## Permutations and the number 9

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Lately I stumbled over permutations and noticed a funny fact: When you take any integer, produce a permutation and subtract one form the other, you always get a multiple of 9.

It works with a any length. Some examples:

Length 2
23 – 32 = -9
84 – 48 = 36
60 – 06 = 54

Length 3
123 – 132 = -9
and so on…

It still works for changing the front position:
123 – 213 = -90

In fact it works for changing any position:
123 – 321 = -198 (-22*9)

This leads to the conclusion I stated in the beginning. The difference of any permutation of the same number is a multiple of 9. Feel free to check it with you favorite numbers.

It’s not a great mathematical invention (if it is, please let me know), but I didn’t know it, didn’t expect it und find it kind of fascinating. I asked a couple of friends and no one was aware of this.

I still try to find a way to use these fact, maybe for spell-check-like for numbers or a fast calculation of permutations, but so far it seems just useless

Instead I can provide a proof for numbers of any length and I’ll post it here in ten days from today. Meanwhile I want you to try it yourself. The proof is really, really simple, almost trivial and it won’t cost you more than five minutes to understand it all.

### 3 Responses to “Permutations and the number 9”

1. diryboy Says:

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?

2. Martin Bennedik Says:

Please generalize for bases other than 10.

3. Wolfram Bernhardt Says:

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 .-)