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Simple Proof

  1. Sep 8, 2005 #1
    Where [tex] \mathbb{Z}^{+}[/tex] represents the set of all positive integers,
    How do I prove that

    [tex] \begin{gathered} \forall \left\{ {a_0 ,a_1 ,a_2 , \ldots ,a_n } \right\} \subset \mathbb{Z}^ + \;{\text{where}}\;\max \left\{ {a_0 ,a_1 ,a_2 , \ldots ,a_n } \right\} \leqslant 9, \hfill \\
    \left( {\sum\limits_{k = 0}^n {a_k 10^k } } \right)\;{\text{is divisible by }}3{\text{ iff }}\left( {\sum\limits_{k = 0}^n {a_k } } \right)\;{\text{is divisible by 3}} \; {?} \hfill \\ \end{gathered} [/tex]
    Last edited: Sep 8, 2005
  2. jcsd
  3. Sep 8, 2005 #2


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    Generally, you would want to look at problems like this modulo 3. Or, equivalently, to see if 3 divides their difference.
  4. Sep 8, 2005 #3
    :redface: Do you know any good books or sources on modulus?
    (I'm only a HS student, just started CalcIII)
  5. Sep 9, 2005 #4


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    I used induction and the fact that if an integer z is divisible by 3, then there exists (a unique) integer m such that z = 3m. Strangely though, I did not use the fact that the a_i's are smaller or equal to 9...
    Last edited: Sep 9, 2005
  6. Sep 9, 2005 #5
    Consider that a number is divisible by 3 iff the digits of the number add up to a multiple of three and that multiplying a number by a multiple of 10 just adds 0s to it and thus doesn't affect the sum of its digits.
  7. Sep 9, 2005 #6
    This is equivalent to bomba923's question.
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