<pre>
<pre>
 2 105263157894736842
 3 1034482758620689655172413793
 4 102564 153846 179487 205128 230769
 5 142857 102040816326530612244897959183673469387755
 6 1016949152542372881355932203389830508474576271186440677966
   1186440677966101694915254237288135593220338983050847457627
   1355932203389830508474576271186440677966101694915254237288
   1525423728813559322033898305084745762711864406779661016949
 7 1014492753623188405797 1159420289855072463768 1304347826086956521739
 8 1012658227848 1139240506329
 9 10112359550561797752808988764044943820224719
</pre>

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

<pre>
    (A-D)/B + D B^(N-1) = K A .
</pre>

Solving this for A we have

<pre>
        D (B^N - 1)
    A = ----------- .
          B K - 1
</pre>

In order for A ><code> B^(N-1) we must have D ></code> 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)<code>N0(S,B)N0(T,B) when (S,T)</code>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.
</pre>
