I have been trying to solve this Python "Depending on Parity" problem for two days straight and have no idea what to do now.

Question

The problem :

elif Statements

Write a program that takes a number as input and
- returns its double, if the number is even
- returns its triple, if the number is odd
- returns 0, if number is 0

Sample Input:
1

Sample Output:
3

An integer is even if it is divisible by two and odd if it is not even.

My codes :

1).

number = input()
#your code goes here
even = number % 2 == 0
odd = number % 2 != 0
if even:
print(number * 2)
elif odd:
print(number * 3)
else:
print(0)

2).

number = input()
#your code goes here
if number % 2 == 0:
print(number * 2)
elif number % 2 != 0:
print(number * 3)
elif number == 0:
print(0)
else:
print(0)

Any suggestions will be greatly appreciated.

Answer 1

Кирилл Чернятин input() function default return type in string , convert the value to integer first by using int(input())

Answer 2

Here's a suggestion :
You don't even need to evaluate if number is equal to zero or not as it will always pass the first *if* condition and will print (0*2) which is also zero.

Answer 3

Abhay

Holy Toledo. It worked ! I have realized my mistake , thank you for information & have a great day ! :D.

Answer 4

You are welcome , have a nice day as well :)

Answer 5

Abhay

Thank you ! :D.

Answer 6

number = input()
#your code goes here
if number % 2 == 0:
print(number * 2)
elif number % 2 != 0:
print(number * 3)
elif number == 0:
print(0)
else:
print(0)

easiest way to solved is there !!!!!!! Look here man !

Answer 7

Arsenic

Huh. Yeah , that's actually a really solid advice. Thank you for information ! :).

Answer 8

num=int(input('Enter the number: '))
if(num%2==0 and num!=0):
num=num*2
print('Double of given number is ',num)
elif(not num%2==0 and num!=0):
num=num*3
print('Triple of given number is ',num)
else:
print('0')