(I)Solution:We first show that f(A)≤A,the equation holds iff A=1,2,3,4,5,6,7,8,9,19.
A=(anan−1…a0),(an≠0).
If n≥2,A≥10n≥25×2n,f(A)≤9(1+2+…+2n)=9(2n+1−1)<18×2n<25×2n.∴.
If n=0,it's clear that f(A)=A.A=1,2,3,4,5,6,7,8,9.
If n=1,A=10a_1+a_0,f(A)=2a_0+a_1,(0\leq a_0\leq9,1\leq a_1\leq9).
Then f(A)-A=a_0-9a_1\leq a_0-9\leq0.The equation holds if and only if a_0=9,a_1=1.
So f(A)\leq A,the equation holds iff A=1,2,3,4,5,6,7,8,9,19.
If there doesn't exist k such that A_{k+1}=A_k,then since A_{k+1}=f(A_k)\leq A_k,A_{k+1}<A_k.
So A_{A+1}\leq A-A=0,a contradiction.
So there must exist k,A_{k+1}=A_k.
(II)For A=19^{86},it's easy to know that A_k\in\{1,2,3,4,5,6,7,8,9,19\}.
We show that if 19|A,then 19|f(A).
A=(a_na_{n-1}\ldots a_0),(a_n\neq0).
A=a_n10^n+a_{n-1}10^{n-1}+\ldots+a_0=\sum_{i=0}^na_i10^i
f(A)=\sum_{i=0}^n2^{n-i}a_i
So 2^nA-f(A)=\sum_{i=0}^na_i20^i\times2^{n-i}-\sum_{i=0}^n2^{n-i}a_i=\sum_{i=0}^na_i2^{n-i}(20^i-1)
And for i=0,1,\ldots,n,19|20^i-1,so 19|2^nA-f(A),and \gcd(19,2^n)=1,so if 19|A,19|f(A).
And 19|A=19^{86},so 19|A_k,and A_k=19.