...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 <[email protected]> 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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FRIAM Applied Complexity Group listserv
Meets Fridays 9a-11:30 at cafe at St. John's College
lectures, archives, unsubscribe, maps at http://www.friam.org
============================================================
FRIAM Applied Complexity Group listserv
Meets Fridays 9a-11:30 at cafe at St. John's College
lectures, archives, unsubscribe, maps at http://www.friam.org
============================================================
FRIAM Applied Complexity Group listserv
Meets Fridays 9a-11:30 at cafe at St. John's College
lectures, archives, unsubscribe, maps at http://www.friam.org
============================================================
FRIAM Applied Complexity Group listserv
Meets Fridays 9a-11:30 at cafe at St. John's College
lectures, archives, unsubscribe, maps at http://www.friam.org
