# [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.

## 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
```

```1
2
Fizz
4
Buzz
Fizz
7
8
Fizz
Buzz
11
Fizz
13
14
FizzBuzz```

## Explanation:

• 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):
i=1;
while i<=n:
if i%3==0 and i%5==0:
print('FizzBuzz')
elif i%3==0 and i%5!=0:
print('Fizz')
elif i%3!=0 and i%5==0:
print('Buzz')
elif i%3!=0 and i%5!=0:
print(i)
i+=1```

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