Thread: Coset conundrum
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Old 2021-08-29, 13:26   #3
Dr Sardonicus
 
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Feb 2017
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Quote:
Originally Posted by uau View Post
<snip>
The "and also that one of the ri 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(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*(bh - 1)/n, and sb(A) the sum of the base-b digits of A. Then m = sb(A)/(b-1).

Last fiddled with by Dr Sardonicus on 2021-08-29 at 13:29
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