Nines (Was Re: yuga, VarNa and colour)

[Kengo Harimoto]

I remember having learnt the proof for this (that any number divisible
by 9 has its sum of each digits divisible by 9) at my junior high
school.  I tried to remember it and failed :-)

However, I came up with the following:

Any integer can be described as a sum of (9+1))^n x a (n is an integer
larger than 0 and a is an integer between 0 to 9).  Note that
rightmost digit is also described as (9+1)^0 x a.  So, the number
1,728,000 can be described as ((9+1)^6 x 1) + ((9+1)^5 x 7) + ((9+1)^4
x 2) + ((9+1)^3 x 8) + ((9+1)^2 x 0) + ((9+1)^1 x 0) + ((9+1)^0 x 0).

Now there is a number that one wants to know if it is divisible by
9. That is:

        X = ((9+1)^n x a) + ((9+1)^(n-1) x b) + ((9+1)^(n-2) x c)
          ... + ((9+1)^0 x p)

As (9+1)^n etc. can be expanded to:

        9^n + (bunch of numbers that are multiple of 9) + 1

(note that 1^n is always 1) X becomes:

        X = (((9^n) x a) + ((bunch of numbers all divisible by 9) x a) + a)
          + (((9^(n-1)) x b) + ((another bunch of numbers all divisible by 9)
          x b ) + b) ...  + p

Then the only part of X that may not be divisible by 9 (remainder) is:

        Y = a + b+ c + ... p

If Y is divisible by 9, then X is divisible by 9 and Y is the sum of
all digits.  (On a side note, the same thing can be said about 3.  If
the sum of all digits of an integer is divisible by 3, the integer is
divisible by 3.  This is easily deduced from the above as 9 is the
square of 3.)

If Y is not a single digit, repeat the above until you get a single
digit number.  As only single digit number that happens to be
divisible by 9 is 9, if a number is divisible by 9, the sum of all the
digits ultimately reaches 9....

(Here comes a disclaimer:) I don't really remember this was the right
proof, and I was a bad student of math up to high school.  And I
haven't studied math since then.  Also, I did not learn math in
English, I am not familiar with those terms in English.

Now if we go back to Indological topic, that so many numbers are
divisible by 9 and therefore 3 seems rather to have something to do
with Babylonian arithmetics.  Didn't they have a system based on 36 or
60?  (It seems natural that 36 or 60 is important numbers in ancient
cultures because they are divisible by so many numbers including 12.
If one counts one's palm as one, he/she now has 6 fingers to count and
12 fingers in total.  Therefore counting up to 12 may not be so
strange.  Why in English etc. a unit of number ends at 12?  Shouldn't
one say "two-teen" instead of "twelve"?  I know Beavis sort of does
:-)

-- Kengo