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Prove the following

  1. Sep 18, 2012 #1
    1. The problem statement, all variables and given/known data
    In the link

    2. Relevant equations
    In the link


    3. The attempt at a solution
    First I tried expanding the sum as a power of 2 over a power of 3, but I failed.
     

    Attached Files:

  2. jcsd
  3. Sep 19, 2012 #2

    Mark44

    Staff: Mentor

    I was not able to open the file. Why not just put the problem and your work directly into the form?
     
  4. Sep 19, 2012 #3

    Ray Vickson

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    The statement was: prove that every positive integer can be represented in the form
    [tex] \frac{2^a}{3^b} -\sum_{k=0}^b \frac{2^{c_k}}{3^k}, \\
    \text{where } a, b, c_k \text{ are integers },1 = c_0 < c_1 < c_2 < \cdots . [/tex]

    It does not say whether or not the a and b are allowed to vary with the integer n to be represented, or whether a and/or b are supposed to be fixed.

    RGV
     
    Last edited: Sep 19, 2012
  5. Sep 19, 2012 #4

    HallsofIvy

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    Start by writing the number in base three. Then you have the number as a sum of negative powers of three with all numerators either 1 or 2. Those with are what you want. If there is a numerator of 1, combine fractions.
     
  6. Sep 19, 2012 #5
    @Ray a and b are allowed to vary, but they have to be integral.
    @Mark I do not know how to use the codes
    @HallsofIvy I am pretty sure that the answer does not involve different bases. The expression is merely a subtraction of powers of 2 over powers of 3.
     
  7. Sep 20, 2012 #6
    Truth be told, this isn't a homework question for school. This is merely a question on a recent math competition that I have not answered and I want to know.
     
  8. Sep 20, 2012 #7

    HallsofIvy

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    Which is a question of base 3, pretty much by definition!
     
  9. Sep 20, 2012 #8
    However it will get just as complicated as in base ten because of all the powers of 2
     
  10. Sep 23, 2012 #9
    Suppose we can approach it in a different way. We can just sum up the terms in the summation and define a "simpler" term, which we can then subtract from the first term.
     
  11. Sep 24, 2012 #10

    HallsofIvy

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    No, it won't. Every integer, written is base three, is, by definition, of the form [itex]\sum_{i=0}^N a_i/3^i[/itex] where each [itex]a_i[/itex], because it is a base 3 digit, is either 0, or 1= 20, or 2= 21.
     
  12. Sep 25, 2012 #11
    I am just looking for an answer, nothing more.
     
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