Define $D(m)$ to be the (ordered) set of divisors of $m \in \Bbb{N}$.
Let $m = p_n\# = 1 \cdot 2 \cdot 3 \cdot 5 \cdots p_n$. Clearly, $|D(m)| = $ the number of divisors of $p_n\# = 2^n$.
Now define a divisor indicator vector at $x$ to be $\langle (d \mid x) : d \in D(m) \rangle$ where $(d \mid x) = \begin{cases} 1, \text{ if } d \mid x \\ 0, \text{ if } d \nmid x \end{cases}$; for any given $x \in \Bbb{Z}$.
Conjecture 1. The number of possible divisor indicator vectors is precisely $2^n$.
For example:
1 2 3 5 6 10 15 30
1 0 0 0 0 0 0 0
1 1 0 0 0 0 0 0
1 0 1 0 0 0 0 0
1 0 0 1 0 0 0 0
1 1 1 0 1 0 0 0
1 1 0 1 0 1 0 0
1 0 1 1 0 0 1 0
1 1 1 1 1 1 1 1
I.e. there are $8$ divisors so the number of divisor indicator vectors is also $8$. Clearly it's since if $15 \mid x$ then both $3 \mid x$ and $5\mid x$ and vise-versa, since $\gcd(3,5) = 1$, but how would the argument go for the general case?
My next question which I think is better is about this conjecture:
Conjecture 2. The set of all possible divisor indicator vectors is linearly independent over any field.
I say any field because every field by definition contains the values of the divisor indicator $(d \mid \cdot)$ or $0$ and $1$. So for each field $F$ you'd define $(d\mid \cdot) : \Bbb{Z} \to \{0, 1\} \subset F$.
Now this might have something to do with the first conjecture since our degrees of freedom are precisely determined by the $n$ primes involved and taken together (just the prime components):
2 3 5
1 0 0
0 1 0
0 0 1
we have clear linear independence.
What is the easiest & most elegant way you can think of to prove both of these conjectures?
Proof of 1. For any given $x \in \Bbb{Z}$ there are $2^n$ possibilities: either $2 \mid x$ or it doesn't, either $3\mid x$ or it doesn't, $\dots \ $, either $p_n \mid x$ or it doesn't. The answer to the divisibility by each of the $p_i$ completely determines the rest of the vector's entries, by CRT you could say.
Proof of 2. Let $X$ be a finite set of $2^n$ representatives $x \in X$ so that $\langle (d \mid x) : d \in D(p_n\#)\rangle, \ x \in X$ ranges over all possible divisor indicator vectors. Suppose that:
$$ \sum_{d \mid p_n\#} c_d \cdot (d \mid x) = 0, \forall x \in X $$
Note this is the same thing as saying $A\overline{c} = 0$ where $A_{i,j} = (d_i \mid x_j)$ where $d_i$ is the $i$th divisor in the (ordered) set $D = D(p_n\#)$ and $x_j =$ the $j$th entry in $X$ (in any fixed order). And where $\overline{c} = \langle c_d : d \in D\rangle^T$. So, what's left to prove is that $c_d = 0$ for all $d \in D$. Let's also index the entries of $X$ by elements of $D$. Clearly, there exists some $x_1 \in X$ such that: $c_1(1\mid x_1) = c_1 = 0$ by definition of $X$ as a full set of representatives.
Now, if for any divisor $d \in D$ we have that $c_1(1\mid x_1) + c_d(d\mid x_d) = 0$, and $(d\mid x_d) = 1$, then we have $c_d = -c_1 = 0$. Thus for each prime $p \in D$ we have that $c_p = 0$. Now for all $d \in D$ such that $\Omega(d) = 2$ we have (for some $x \in X$ by definition):
$$ c_1(1\mid x) + c_2(2 \mid x) + c_3(3\mid x) + c_6(6 \mid x) = 0 $$
for example. But the first $3$ terms have already been shown to zero out, implying that $c_6 = 0$ as well. Clearly this generalizes to any $\Omega(d) = 2$. Next perform a similar proof for when $\Omega(d) = 3$ and so on...
You might make it more elegant via formal induction, but I believe this proof is good enough. $\blacksquare$