[DEADBEEF]

Something like... (Pseudocode)

    for( i= 1,100 )
    {
      if( i%3==0 ) { print( 'Fizz' ) }
      if( i%5==0 ) { print( 'Buzz' ) }
      print( '\n' )
    }

There's no need to treat FizzBuzz as a special case, as if the number is divisible by both it will have already met the conditions of the previous two statements.

I wonder if there's some magic you could sprinkle in to the iterator so it only iterates through numbers which are divisible by 3/5, ignoring the rest?

It probably wouldn't speed the operation up much (if at all) in this case, in a more complex real world scenario though it's best to look at every angle.

[RiderOfGiraffes]

You haven't printed any of the other numbers. Read the spec more carefully. You still need to print 1, 2, 4, etc.