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 highe

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.