>Let $M$ be a natural number $M>1$. For  every prime $p_i$ not dividing $M$ take an arithmetic progression $A_i=k_i+np_i$  , $n\geq 0$ such that $k_i>p_i^2/M$. Is there any $M$ and some choice of the $k_i's$ such that  $\cup A_i= \mathbb{N}$ ?

> CONJECTURE: There is not such choice for any $M$

- MOTIVATION: It is not hard to see that if there is not such a choice for any $M$ Dirichlet's theorem in arithmetic progressions is an easy consequence. Of course this is much more general. We can modify many well known problems in this way. 

>For example ,to the previous question, if we take $2$ arithmetic progressions for each $p_i>M$, instead of $1$ with the same condition then we have something stronger than the twin prime conjecture and polignac's conjecture in general,
If we take $3$ arithmetic progressions we have something stronger than the problem that infinitely many times exists every logical form with $3$  primes ( for example p, p+2,p+6), etc. To these problems it is enough to take infinitely many $M$, not all...
 

- Secondly it is one of the easiest questions to this spirit and i want to know if we can say something about this kind of questions with known techniques. So

>What kind of mathematical techniques we could use to reach this kind of problems?

-related to https://mathoverflow.net/questions/80494/covering-mathbbn-with-prime-arithmetic-progressions

>Can we compute the function of the length  of the intervals that we can cover using the primes not dividing $M$ below some $x$ and compare it to the growth function of the number of the primes below some $x$ ? 

- Notice that even for the primes below $M$ the arithmetic progressions can start from $1$ , the first term of every $A_i$ for $p_i < M$ can be less than $p_i$ this changes and the first term of the $A_i's$ becomes larger and larger comparing with the $p_i's$    

I am waiting for any help to this direction, thank you.