Group theory: Let $H=\{0,\pm 3, \pm 6, \pm9,\ldots\}$ Find all the left cosets of $H$ in $\Bbb Z$. I am having trouble understanding the following homework question,
Let $H=\{0,\pm 3, \pm6, \pm9,\ldots\}$ Find all the left cosets of $H$ in $\Bbb Z$.
I know the answer is $H$, $1+H$, and $2+H$ but I am having difficulty understanding why.
Thank you!

Consider the three cosets $\color{Blue}{3\Bbb Z}$, $\color{Green}{1+3\Bbb Z}$, $\color{Red}{2+3\Bbb Z}$ as they sit inside $\color{Black}{\Bbb Z}$ itself:
$~~~$ 
Do you see any numbers that were skipped? Notice that the pattern repeats. How can we justify this though, formally and algebraically? We need to be able to take any coset and show that it is actually equal to one of these three. It suffices to show that every integer is either $0$, $1$ or $2$ plus a multiple of $3$. This leads us to consider representations of integers of the form $n=3q+r$ where the residue $r$ is $0\le r<3$. Know any theorems or identities or whatnot that deal with this?
