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

I don't understand the question really...

I suppose I don't understand the question. I thought I was following the directions.

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.
    start = 5

    while true:
        random_int = int(random.randint(1,99))

        if random_int % 2 == 0:
            print("{} is even".format(random_int))
        else:
            print("{} is odd".format(random_int))

        start -= 1
        return not num % 2

1 Answer

  • when start gets to 0 it will be false
  • you can leave the even odd function just like it is
  • your solution works if you make True false at the end point
  • True needs to be capitalized if you use it
  • your solution passes w/o the even odd function
  • I think your indents were off
  • yes this challenge can be a bit confusing
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: # Make a while loop that runs until start is falsey
    num = random.randint(1, 99) # Inside the loop, use random.randint(1, 99)
    if even_odd(num): # If that random number is even (use even_odd to find out)
        print("{} is even".format(num)) # putting the random number in the hole
    else: # Otherwise
        print("{} is odd".format(num)) # again using the random number
    start -= 1 # Finally, decrement start by 1