Proving Homework Statement: Expanding Power of 2 over 3

[eddybob123]

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.

 

[Mark44]

I was not able to open the file. Why not just put the problem and your work directly into the form?

 

[Ray Vickson]

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
$$\frac{2^a}{3^b} -\sum_{k=0}^b \frac{2^{c_k}}{3^k}, \\<br /> \text{where } a, b, c_k \text{ are integers },1 = c_0 < c_1 < c_2 < \cdots .$$

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

 

[HallsofIvy]

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.

 

[eddybob123]

@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.

 

[eddybob123]

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.

 

[HallsofIvy]

@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!

 

[eddybob123]

However it will get just as complicated as in base ten because of all the powers of 2

 

[eddybob123]

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.

 

[HallsofIvy]

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 $\sum_{i=0}^N a_i/3^i$ where each $a_i$, because it is a base 3 digit, is either 0, or 1= 2⁰, or 2= 2¹.

 

[eddybob123]

I am just looking for an answer, nothing more.