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.