Fizz-Buzz-Fail

So, I just learned about the Fizz Buzz test that is designed to weed out CS applicants that don't know anything.  It's pretty simple:

Write a program to do the following:
1.  For any number divisible by 3, print "Fizz"
2.  For any number divisible by 5, print "Buzz"
3.  For any number divisible by 3 and 5, print "FizzBuzz"
4.  Do this in order for all numbers between 1 and 100, inclusive

Simple, right?

Well, it turns out that a shocking number of applicants can't do this at all, and even more can't do it right.  Ok, I believe that.  There are a lot of people that get through CS programs that aren't cut out for the discipline.

What is most shocking to me, however, is how many different websites out there offer a solution that follows the following format:

For each int i from 1 to 100 (including 100):
    if i is divisible by 3:
        print "Fizz"
    else if i is divisible by 5:
        print "Buzz"
    else if i is divisible by 3 and i is divisible by 5:
        print "FizzBuzz"

/facepalm, am I right?  If you can't see what's wrong with this, and you are a CS person, you need to quit your job/school and go learn groundskeeping.

By the way, my solution in C:

for( int i=1; i<=100; i++ ) printf( "%s%s\n", i%3?"":"Fizz", i%5?"":"Buzz" );

C++

for( int i=1; i<=100; i++ ) cout << i%3?"":"Fizz" << i%5?"":"Buzz" << endl;

Python:

print '\n'.join( [ ( ( i%3 and ' ' or 'Fizz' ) + ( not i%5 and 'Buzz' or ' ' ) ).strip() for i in range(1,101) ] )

Python (second method)

print '\n'.join( [ '%s%s' % ( not i%3 and 'Fizz' or '', not i%5 and 'Buzz' or '' ) for i in range(1,101) ] )

Python ( third method)

print '\n'.join( [ '%s%s' % ( '' if i%3 else 'Fizz', '' if i%5 else 'Buzz' ) for i in range(1,101) ] )

P.S.  The python expression was trickier, so I had fun with three different approaches.

P.P.S  Python 3.0 uses print as a function, so it could be included in the list comprehension and eliminate the needed join() call