I am trying to construct (or disprove the existence of) some finite set, $S=\left\{(y,k)\mid y,k\in\mathbb{Z},y\ge2\right\}$, where no two $y$ are the same, and every integer $x\in\mathbb{Z}$ may be written as $x=n\cdot y_i + k_i$, where $(y_i,k_i)\in S$.
In other words, I am wondering if it might be possible to perhaps construct something like:
$$x+k_1\equiv r_1\pmod2$$
$$x+k_2\equiv r_2\pmod3$$
$$x+k_3\equiv r_3\pmod4$$
$$\vdots$$
$$x+k_i\equiv r_i\pmod {i+1}$$
where, for all $x\in\mathbb{Z}$, there is at least one $0 < j\le i$ where $r_j=0$.
So far I have written a greedy algorithm in python to try and find an example, with no luck. For example, the best case I have found for $i=4$ gives a covering of $\frac{56}{60}=\frac{14}{15}$ of the integers. I have a feeling it is not possible.
Yes! What you're asking for, in Wikipedia's terminology, is a distinct covering system. Here's a nice example:
$$\{0(\mathrm{mod}\ {2}),\ 0(\mathrm{mod}\ {3}),\ 1(\mathrm{mod}\ {4}), \ 5(\mathrm{mod}\ {6}),\ 7(\mathrm{mod}\ {12}) \}$$
According to arXiv:math/0412217v2, the above was found by Erdős, as well as this one:
$${0(2), 0(3), 0(5), 1(6), 0(7), 1(10), 1(14), 2(15), 2(21), 23(30), 4(35), 5(42), 59(70), 104(105)}$$