Thread: Coset conundrum
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Old 2021-08-28, 21:28   #2
uau
 
Jan 2017

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Quote:
Originally Posted by Dr Sardonicus View Post
A) Prove that \sum_{i=1}^{h}r_{i}\;=\;m\times n for a positive integer m.
If you multiply the elements of the set {b^i for all i} by b, you get the same set with permuted elements. Thus multiplying the sum by b does not change it mod n. Thus S*b = S mod n, S*(b-1)=0 mod n, and S must be 0 mod n.
Quote:
B) Prove that m = 1 if and only if n is a repunit to the base b, and also that one of the ri is equal to 1.
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
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