Permutations and the number 9 (proof)

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Ten days ago I promised a proof why the differences of any permutation of the same digits is always a multiple of 9. Here it is

Let’s say x and y are digits and ‘xy’ is not ‘x*y’, but the number that consist of the digits x and y.

xy – yx = multiple of 9.

Is it always true? Yes, it is:

(10x + 1y) – (10y + 1x) = multiple of 9?
9x – 9y = multiple of 9?
9(x-y) = multiple of 9? Yes, for sure.

I calculated the numeric values of xy and yx by "10 times the higher position + 1 time the lower position", just like we all do every day in our beloved decimal-system. More formally we describe the factor for the position (1, 10, 100…) by

10^position, (position is zero-based, counting from right to left, of course)

Now, when we change the position of a single digit within a number we change its numeric value from

x*10^(old position) to x*10^(new position).

We can neglect x here, because it’s a factor that occurs in both values, so we can get rid of it by division.
The remaining part for building the difference is

10^p1 – 10^p2, a formula that always produces multiples of 9: 10-1, 1000-100, 10-1000000.

Conclusion

Since the difference of a number and one of it’s permutations is a sum of (10^p1-10^p2)-parts, which are all multiples of 9, the total also can be divided by 9.

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