Comments on: Permutations and the number 9 https://www.ticklishtechs.net/2009/02/17/permutations-and-the-number-9/ a mostly .NET but also some other cool techs blog Thu, 13 Aug 2020 18:46:34 +0000 hourly 1 https://wordpress.org/?v=5.7.11 By: Wolfram Bernhardt https://www.ticklishtechs.net/2009/02/17/permutations-and-the-number-9/comment-page-1/#comment-2000 Wed, 18 Feb 2009 12:46:56 +0000 http://www.ticklishtechs.net/2009/01/21/permutations-and-the-number-9/#comment-2000 In reply to Martin Bennedik.

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 https://www.ticklishtechs.net/2009/02/17/permutations-and-the-number-9/comment-page-1/#comment-1999 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. 😉

]]> By: diryboy https://www.ticklishtechs.net/2009/02/17/permutations-and-the-number-9/comment-page-1/#comment-1995 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?

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