Div 3

let F[i] = 1, 12, 123, 1234, ..., 12345678910, ...
we see F[i] = 10^k * F[i-1] + i
F[i] Mod 3 = (10^k Mod 3) *(F[i-1] Mod 3) + (i Mod 3)
F[i] Mod 3 = (F[i-1] + i) Mod 3
after found that F[i] Mod 3 is cyclic,it's easy to solve.