Outside of the mod operator, you could always use a combo of floor() and ceiling() (which I assume most languages have in a standard library). e.g.:

  from math import floor,ceil
  def divisible_by_3(n):
    result = n / 3.0
    return ceil(result) == floor(result)

  def divisible_by_3(n):
    result = n / 3.0
    return int(result) == result

If your language doesn't do integer division that way, the naive approach is even easier if you know how to round: x / y == round(x / y) iff x % y == 0.

There are many, many other approaches which will work, too. I saw one guy hand-build an array of ints, initialize to zero, loop over it once with arr[i++] = 0; arr[i++] = 0; arr[i++] = 3 to set the multiples of 3, then do the same thing with the fives except checking to see if there was already a 3 there, then looping over the array a fourth time to handle the actual printing.

That is the kind of competent, worksmanlike programming that runs the world while the can't-do-FizzBuzz guys are hopefully not touching the code too much.

    var s = require('sys');
    for (var i = 1; i < 101; i++) {
      if (i%3 == 0 && i%5 == 0) s.puts('fizzbuzz');
      else if (i%3 == 0) s.puts('fizz');
      else if (i%5 == 0) s.puts('buzz');
      else s.puts(i);
    }

    node.fizzbuzz.js