x | y

So, you're writing your conjecture again, and you want to say that any amount $y$ of apples can be roughly distrubuted amongst $x$ baskets. You go on to say that at least one basket will have $$\frac{y}{x}$$ apples, annd this generally seems to hold! Except, it only works when $$x\mid y.$$ Dang! How are you supposed to battle with that? How do we even figure out if an arbritary number divides another?  

Here's one way.

Whenever $x\mid y,$ we will have that $\frac{y}{x}$ is an integer.  We could simply compute that, and if it is an integer, then bingo! But, our function does not know how to test for an integer. So, we'll explicitly have to output $0$ or $1.$

There's a nice way we can get   $0:$

$$x\ \mid\ y\ =\ \frac{y}{x}-\left\lfloor\frac{y}{x}\right\rfloor\Rightarrow\left\{\text{true}:\ 0,\ \text{false}:\ \left(0<?<1\right)\right\}$$

Right now, when it's false, we get whatever the residue of the floor is.  That isn't very pretty - we'd ideally output the same thing every time it's false.  We can do this easily with a ceiling function, but we'd much rather use a floor function since we've already used it once.

$$x\ \mid\ y\ =\ \frac{y}{x}-\left\lfloor\frac{y}{x}\right\rfloor-1\Rightarrow\left\{\text{true}:\ -1,\ \text{false}:\ \left(-1<?<0\right)\right\}$$

$$x\ \mid\ y\ =\ -\left(\frac{y}{x}-\left\lfloor\frac{y}{x}\right\rfloor-1\right)\Rightarrow\left\{\text{true}:\ 1,\ \text{false}:\ \left(1>?>0\right)\right\}$$

$$x\ \mid\ y\ =\left \lfloor-\left(\frac{y}{x}-\left\lfloor\frac{y}{x}\right\rfloor-1\right)\right\rfloor\Rightarrow\left\{\text{true}:\ 1,\ \text{false}:\ 0\right\}$$

$$x\ \mid\ y\ =\left \lfloor\left\lfloor\frac{y}{x}\right\rfloor-\frac{y}{x}+1\right\rfloor\Rightarrow\left\{\text{true}:\ 1,\ \text{false}:\ 0\right\}$$

And this is perfect - if we want to add something only if $x\mid y,$  then we just multiply by this function - if it isn't true, it will just zero out!