What Repunits are Divisible by Factors of b+1 and b-1?

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Homework Help Overview

The discussion revolves around determining the divisibility of base b repunits by factors of b-1 and b+1. A base b repunit is defined as an integer with a base b expansion consisting entirely of 1's. Participants are exploring the conditions under which these repunits are divisible by the specified factors.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to identify specific values of n for which base b repunits are divisible by b-1 and b+1. There is discussion about applying the remainder theorem, although some participants note that they have not covered this topic yet. Questions arise regarding the implications of substituting values into polynomial expressions related to the divisibility conditions.

Discussion Status

The discussion is ongoing, with participants sharing insights and attempting to clarify their understanding of the problem. Some have proposed conditions for divisibility based on the number of digits in the repunit and the alternating sum of digits, while others are exploring the implications of specific cases such as b=1 and b=-1. There is no explicit consensus, but several productive lines of reasoning are being explored.

Contextual Notes

Participants mention that they have not yet covered the remainder theorem, which may limit their ability to fully engage with some of the proposed approaches. There is also a suggestion to consider different cases for even and odd n when discussing b=-1.

pupeye11
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Homework Statement



A base b repunit is an integer with base b expansion containing all 1's.

a) Determine which base b repunits are divisible by factors b-1

b) Determine which base b repunits are divisible by factors b+1

Homework Equations



R_{n}=\frac{b^{n}-1}{b-1}

The Attempt at a Solution



So a repunit of base 3 would be n=2 since we get 1. I don't get how to find any beyond this point though. I tried using other n numbers but I didn't get any others that come out to all 1's.
 
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pupeye11 said:

Homework Statement



A base b repunit is an integer with base b expansion containing all 1's.

a) Determine which base b repunits are divisible by factors b-1

b) Determine which base b repunits are divisible by factors b+1

Homework Equations



R_{n}=\frac{b^{n}-1}{b-1}

The Attempt at a Solution



So a repunit of base 3 would be n=2 since we get 1. I don't get how to find any beyond this point though. I tried using other n numbers but I didn't get any others that come out to all 1's.

So if the repunit is divisible by (b+1) then b^{n}-1 is divisible by b^2-1.

Similarly if the repunit is divisible by (b-1) then b^{n}-1 is divisible by (b-1)^2.

Try applying the concepts of the remainder theorem to each of the above.
 
We haven't covered the remainder theorem yet...
 
pupeye11 said:
We haven't covered the remainder theorem yet...

Ok, well imagine that you divided b^n-1 by (b-1)(b+1). Then if "Q" was the quotient and "R" the remainder of this division we could write :

b^n-1 = (b-1) (b+1)\,Q+ \alpha b + \beta

Where R(b) = \alpha b + \beta is the remainder.

Try substituting first b=1 and then b=-1 into the above equation and see what conditions are required to make R=0.
 
Last edited:
Well for b=1 then \alpha will be positive so \beta will have to be the negative of \alpha

The opposite is true for b=-1, \alpha will be negative and \beta is going to be the positive of that number...
 
I figured out that for a its if and only if the number of digits is a multiple of d, where d divides b-1.

For b, would it just be if and only if the alternating sum comes out to a multiple of d, where d divides b+1?
 
pupeye11 said:
Well for b=1 then \alpha will be positive so \beta will have to be the negative of \alpha
Yeah do part "b)" first, it's the easiest.

Yes that first part is correct, when b=1 we must have \alpha + \beta = 0 to get zero remainder.

The opposite is true for b=-1, \alpha will be negative and \beta is going to be the positive of that number...

No not quite. For b=-1 you should consider the case of even n and odd n separately. Do that and it will be very easy so find the conditions required for zero remainder.
 

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