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

Mark Kohner
Mark Kohner
1,760 Points

python basics even_odd check

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. return not num % 2

while start: num = random.randint(1, 99) if even_odd(num): print("{} is even.".format(num)) else: print("{} is odd".format(num)) start -= 1

Can you please show us the code challenge's question?

Mark Kohner
Mark Kohner
1,760 Points

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!

2 Answers

You did a great job! It took me a while to find out why it isn't working, and I found that the part where the start variable decreasing by one shouldn't be in the if statement. It should be outside of the if statement but inside the while loop. Try this:

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.
    return not num % 2

while start:
    num = random.randint(1, 99)
    if even_odd(num):
        print("{} is even.".format(num))
    start -= 1

Hope this helps! ~Alex :smile:

You are missing the other case here. You have to implement the num is odd case as well or you will get an infinite loop if the number is odd. It would also be better to take the decrement of your start value out of the if block and put it at the end of the while block so you avoid the infinite loop here. You're also avoiding duplicate code doing this :D.