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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: a = random.randint(1, 99)

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

It says Wrong number of prints.

even.py
```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 == True:
a = random.randint(1, 99)
b = str(a)
if even_odd(a) == 1:
print("{} is even".format(b))
if even_odd(a) == 0:
print("{} is odd".format(b))
start-=1
``` Since `start` is initially `5`, the while condition will be false since 5 is not equivalent to `True`. You can use the " truthiness" of `start` instead as `while start:`
Also, since `even_odd()` returns a Boolean value, the return value can be used directly as the if condition without comparing to 1 or 0.
You don't have to creat `b` for use in the format method. `a` can be used directly. Format will convert to str for you.