What you try to demonstrate is true, but your demonstration, even if it is a good try, is wrong.Philosophy Explorer wrote:First, in regards to Rilx, he was wrong on several accounts. First he did the algebra wrong. Then he left out some steps. Then he indicated that an infinity of digits would be involved which isn't true (and I'm going to prove it).
To start this off, let's suppose we have these two three-digit numbers: abc and cab where a, b and c can take on any values from 0 to 9 (this is sufficient to guarantee we're covering any three-digit number that you like in base 10). Now rewrite those anagram numbers as follows:
100a + 10b + c
100c + 10a + b
Taking the difference, we get:
100(a - c) + 10(b - a) + (c - b)
Next, we break down this expression as follows:
(a - c) + 99(a - c) + (b - a) + 9(b - a) + (c - b)
Rearranging terms, we get:
(a - c) + (b - a) + (c - b) + 99(a - c) + 9(c - b)
Cancelling like terms, we're left with:
99(a - c) + 9(c - b)
This proves that my theorem works for any three-digit anagram numbers as the last expression is evenly divisible by nine.
In my next post I will extend this proof to any two anagram numbers with any amount of digits.
You demonstrate that 9 divide the difference, of anagram of the form (abc, cab).
but anagram of 3 digits, could also be of the form (abc, abc), (abc, acb), (abc, bca), (abc, bac),(abc, cba).
The rest of your demonstration is wrong for the same kind of reason.
In final, you only demonstrate that 9 divide the difference of anagram of a particular form.