Nines (Was Re: yuga, VarNa and colour)

[jacob.baltuch at euronet.be]

>>I think what was meant was that if you start with 99, then 9 + 9 = 18 and 1
>>+ 8 = 9. I.e., supposedly all numbers that are multiples of nine will "add
>>down" to nine if you make sums of their component digits till you get to a
>>single-digit number.
>
>That is true, because the sum of the digits of a number that's a multiple
>of 9 will also be a multiple of 9 (not too hard to see by induction, assume
>true for N = 9 * p and prove for N + 9)

I'm sorry, don't try this at home :) By trying to keep things simple
I'm afraid I made things more complicated. If you accept to talk about
remainders, it's much easier: the remainder of (a0 + ... + aN * 10^N) by
9 is the same as the remainder of a0 + ... + aN by 9, the proof of which
is very easy.