FizzBuzz in Ruby without conditionals

gistfile1.rb

def fizzbuzz limit
  i = 1
  answers = [['fizzbuzz', 'buzz'], ['fizz', i]]
  loop do
    puts answers[(i % 5) >> (Math.log2((i % 5) + 1).ceil - 1)][(i % 3) >> (Math.log2((i % 3) + 1).ceil - 1)]
    i += 1
    1/((limit + 1) - i)
    answers[1][1] = i
  end
end

def fizzbuzzWat limit
  answers = [['fizzbuzz', 'buzz'], ['fizz', 1]]
  loop do
    puts (answers[(answers[1][1] % 5) >> (Math.log2((answers[1][1] % 5) + 1).ceil - 1)][(answers[1][1] % 3) >> (Math.log2((answers[1][1] % 3) + 1).ceil - 1)] * ((answers[1][1] += ((limit + 1) - answers[1][1])/((limit + 1) - answers[1][1])) / answers[1][1]))
  end
end

A readable version, and a shortened, "fun" version