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Prove the following

  • Thread starter eddybob123
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  • #1
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Homework Statement


In the link

Homework Equations


In the link


The Attempt at a Solution


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

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  • #2
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I was not able to open the file. Why not just put the problem and your work directly into the form?
 
  • #3
Ray Vickson
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I was not able to open the file. Why not just put the problem and your work directly into the form?
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
 
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  • #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.
 
  • #5
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@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.
 
  • #6
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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.
 
  • #7
HallsofIvy
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@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.
Which is a question of base 3, pretty much by definition!
 
  • #8
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However it will get just as complicated as in base ten because of all the powers of 2
 
  • #9
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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.
 
  • #10
HallsofIvy
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However it will get just as complicated as in base ten because of all the powers of 2
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.
 
  • #11
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I am just looking for an answer, nothing more.
 

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