How can one find $c \in \mathbb{N}$, so that $c \cdot 11 = 23 \mod 103$?
I know that $a \cdot b \mod n = (a \mod n \cdot b \mod n) \mod n$.
Furthermore, Fermat's little theorem says that for all prime $p$ and if $p$ does not divide $a$
$a^{p-1} \equiv 1 \pmod p$
And I know that here all numbers $11, 23, \text { and } 103$ are prime numbers.
That still doesn't help me though...
On
By Gauss's algorithm $\!\bmod 103\!:\, \ c\equiv \dfrac{23}{11}\equiv \dfrac{9\cdot 23}{9\cdot 11}\equiv\dfrac{1}{-4}\equiv \dfrac{104}{-4}\equiv -26\equiv 77$
Fermat's little theorem provides a solution:
$$ c \cdot 11 \equiv 23 \bmod 103 \implies c \equiv c \cdot 11^{102} \equiv 23 \cdot 11^{101} \bmod 103 $$
Thus, $c = 23 \cdot 11^{101}$ works. It is not the smallest solution, but the question does not ask for that...