# MADE A COUPLE OF CORRECTIONS. STILL TELLING ME TASK ONE IS NO LONGER PASSING.

HERE IS THE ORIGINAL QUESTION

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!

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.
return not num % 2
start = 5
while start > 0:
num = random.randint(1, 99)
if even_odd(num) == 0:
print("{} is even".format(num))
else:
print("{} is odd".format(num))
start -+ = 1
```

this should work

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

def start(): start = 5

while start: num = random.randint(1, 99) if even_odd(num): print('{} is even'.format(num)) else: print('{} is odd'.format(num)) start -=1 The below syntax is not correct:

```start -+ = 1
```

You want to use:

```start -= 1
```

You need no spaces between minus and equal and the plus sign should not be there. You will then have to check the logic in the if / else statements, as it is printing the opposite of what is required:

```  if even_odd(num) == 0:
print("{} is even".format(num))
```
• the above is the same as saying to python: if even_odd() is False...print out that the number is even. Remember that 0 is falsey. The even_odd() function checks if the number is even and returns True if it is even.