[Practice Python Question] FizzBuzz | Python-3 Solution by APDaga

[Practice Python Question] FizzBuzz | Python-3 Solution by APDaga

I came across a Python 3 practice question. 

Here, I am providing my solution to the problem "FizzBuzz" with the intention to help you learn and understand Python 3 in a better way.

Please make use of my blog posts for learning purposes only and feel free to ask your questions in the comment box below in case of any doubt.

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Python Problem Statement:

Given a number n, for each integer i in the range from 1 to n inclusive, print one value per line as follows:

  • If i is a multiple of both 3 and 5, print FizzBuzz.

  • If i is a multiple of 3 (but not 5), print Fizz.

  • If i is a multiple of 5 (but not 3), print Buzz.

  • If i is not a multiple of 3 or 5, print the value of i.

Function Description:

Complete the function fizzBuzz in the editor below.

fizzBuzz has the following parameter(s):
  • int n: upper limit of values to test (inclusive)

  • Returns: NONE

  • Prints:The function must print the appropriate response for each value i in the set {1, 2, ... n} in ascending order, each on a separate line.


0 < n < 2 × 10^5

Sample Input:

STDIN    Function
-----    --------
15    →  n = 15

Sample Output:



  • The numbers 3, 6, 9, and 12 are multiples of 3 (but not 5), so print Fizz on those lines.

  • The numbers 5 and 10 are multiples of 5 (but not 3), so print Buzz on those lines.

  • The number 15 is a multiple of both 3 and 5, so print FizzBuzz on that line.

  • None of the other values is a multiple of either 3 or 5, so print the value of i on those lines.

Python 3 Solution:

#!/bin/python3 ...
# Complete the 'fizzBuzz' function below.
# The function accepts INTEGER n as parameter.

def fizzBuzz(n):
	# Write your code here
	while i<=n:
		if i%3==0 and i%5==0:
	        elif i%3==0 and i%5!=0:
	        elif i%3!=0 and i%5==0:
	        elif i%3!=0 and i%5!=0:

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Feel free to ask doubts in the comment section. I will try my best to answer it.
If you find this helpful by any means, like, comment and share the post.
This is the simplest way to encourage me to keep doing such work.

Thanks & Regards,
-Akshay P Daga

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