Inverse of $3$ in $\mathbb{Z}_7$ using Fermat's Theorem or its corollaries.

Question

What is $3^{-1}$ , the multiplicative inverse of $3$ in $\mathbb{Z}_7$. Use Fermat's Theorem or its collaries.

How do I make use of the Fermat's theorem to solve this? I know how to solve it using Linear Diophantine Equations and the EEA only

Answer 1

Hint: Fermat states $3^6\equiv 1 \pmod 7$, therefore $3\cdot 3^5 \equiv 1 \pmod 7.$ Can you continue?

Answer 2

Hint:

By Fermat's if $a\nmid p$

$a^{p-1}\equiv 1\pmod p$

So, $3\cdot 3^5=3^6\equiv 1\pmod 7$