Addition and subtraction of binary numbers are in $\mathsf{AC^0}$.
For any constant number $c$,
$x \bmod c$ is $\mathsf{AC^0}$ reducible to division by $c$ ($\lfloor x/c \rfloor$):
$$x \bmod c = x -
(\overbrace{\lfloor x/c \rfloor + \cdots + \lfloor x/c \rfloor}^{c \text{ times}}) $$
It is known that $x \bmod c$ is hard for $\mathsf{AC^0}$ for any $c$
which is not a power of $2$.
Thus $\lfloor x/c \rfloor$ is hard for $\mathsf{AC^0}$ for any $c$
which is not a power of $2$.
As noted by Emil in the comments there is an easy reduction for odd prime $c$
from $\mathit{MOD}_c$ (that is, $\sum_ix_i\bmod c$ with $x_i\in\{0,1\}$) to
$x\bmod c$ with binary input:
we use only input bits which are multiples of $p-1$ and use FLT ($2^{(p-1)i} \bmod p = 1$).