Errrrh,

I had a look in an old maths book I have, and theres an easier way than
all that "casting out nines" and base 4 stuff I told you about earlier:

Apparently if you add the digits of the number to be tested, sucessively
until theres only 1 digit left, then if, and only if, the result is 3 or
6 or 9 then the original number was divisible by 3.

e.g. Is 12345 divisible by 3?
1+2+3+4+5 = 15
1+5 = 6
so yes 12345 is divisible by 3.

Now, about that code...

--
erik