>     Bill Cornutt's answer ("It is the classic solution to the 'one
>     potato, two potato' problem") would, of course, be the USUAL
>     reason, but my need for the routine is a little different:
>
>     The "divisible by 3" routine is only the first step of a much
>     larger process:  I'm trying to see if the input number is EQUAL
>     to 3.

Hmm... this sounds similar in concept to the Leeland (sp?) sort:

[1] Check to see if first item is less than second, second less than third,
    etc.

[2] If all checks pass, sort is complete.  Otherwise...

[3] Randomly shuffle the numbers and go back to step [1].