def find(num): # password logic here if num%2 == 0: numtype="even" else: numtype = "odd" return numtypenum = int(input("Enter the number: ")) # 1. Take her inputnumtype = find(num) # 2. Speak to the find functionprint("Given number is",numtype). # 3. Publish if the number is also or odd


You are watching: How to check if a number is even in python

coder# python challenge07.py enter the number: 5Given number is oddcoder# python challenge07.py get in the number: 8Given number is even
Explanation:-input() duty is provided to take user inputfind() function is called to to examine if a number is off/even. This duty returns numtype together odd/evenAt last, publish if the offered number is odd/even

solution #2

Avoid usage of else block by assigning a default value (odd) to numtype


def find(num): # code logic right here numtype = "odd" if num%2 == 0: numtype="even" return numtypenum = int(input("Enter the number: ")) # take your inputnumtype = find(num) # call the find functionprint("Given number is",numtype) # print if the number is also or odd
coder# python challenge07.py enter the number: 5Given number is oddcoder# python challenge07.py go into the number: 8Given number is even

def find(num): # code logic right here if num%2 == 0: return "even" return "odd"num = int(input("Enter the number: ")) # take your inputnumtype = find(num) # contact the find functionprint("Given number is",numtype) # print if the number is even or odd
coder# python challenge07.py go into the number: 5Given number is oddcoder# python challenge07.py get in the number: 8Given number is even

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Tim West
The bitwise-and operator & behavior is trivia, yet it"s useful due to the fact that the bitwise operators tend to be consistent across many programming languages.


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