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 trialPeter Falcone
Courses Plus Student 1,108 Pointsa little confusion with even.py
this is my best shot, haven't submitted it yet, because for one thing, I really don't know what to do with the "return not num % 2" bit, so as you can see, I just wrote what made sense to me
import random
start = 5
while start:
num = random.randint(1, 99)
if start = False:
break
def even_odd(num):
if num % 2 == 0:
print("{} is even").format(num)
elif num % 2 != 0:
print("{} is odd").format(num)
# If % 2 is 0, the number is even.
# Since 0 is falsey, we have to invert it with not.
return not num % 2
start -= 1
1 Answer
Christopher Shaw
Python Web Development Techdegree Graduate 58,248 PointsUnfortunately, not on track. You should have left even_odd as it was and called the function as you needed it to test a value. So after the original function, start the while loop.
if not even_odd(num):
# matches false
#or
if even_odd(num):
#matches true
Peter Falcone
Courses Plus Student 1,108 PointsPeter Falcone
Courses Plus Student 1,108 Pointsjust gonna keep working at it then, thanks