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10-adic number proof: A^10 has the same n+1 last digits as 1

  1. Oct 24, 2015 #1
    The question at hand: Let X be a 10-adic number. Let n be a natural number (not 0). Show that A^10 has the same n+1 last digits as 1 if A has the same n last digits as 1 (notation: A =[n]= 1)

    My work so far:

    (1-X)^10 = (1-X)(1+X+X^2+...+X^10)
    A =[n]= 1
    1-A =[n]= 0.
    I think I can also say that 1+X+X^2+...X^10 =[n]= 1.
    But now I don't know how to continue. Could anybody help me out?

    Thanks in advance.
  2. jcsd
  3. Oct 24, 2015 #2
    Of course A = X, my bad!
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