What is the difference between $\mathbb Z_n$ and $\mathbb Z / n \mathbb Z$? I'm having trouble telling these apart. As far as I was aware $\mathbb Z /n \mathbb Z$ is the quotient ring where all multiples of $n$ will be zero - i.e. everything is done modulo $n$. But then what is $\mathbb Z_n$? Is it the set where every element in $\mathbb Z$ is multipled by $n$?

This answer expands Bill Dubuque's comment:
Define $\mathbb{Z_n}$ as the quotient set of the congruence relation defined on $\mathbb{Z}$ as the following:
$$a \equiv b \pmod n \iff \exists q \in \mathbb{Z}: nb = a-b \iff n \mid a-b$$
Prove that it's a congruence relation. If we show each equivalence class as $[x]$ show that the following operations are well-defined:
$$ [a] \oplus [b] = [a+b] $$
$$ [a] \odot [b] = [a \cdot b]$$
$(\mathbb{Z_n}, \oplus, \odot)$ forms a ring, now consider the mapping $\varphi: \mathbb{Z} \to \mathbb{Z_n}$ given by $\varphi(n) = [n]$. Since this is a canonical mapping that sends an element in $\mathbb{Z}$ to its equivalence class it is surjective.
The way we've defined our binary operations on $\mathbb{Z_n}$ show that $\varphi$ is a ring homomorphism.
Now find the kernel. Show that $\ker{\varphi} = n\mathbb{Z}$.Then by applying the first isomorphism theorem we get:
$$\frac{\mathbb{Z}}{n\mathbb{Z}} \cong \mathbb{Z_n}$$
Therefore both rings are algebraically identical.
A better construction is to look at all this from the point of view of ideal theory:
$I=n\mathbb{Z}$ is an ideal of $\mathbb{Z}$ for any $n \in \mathbb{N}$. Now the quotient ring $R/I$ gives $\mathbb{Z_n}$
