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 >>> >>> >>> >>> ============================================================ >>> 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
