> Also, to tell if a number is divisible by three, add the digits.  If the
> sum is divisible by three, so is the original number.

I've never quite understood how to prove this sort of thing.  Entirely too
discrete, or something.  Furthermore...  Does this HELP any if you have a
binary number in a computer?  After all, you'd have to derive the decimal
digits of the number, which (usually) involves successive and multiple
divisions by 10, which is a lot harder than just dividing by three in
the first place...

I suppose that if you have an easy way to convert to base 9, then all
the numbers divisible by 9 will end in zero, right?

BillW

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