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Direct Proof

  1. Jan 17, 2010 #1
    1. The problem statement, all variables and given/known data
    Prove that 10^n leaves remainder 1 after dividing by 9.

    3. The attempt at a solution

    There is an integer K, such that 10^n = 9k + 1

    Where do i go from here if I want to do it just directly?
     
  2. jcsd
  3. Jan 17, 2010 #2
    Do you know modular arithmetic?
    [tex]10^n \equiv 1^n =1 \pmod 9[/tex]
    Alternatively use the binomial theorem by writing:
    [tex](9+1)^n = \sum_{i=0}^n \binom{n}{i} 9^i = 1 + 9\sum_{i=1}^n \binom{n}{i}9^{i-1}[/tex]
    Finally you could use induction by noting that if [itex]10^n = 9k+1[/itex], then,
    [tex]10^{n+1} = 10^n 10 = (9k+1)(9+1) = 9^2k + 9k + 9 + 1[/tex]
    I would call all approaches direct though induction may not qualify depending on your definition.
     
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