Using brute force on the end 5dpf's:
~ 50000 odds less than 100000
~ 40000 with 5's removed, there should be
~ 20000 pairs, but we lose those matched with the missing 5's, so actual pairs
= 11145.
So the plan could be to sieve a block of k*10^5 + 45903 and k*10^5 + 46303. Removing a number from the low end of one group should also remove one from the high end of the other.
If handled properly, that will reduce the work by a factor of 7.
