Let $M$ be a natural number $M>1$. For  every prime $p_i$ not dividig $M$ 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  $\cup A_i= \mathbb{N}$ ?

- 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. Secondly it is one of the easiest questions to this spirit and i want to know if we can say something to this kind of questions or it is too difficult using our known techniques. We can modify many well known problems in this way.

-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 $x$ ? We can 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 $p_i's$    

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