[Dr Sardonicus]

Originally Posted by uau

<snip>
The "and also that one of the r_i is equal to 1" part seems ambiguous or wrong. For k != 1, m may or may not equal 1?

n = 1111, b = 10, k = 2: m = 1
n = 1111, b = 10, k = 21: m = 2

Mea culpa. I made a huge blunder. Actual characterization of the multiplier m follows, proof left as exercise.

Conditions restated for ease of reference:

Let n > 2 be an integer, b an integer, 1 < b < n, gcd(b, n) = 1, gcd(b-1,n) = 1, and let h be the multiplicative order of b (mod n).

Let k be an integer, 1 <= k < n.

(B) Let A = k*(b^h - 1)/n, and s_b(A) the sum of the base-b digits of A. Then m = s_b(A)/(b-1).