Prove $n\in \mathbb{Z}^+$ is divisible by $d \in \mathbb{N}$ such that $n \equiv 0\pmod d$

Question

I'm not sure how to prove $n\in \mathbb{Z}^+$ is divisible by $d \in \mathbb{N}$ such that $n \equiv o \pmod d$

I get that I have to prove the cases of (1) if $d\mid n$ then $n\equiv 0 \pmod d$, and (2) if $n\equiv 0 \pmod d$, then $d\mid n$. But I don't know how to prove them. So, what's the proof for case 1? I think I can figure out 2 from there.

Answer 1

One says that $n \pmod{d} = r$ when $ n = q d + r $ for some integer $q$ and $r \in \{0,\dots, d-1\}$. Thus, $r=0$ means $ n = q d $, which is just another way of saying that $d$ divides $n$.