let 100a+10b+c = m and (100b+10c+a)=n
m-n = (100a+10b+c) - (100b+10c+a) = 99a-90b-9c
its obvious that every such (m-n) will be divisible by 9. my question is : is there any theorem i could use to make this proof more presentable?
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No, there is no need to use any "theorem". You should write $99a-90b-9c=9(11a-10b-c)$ to make it "obvious". Of course, what you have is not a full proof. It only concerns $3$-digit numbers. – Dietrich Burde Jul 24 '23 at 16:26
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A very useful, so to speak "theorem" is that every natural number is congruent $\pmod 9$ with the sum of its digits (see the linked "duplicate": https://math.stackexchange.com/q/99725/700480). Which is not hard to prove (proof similar to the proof you would produce anyways) but it has a bonus (you see why rearranging does not matter: because, for the plain sum of digits, the order of digits doesn't matter) and another bonus (immediately proves the criteria for divisibility by $3$ and $9$). – Jul 24 '23 at 16:49