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First solution in Clear category for Fizz Buzz by Rafal__Kotas
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. 1, 2016