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Second solution in Clear category for Fizz Buzz by rogos
def fizzbuzz(number):
if(number % 3 == 0 and number % 5 ==0):
return "Fizz Buzz"
elif(number%3==0):
return "Fizz"
elif(number%5==0):
return "Buzz"
elif(number%3!=0 and number%5!=0):
return number
def checkio(number):
res = fizzbuzz(number)
return str(res)
Jan. 20, 2017
