[patio11]

Let's see: x / y * y == x iff x % y == 0, if your language handles integer division like most do.

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.