# Is it possible to construct modulo classes that cover the integers?

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.

$$\{0(\mathrm{mod}\ {2}),\ 0(\mathrm{mod}\ {3}),\ 1(\mathrm{mod}\ {4}), \ 5(\mathrm{mod}\ {6}),\ 7(\mathrm{mod}\ {12}) \}$$
$${0(2), 0(3), 0(5), 1(6), 0(7), 1(10), 1(14), 2(15), 2(21), 23(30), 4(35), 5(42), 59(70), 104(105)}$$