Calculating $3/10$ in $\mathbb{Z}_{13}$

Question

I'm trying to calculate $\frac{3}{10}$,working in $\mathbb{Z}_{13}$. Is this the correct approach?

Let $x=\frac{3}{10} \iff 10x \equiv 3 \bmod 13 \iff 10x-3=13k \iff 10x=13k+3$ for some $k \in \mathbb{Z}$ $\iff 10x \in \{16,29,42,55,68,81,94,107,120,133,146,159\}$, and the only integer solution to this is $x=9$, therefore $3/10=9$ in $\mathbb{Z}_{13}$.

Where have I gone wrong?

Answer 1

Note that $10\times 4 = 40 \equiv 1$ mod $13$

Then multiply the original equation by $4$ to obtain $40x\equiv 12$ so that $x\equiv 12$

Answer 2

As $(10,13)=1$ using Fermat's Little Theorem, $10^{12}\equiv1\pmod{13}$

$$\implies10^{-1}\equiv10^{11}$$

Again $10\equiv-3,10^{11}\equiv(-3)^{11}\equiv-3^{11}$

and $3^3=27\equiv1\implies3^9\equiv1^3\equiv1$

$\implies-3^{11}\equiv-3^2\equiv-9\equiv4$

Answer 3

This does work, but it's a little awkward to introduce decimals here.

Instead, note that since $$10 \cdot 4 = 40 \equiv 1 \bmod 13,$$ in $\mathbb{Z}_{13}$ we have $10^{-1} = 4$ and thus $$3 \cdot (10^{-1}) = 3 \cdot 4 = 12.$$

Answer 4

Since $10\equiv -3 \mod 13$, we have that $\dfrac{3}{10}\equiv\dfrac{3}{-3}=-1\equiv 12\mod 13$.