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

Posted by fionaquinn2
fionaquinn2
2,032 Points

Even or Odd

I'm stuck with this task.
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'm getting the error message "too many prints".

Please see my code attached. How do I move on?

even.py 
import random
start = 5
def even_odd(num):
    if (num % 2) == 0:
        print("{0} is Even".format(num))
    else:
        print("{0} is Odd".format(num))

while start == 5:
    num =random.randint(1,99)
    even_odd(num)
    start = start -1

1 Answer

Sneha Nagpaul
Sneha Nagpaul
10,124 Points
Sneha Nagpaul
Sneha Nagpaul
10,124 Points

At first glance, the while loop condition looks odd. start is 5 when we get to it the first time and then it gets decremented. Which means, the while condition start == 5 gets violated after the first execution.

I guess it's not such a big deal, but in my approach I chose not to modify the even_odd() function provided and tried to use it with print statements inside the while loop.