# Prove the following

In the link

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.

#### Attachments

• Prove that every positive integer can be represented in the form.doc
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## Answers and Replies

Mark44
Mentor
I was not able to open the file. Why not just put the problem and your work directly into the form?

Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
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
$$\frac{2^a}{3^b} -\sum_{k=0}^b \frac{2^{c_k}}{3^k}, \\ \text{where } a, b, c_k \text{ are integers },1 = c_0 < c_1 < c_2 < \cdots .$$

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

Last edited:
HallsofIvy
Science Advisor
Homework Helper
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.

@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.

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
Science Advisor
Homework Helper
@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!

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

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
Science Advisor
Homework Helper
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.

I am just looking for an answer, nothing more.