Welcome to the Treehouse Community

Want to collaborate on code errors? Have bugs you need feedback on? Looking for an extra set of eyes on your latest project? Get support with fellow developers, designers, and programmers of all backgrounds and skill levels here with the Treehouse Community! While you're at it, check out some resources Treehouse students have shared here.

Looking to learn something new?

Treehouse offers a seven day free trial for new students. Get access to thousands of hours of content and join thousands of Treehouse students and alumni in the community today.

Start your free trial

Python Python Basics (2015) Letter Game App Even or Odd Loop

Posted by Levi Perlei
Levi Perlei
1,876 Points

I'm stucked :(

I'm kinda confused here, how should i check if the number is even or odd?

even.py 
import random
start = 5
def even_odd(num):
    # If % 2 is 0, the number is even.
    # Since 0 is falsey, we have to invert it with not.
    return not num % 2
while True:
    num = random.randint(1, 99)
    even_odd

    if num :
        print("{} is even".format(num))
    else:
        print("{} is odd".format(num))
start - 1

1 Answer

Vidhya Sagar
Vidhya Sagar
1,568 Points
Vidhya Sagar
Vidhya Sagar
1,568 Points

You got to define it inside even_odd and syntax is missing a bit. Since you use %2 it gives us the remindder of the num you do the operation with . So if num%2 gives 1 it is odd ,if it gives 0 it is even ,use the modulo in the if statements and you're good to go.Try this 

import random
start=5

def even_odd(num):
    # If % 2 is 0, the number is even.
    # Since 0 is falsey, we have to invert it with not
    if num%2:
        print ("{} is odd".format(num))
    else:
        print ("{} is even".format(num))

    return not num % 2
while start:
    num=random.randint(1,99)
    even_odd(num)
    start=start-1