You have the theorem.
Guide to the proof. For a four digit number abcd the value in positional
notation is a1000 + b100 + c10 + d. If a'b'c'd' is the rearranged number,
the value is a'1000 + b'100 + c'10 + d', and the difference is
(a-a')1000 + (b-b')100 + (c-c')10 + (d-d') =
(a-a')(999 +1) + (b - b')(99 +1) + (c-c')(9+1) + (d-d') =
(a-a')(999) + (b-b')(99) + (c-c')(9) + (d-d') + (a-a') + (b- b') + (c-c')
Clearly the last 4 terms sum to zero, because the primed numbers are just a
rearrangement on the unprimed ones, and the first three terms are divisible by 9
Dean Gerber
________________________________
From: Robert J. Cordingley <[email protected]>
To: The Friday Morning Applied Complexity Coffee Group <[email protected]>
Sent: Monday, October 8, 2012 10:20 AM
Subject: [FRIAM] Nines: Trivia Question?
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
