Welcome to the Treehouse Community

The Treehouse Community is a meeting place for developers, designers, and programmers of all backgrounds and skill levels to get support. Collaborate here on code errors or bugs that you need feedback on, or asking for an extra set of eyes on your latest project. Join thousands of Treehouse students and alumni in the community today. (Note: Only Treehouse students can comment or ask questions, but non-students are welcome to browse our conversations.)

Looking to learn something new?

Treehouse offers a seven day free trial for new students. Get access to thousands of hours of content and a supportive community. Start your free trial today.

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

What's wrong?

I just don't know, I'm out of ideas, please help me.. :(

even.py
import random

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

start = 5

while start == True:
    numb = random.randint(1, 99)
    if numb % 2 == 0:
        print("{} is even".format(numb))
    else:
        print("{} is odd".format(numb))
    start -= 1

1 Answer

hamsternation
hamsternation
26,577 Points

The problem here is with your while statement:

while start == True:

In the console, try assigning start to a number, say 5. Then type in >>> start == True

What do you get?

You'd get False

That's because while ints are truthy they're not exactly == True. That's why the comparison returns false.

To revise your statement, you could change your while statement to:

while start:

I know this looks the same, but in this case, the while loop evaluates the truthiness of the variable start. And when it goes down to 0, it becomes falsey.

Hope this helps!

:)