Enable Javascript in your browser and then refresh this page, for a much enhanced experience.
Easy Fizz Buzz solution in Clear category for Fizz Buzz by dominieq
def checkio(number):
if number%15==0:
return "Fizz Buzz"
elif number%3==0:
return "Fizz"
elif number%5==0:
return "Buzz"
else:
return str(number)
Nov. 13, 2016