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A fortunately timed trip

July 18, 2009

 

So where are we?

I’m looking at repeated sets of 1s that are divisible by the number of 1’s.

So:

111 is divisible by 3

111,111,111 is divisible by 9

Whereas

11 isn’t divisible by 2

1,111,111 isn’t divisible by 7

..and generalized to different number bases.

e.g. 1,111 in base 3 = 27+9+3+1 = 40 which is divisible by 4.

I called these Saturday numbers. The proper name is a Niven Repunit. Niven because it’s divisible by the sum of its digits and repunit because it’s a repeated unit – repeated 1s.

Defining repunits

Okay – in any number base a repunit is:

R_{n}^{(b)}=\sum_{i=0}^{n-1}b^i=\frac{b^n-1}{b-1}

Where 

b

is the number base and 

n

is the number of digits.

And we want:

R_{n}^{(b)}\pmod{n}\equiv0

Which is a fancy way of saying if you divide it by 

n

you get 0 remainder (so it exactly divides by

n

). Which is exactly what we’re after.

Enter the mathematicians (Peter and Paul)

@PeterRowlett and @paulhertz swim in a sea of mathematics (whereas I swim in a sea of software development), they know how to do this stuff – and importantly where to look things up.

Paul did a search and found this:

Niven Repunits

Which looks promising.

Niven Repunits.  That’s exactly what I’m looking for!

Peter did a search and found that Fibonacci Quarterly is at the British Library:

British library

My trip

As it happens, I’m in London on business next week.  I can quite easily make a detour to the British Library to look this up.

But it is helpful to define success/fail criteria before doing things, so these are my criteria for success/failure next week (wrt library trip – success for the business trip will be defined by the quality of the business requirements!).

 

What I hope I find

I hope to get a better understand of numbers, and why (whether?) my original conjectures are true.

What I don’t hope to find

The title of the article is Niven Repunits and

10^n\equiv1

Now if all the article tells me is that Niven Repunits are values of 

n

that satisfy

10^n\equiv1

then it’s just restating the problem.

 

Why?

 

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

 

Well if

\frac{b^n-1}{b-1}\pmod{n}\equiv0

Then

a\frac{(b^n-1)}{b-1}\pmod{n}\equiv0

Where

a\in\mathbb{N}_0

Let

a=b-1

Then

b^n-1\pmod{n}\equiv0

or

b^n\pmod{n}\equiv1

 

Neater (but still obvious)

So saying a repunit is a Niven number (in base b) when:

b^n\pmod{n}\equiv1

is really just restating the original problem.

I hope I don’t find this – because even I can already see that!

 

Undue pessimism?

I’m fairly sure that getting something published in any maths journal takes more than a fairly obvious restatement of the problem.

I’m probably reading too much into the title. I don’t expect they publish the question and the conclusion on the same line!

I’ll post an update next week.

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