(i) Let $p$ be prime and $p \not \mid n$. Prove that the solutions of $nx \equiv a \pmod p$ is
given by
$$ x \equiv n^{p−2}a \pmod p $$
(ii) Solve the linear congruence
$$ 10x \equiv 11 \pmod {17}$$
(i) Since $p$ is prime, and does not divide $n$, we can use FlT
$ n^{p-1} \equiv 1 \pmod p $
Multiplying both sides by $a$
$ n^{p-1}a \equiv a \pmod p $
That is
$ n ( n^{p-2}a) \equiv a \pmod p $
Comparing this to $nx \equiv a \pmod p$ tells us $x \equiv n^{p-2}a \pmod p$.
(ii) Using the result from (i) we have
$ x \equiv 10^{15} \times 11 \pmod {17} $
Using $10^5 \equiv 100000 \equiv 6 \pmod {17}$, we have
$ x \equiv ({10^ 5})^3 \times 11 \equiv 6^3 \times 11 \equiv 2376 \equiv 13 \pmod {17} $
That is,
$ x \equiv 13 \pmod {17} $