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

Posted by rgxtfolgir
rgxtfolgir
4,249 Points

Help with letter game part 3 code challenge

The challenge is:

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!

With my code I get an error saying: "Wrong number of prints"

What is wrong?

even.py 
import random

start = 5

def even_odd(random_num):
    while start == True:

        random_num = random.randint(1, 99)
        new_num = random_num % 2

        if new_num != 0:
            print("{} is odd".format(new_num))

        else:
            print("{} is even".format(new_num))

    else:
        start -1


    return not num % 2
artkay
artkay
5,401 Points
artkay
artkay
5,401 Points

I would start with fixing errors.

  1. I don't think there is a need for a parameter in the function, just even_odd() should be enough
  2. I don't think you need this return (and - also - you returning a var that has no value at all)
  3. start -1 will never occur while it is in else: part - you want to delete the else: line
  4. 5 != True, so you should check if start greater than zero, i.e. while start > 0:

Hope this helps.

Art.