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Python Python Basics (2015) Number Game App Even or Odd

Laknath Gunathilake
Laknath Gunathilake
1,860 Points

can't seem to pass the challenge

when I run the code it says TypeError: even_odd() takes 0 positional arguments but 1 was given

even.py
def even_odd():
    number=int(input("enter number:"))
    if number%2==0:
      return True 
    else:
      return False 

1 Answer

Grigorij Schleifer
Grigorij Schleifer
10,365 Points

Hi Laknath,

you can use this code:

you donΒ΄t need to ask the user for the number . Give the number as a method argument for the even_odd method.

def even_odd(number):
# number is the methods argument

The rest ist fine :)

Grigorij