4 102564 153846 179487 205128 230769
5 142857 102040816326530612244897959183673469387755
7 1014492753623188405797 1159420289855072463768 1304347826086956521739
8 1012658227848 1139240506329
In base B, suppose you have an N-digit answer A whose digits are
rotated when multiplied by K. If D is the low-order digit of A, we
(A-D)/B + D B^(N-1) = K A .
Solving this for A we have
D (B^N - 1)
A = ----------- .
B K - 1
In order for A >= B^(N-1) we must have D >= K. Now we have to find N
such that B^N-1 is divisible by R=(BK-1)/gcd(BK-1,D). This always has
a minimal solution N0(R,B)<R, and the set of all solutions is the set
of multiples of N0(R,B). N0(R,B) is the length of the repeating part
of the fraction 1/R in base B.
N0(ST,B)=N0(S,B)N0(T,B) when (S,T)=1, and for prime powers, N0(P^X,B)
divides (P-1)P^(X-1). Determining which divisor is a little more
complicated but well-known (cf. Hardy & Wright).
So given B and K, there is one minimal solution for each
D=K,K+1,...,B-1, and you get all the solutions by taking repetitions
of the minimal solutions.