align} 8^10 & = 77 \times 13944699 +1 \space and \space 8^8 = 77 \times 217886 - 6 \\ (8^10)^5 & = 1^5 (mod 77) \space and \space 8^8 \times 8 = -6 \times 8 (mod 77) \\ 8 ^50 & = 1(mod77) \space and \space 8^9 = -48 (mod77) \\ 8 ^50 & = 1(mod77) \space and \space 8^9 = 29 (mod77) \\ 8 ^50 \times 8^9 & = 1 \times 29 (mod77) \\ 8^59 & = 29 (mod 77) - ---- - - - 2 \end{align}$

using equation 1 and 2

$\begin{align} x &= 8^{-1} (mod 77) \\ &= 29 \end{align}$