I see my error here -- you can in fact eliminate half even for 2^k mod 10, becau...

andrewla 30 days ago | parent | context | favorite | on: Powers of 2 with all even digits

I see my error here -- you can in fact eliminate half even for 2^k mod 10, because both 6 and 8 force a carry; so ending in a 2 or a 6 means that the next higher digit must be odd.