# Practice Makes Perfect

#1

def cube(number):
return number **3

def by_three(number):
if number % 3:
cube(number)
else:
return False

# I cant figure out what is wrong. Plz help.

#2

4. if that number is divisible by 3,
-> If it is divisible by 3, the result would be 0 (number % 3), right? But 0 is False and 1 is True in python, so you have to do it this way, that it actually takes this route if it is divisible by 3:

`if number % 3 == 0`

or

``if not number % 3:``

Second issue is:

You have to return the value, or your function will return None, if it is divisible by 3

``return cube(number)``

Hope it helps

#3

Thank you, It did help

#4

def cube(number):
return number **3

def by_three(number):
if number % 3 == 0:
return cube(number)
else:
return False

#5

Hi
I have this code - and it does not work

f cube(number):
return number**3

def by_three(number):
if number % 3 == 0:
return cube(number)
else:
return False

#6

#7

I finally got it to work using this

def cube(number):
return number ** 3

def by_three(number):
if number % 3 == 0:
return cube(number)
else:
return False

make sure your indentations are correct. I had the code correct for over an hour and kept getting an error due to the return line not being indented