Quote:
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:
Quote:
Let n > 2 be an integer, b an integer, 1 < b < n, gcd(b, n) = 1, gcd(b1,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 baseb digits of A. Then m = s
_{b}(A)/(b1).