very OT but how's it done 656

On Sat, 27 May 2006, Pete

Ah, i'm a sucker for stuff like this.

So far, i've found that the difference between a number and a shuffling of that number is always a multiple of 9; that means it always has a digital sum of 9, and that means that if you know three of its digits (like you tell Fido), you can work out the fourth trivially. I can't prove any of that, but it seems to be true for every number i look at.

This reminds me a bit of the digit-based puzzles you get in places like the Pyrgic Puzzles in the Guardian; the approach that usually works with those is to treat the number as a base-ten polynomial with unknown coefficients. I have a hunch that if you find a way to treat the shuffling in the framework of the polynomials, you'll be able to prove that the difference is always a multiple of nine.

Let's look at two-digit numbers; a number 'ab' can be thought of as the base-10 polynomial 10a + b; the only possible shuffling of it is 'ba', which is 10b + a. If you subtract those and rearrange and simplify:

(10a + b) - (10b + a) 10(a - b) + (b - a) 10(a - b) - (a - b) 9(a - b)

You can prove that the difference will always be a multiple of nine.

Extending this to longer numbers is left as an exercise for the reader!

tom

-- Is this the only way to get through to you?