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Number Theory - Repunits

  • Thread starter pupeye11
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  • #1
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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



[tex]R_{n}=\frac{b^{n}-1}{b-1}[/tex]

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.
 

Answers and Replies

  • #2
uart
Science Advisor
2,776
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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



[tex]R_{n}=\frac{b^{n}-1}{b-1}[/tex]

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 [itex]b^{n}-1[/itex] is divisible by [itex]b^2-1[/itex].

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

Try applying the concepts of the remainder theorem to each of the above.
 
  • #3
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We haven't covered the remainder theorem yet...
 
  • #4
uart
Science Advisor
2,776
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We haven't covered the remainder theorem yet...
Ok, well imagine that you divided [itex]b^n-1[/itex] by [itex](b-1)(b+1)[/itex]. Then if "Q" was the quotient and "R" the remainder of this division we could write :

[tex] b^n-1 = (b-1) (b+1)\,Q+ \alpha b + \beta[/tex]

Where [itex]R(b) = \alpha b + \beta[/itex] 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.
 
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  • #5
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Well for b=1 then [tex]\alpha[/tex] will be positive so [tex]\beta[/tex] will have to be the negative of [tex]\alpha[/tex]

The opposite is true for b=-1, [tex]\alpha[/tex] will be negative and [tex]\beta[/tex] is going to be the positive of that number...
 
  • #6
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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?
 
  • #7
uart
Science Advisor
2,776
9
Well for b=1 then [tex]\alpha[/tex] will be positive so [tex]\beta[/tex] will have to be the negative of [tex]\alpha[/tex]
Yeah do part "b)" first, it's the easiest.

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

The opposite is true for b=-1, [tex]\alpha[/tex] will be negative and [tex]\beta[/tex] 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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