For what c is division by c in AC0? Suppose that our input is a binary $x$ and we have to output $\lfloor x/c \rfloor$, where $c$ is some constant integer.
This is just a shift if $c$ is a power of two, but what about other numbers?
Can we do it with a constant depth circuit for every $c$?
What about $c=3$?

ps. I know that computing $x\bmod c$ is hard, but this seems unrelated.
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 ($2^{(p-1)i} \bmod p = 1$).