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

Minwook Jeong
Minwook Jeong
2,453 Points
Posted by Minwook Jeong
Minwook Jeong
2,453 Points

Help me out the challenge question

Quenstion:

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. I know it's a lot, but I know you can do it!

really confusing how I should express this question with Python language..

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.
    while start == False:
        return not num%2
        num = random.randint(1, 99)
        if num %2 == 0:
            print("{} is even".format(num))

        else:
            print("{} is odd".format(num))

2 Answers

Christopher Shaw
seal-mask
PLUS
.a{fill-rule:evenodd;}techdegree seal-36
Christopher Shaw
Python Web Development Techdegree Graduate 58,248 Points
Christopher Shaw
.a{fill-rule:evenodd;}techdegree seal-36
Christopher Shaw
Python Web Development Techdegree Graduate 58,248 Points

You should not put the while loop inside the function, the function should have been left as it was. You just call the function as you iterate the the while loop to determine if the random number is odd or even.

Khaleel Yusuf
Khaleel Yusuf
15,208 Points
Khaleel Yusuf
Khaleel Yusuf
15,208 Points

Can you please explain further? I don't understand. Thank you.

Minwook Jeong
Minwook Jeong
2,453 Points
Minwook Jeong
Minwook Jeong
2,453 Points 
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 == 0:
        print("{} is even".format(num))

    else:
        print("{} is odd".format(num))
    return not num % 2


while start:
    number = random.randint(1, 99)
    even_odd(number)
    start -= 1

Here is my solution, I referred to what Christopher advised me

def even_odd(num) is a sole tool to judge whether the random number selected by Python is odd or even, it doesnt have any function of loop so as Christopher I put while loop out of even_odd(num)

Then, we need to think about which loop we are going to use, Is it For Loop? or While Loop? I cannot explain it logically but I went with While Loop (based on how we use 'start' before)

while start: -> we set up the ending point of loop as start, while loop will end till the value is False which is 0

so while loop would begin from 5, and stop at 1 since we set up the decrement -1

Thus, it will provide 5 value