Comments on: Permutations and the number 9 http://www.ticklishtechs.net/2009/02/17/permutations-and-the-number-9/ a mostly .NET but also some other cool techs blog Thu, 05 Jan 2012 12:01:34 +0000 hourly 1 http://wordpress.org/?v=3.0.4 By: Wolfram Bernhardt http://www.ticklishtechs.net/2009/02/17/permutations-and-the-number-9/comment-page-1/#comment-2000 Wolfram Bernhardt Wed, 18 Feb 2009 12:46:56 +0000 http://www.ticklishtechs.net/2009/01/21/permutations-and-the-number-9/#comment-2000 I will. It's pretty simple. It works for any base. The 'magic' number is always base-1. If you permute hexadecimals you alway get multiples of 15 as a difference .-) I will. It’s pretty simple. It works for any base. The ‘magic’ number is always base-1. If you permute hexadecimals you alway get multiples of 15 as a difference .-)

]]> By: Martin Bennedik http://www.ticklishtechs.net/2009/02/17/permutations-and-the-number-9/comment-page-1/#comment-1999 Martin Bennedik Wed, 18 Feb 2009 12:04:58 +0000 http://www.ticklishtechs.net/2009/01/21/permutations-and-the-number-9/#comment-1999 Please generalize for bases other than 10. ;-) Please generalize for bases other than 10. ;-)

]]> By: diryboy http://www.ticklishtechs.net/2009/02/17/permutations-and-the-number-9/comment-page-1/#comment-1995 diryboy Wed, 18 Feb 2009 10:05:24 +0000 http://www.ticklishtechs.net/2009/01/21/permutations-and-the-number-9/#comment-1995 not a strict proof: 1. if a number is |abc| and (a+b+c)%9=0, then |abc|%9=0 because 100a+10b+c = 100(a+b+c)-90b-99c 2. suppose we are doing |abc|-|cba|, the sum of every digit of the new number will be (a-c)+(b-b)+(c-a)=a-a+b-b+c-c=0, and 0%9=0 see? not a strict proof:
1. if a number is |abc| and (a+b+c)%9=0, then |abc|%9=0 because 100a+10b+c = 100(a+b+c)-90b-99c
2. suppose we are doing |abc|-|cba|, the sum of every digit of the new number will be (a-c)+(b-b)+(c-a)=a-a+b-b+c-c=0, and 0%9=0

see?

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