Not sure about that.  This whole thing (at least to begin with) has to do with 
writing numbers in the place notation. Not sure what this would mean in 
rational, irrational or imaginary base system. I think integers were implied in 
this. ;)

--joshua

On Oct 8, 2012, at 3:15 PM, Robert J. Cordingley wrote:

> ...and I guess (base) n can be rational, irrational or even imaginary.
> Thanks
> Robert
> 
> On 10/8/12 12:02 PM, Joshua Thorp wrote:
>> I think you just replace '9' with 'n-1' in Dean or Frank's answer and you 
>> have a general proof, for n>=2.
>> 
>> I suppose you may need to convince yourself that a number like n^k - 1 == 
>> (n-1)*n^(k-1) + (n-1)*n^(k-2) + … + (n-1)*(k-k).
>> 
>> --joshua
>> 
>> On Oct 8, 2012, at 11:37 AM, Robert J. Cordingley wrote:
>> 
>>> May be I should reframe the question.
>>> 
>>> How do you prove there isn't a system of numbers to base N where it doesn't 
>>> work?
>>> 
>>> Thanks,
>>> Robert
>>> 
>>> On 10/8/12 11:00 AM, Tom Carter wrote:
>>>> Robert -
>>>> 
>>>>   There's a reasonably good discussion of this here:
>>>> 
>>>>      http://mathforum.org/library/drmath/view/58518.html
>>>> 
>>>>   Thanks . . .
>>>> 
>>>> tom
>>>> 
>>>> On Oct 8, 2012, at 9:20 AM, Robert J. Cordingley <rob...@cirrillian.com> 
>>>> wrote:
>>>> 
>>>>> I probably should know this...
>>>>> 
>>>>> So when you rearrange the digits of a number (>9) and take the 
>>>>> difference, it is divisible by nine.  A result that sometimes points to 
>>>>> accounting errors.  If the numbers are not base 10 the result is 
>>>>> divisible by (base-1).
>>>>> 
>>>>> What is the associated theorem for this?
>>>>> 
>>>>> Thanks
>>>>> Robert
>>>>> 
>>>>> 
>>>>> 
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