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 trialJamel Stringer
Courses Plus Student 906 PointsI need help
Challenge Task 3 of 3
Alright, last step but it's a big one. Make a while loop that runs until start is falsey. Inside the loop, use random.randint(1, 99) to get a random number between 1 and 99. If that random number is even (use even_odd to find out), print "{} is even", putting the random number in the hole. Otherwise, print "{} is odd", again using the random number. Finally, decrement start by 1.
Please I need to understand what I am doing wrong. This is my last exercise before I can move on.
import random
start = 5
while start == 5:
def even_odd(num):
num = random.randint(1, 99)
if randNum % 2 == 0:
print("{} is even".format(num)
else:
print("{} is odd".format(num)
start -1
# If % 2 is 0, the number is even.
# Since 0 is falsey, we have to invert it with not.
return not num % 2
2 Answers
Ryan S
27,276 PointsHi Jamel,
You aren't supposed to modify the even_odd()
function. It was already written for you to be called upon in your while loop. The function will return true if a number is even, and false if it is odd. All your code should be outside of the function.
Good luck.
Jamel Stringer
Courses Plus Student 906 PointsThank you Ryan that makes sense, but now I'm getting "wrong number of prints" error.....SMH!
import random start = 5 while start is True: num = random.randint(1, 99) if num % 2 ==0: print("{} is even".format(num)) else: print("{} is odd".format(num)) start -= 1
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
Ryan S
27,276 PointsYou have the right idea with the logic and it is possible to pass the challenge by going that route, however the challenge prefers that you actually use the even_odd()
` function to determine whether the random integer is even or odd. So your code should be written below the function in order to make use of it.
eg:
if even_odd(num) == True:
# code here
Also, take another look at your while condition. The "is" keyword is not meant for this type of comparison. You should make sure to use the standard comparison operators, like "==" and "!=". Also, remember that the loop will run until start is False, and the integer 0 is evaluated as False.
Hope this helps.