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

[pupeye11]

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.

 

[uart]

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.

 

[pupeye11]

We haven't covered the remainder theorem yet...

 

[uart]

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.

 

[pupeye11]

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...

 

[pupeye11]

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?

 

[uart]

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.