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

Pii Tiimii
Pii Tiimii
3,036 Points
Posted by Pii Tiimii
Pii Tiimii
3,036 Points

I cant get pass this challenge

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 start: a = random.randint(1, 99)

if even_odd(a):
  print("{} is even".format(str(a)))
else:
  print("{} is odd".format(str(a)))
start-=1

It says Wrong number of prints.

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 start == True:
  a = random.randint(1, 99)
  b = str(a)
  if even_odd(a) == 1:
    print("{} is even".format(b))
  if even_odd(a) == 0:
    print("{} is odd".format(b))
  start-=1

1 Answer

Chris Freeman
MOD
Chris Freeman
Treehouse Moderator 68,423 Points
Chris Freeman
Chris Freeman
Treehouse Moderator 68,423 Points

Since start is initially 5, the while condition will be false since 5 is not equivalent to True. You can use the " truthiness" of start instead as while start:

Also, since even_odd() returns a Boolean value, the return value can be used directly as the if condition without comparing to 1 or 0.

You don't have to creat b for use in the format method. a can be used directly. Format will convert to str for you.