# even_odd function

am getting a 'Wrong number of prints ' message after writing the attached code and am stuck

even.py
```import random
start = 5
while start < 5:
secret_num = random.randint(1, 99)

if even_odd(secret_num):
print("{} is even.").format(secret_num)
else:
print("{} is odd.").format(secret_num)
start -= 1

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
```

Your print statement has an error

```print("{} is even.").format(secret_num)
```

The format method is called on the String as show below:

```print("{} is even.".format(secret_num))
``` Hi Clainos,

Your problem is in the loop definition.

```start = 5
while start < 5:
```

Since you made `start` 5 and your program only enters the loop if `start` is less than 5, your loop will never execute.

Note that the task tells you to "Make a while loop that runs until start is falsey." This is telling you that your loop should run until `start` has a value that Python considers False. We know that `start` is an integer, so what types of integer are False? You can check this yourself in the Python interpreter by typing something like:

```5 == True
```

The interpreter will come back and tell you that is True. Similarly:

```1 == True
```

Interpreter will again tell you that is True. What about:

```0 == True
```

This time, Python will tell you that is False. So we know that 0 is a falsey value.

The body of your while loop decrements `start` at the end of every iteration, so your value of `start` will reach 0, at which time it will be False.

Therefore, you can set your while loop to run until `start` is False. This is as simple as writing:

```while start:
```

Hope that clears everything up.

Cheers

Alex

changed that and now getting the message 'Task one is no longer passing'