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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 have

(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.