FizzBuzz Python Solution

FizzBuzz.py

def fizzbuzz(n):

    if n % 3 == 0 and n % 5 == 0:
        return 'FizzBuzz'
    elif n % 3 == 0:
        return 'Fizz'
    elif n % 5 == 0:
        return 'Buzz'
    else:
        return str(n)

print "\n".join(fizzbuzz(n) for n in xrange(1, 21))

#!/bin/python3

import math
import os
import random
import re
import sys

Complete the 'fizzBuzz' function below.

The function accepts INTEGER n as parameter.

def fizzBuzz(n):

for x in list(range(1,n+1)):
output = ""
if(x % 3 == 0):
output += 'Fizz'
if(x % 5 == 0):
output += 'Buzz'
if(output == ""):
output += str(x)

print(output)
# Write your code here

if name == 'main':
n = int(input().strip())

fizzBuzz(n)

In an interval of (1,N+1) the fuction will print Fizz if i value is divisible by 3, Buzz if is divisible by 5, and FizzBuzz if divisible by both. Else if none is true, it prints i.

def fizzbuzz(n):
   for i in range(1,n+1):
   txt=''

   if(x%3==0):
     txt+='Fizz'

   if(x%5==0):
     txt+='Buzz'

   print(txt) if len(txt)>0 else print(i)

   return

if __name__ = '__main__':
   n = int(input().strip())
   fizzbuzz(n)

for n in range(1, 101):
    if not n % 3 and not n % 5:
        print('FizzBuzz')
    elif not n % 3:
        print('Fizz')
    elif not n % 5:
        print('Buzz')
    else:
        print(n)

In 480 bits (60 bytes) or 60 chars:

for i in range(100):print(i%3//2*'Fizz'+i%5//4*'Buzz'or i+1)

def fizzBuzz(n):
# Write your code here
for i in range(1, n + 1):
if i % 5 == 0 and i % 5 == 0:
print("FizzBuzz")
elif i % 3 == 0:
print("Fizz")
elif i % 5 == 0:
print("Buzz")
elif i % 3 != 0 or i % 5 != 0:
print(str(i))

Proof, code at work

def fizzBuzz():
stack = range(1, 100, 1)
for item in stack:
if item % 3 == 0 and item % 5 == 0:
item = "FizzBuzz"
elif item % 3 == 0:
item = "Fizz"
elif item % 5 == 0:
item = "Buzz"

    print(item)

fizzBuzz()