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Simple Proof

  • Thread starter bomba923
  • Start date
732
0
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:

Answers and Replies

Hurkyl
Staff Emeritus
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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.
 
732
0
:redface: Do you know any good books or sources on modulus?
(I'm only a HS student, just started CalcIII)
 
quasar987
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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:
200
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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.
 
694
0
Consider that a number is divisible by 3 iff the digits of the number add up to a multiple of three
This is equivalent to bomba923's question.
 

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