# 6/19 I don't know

#1

def cube(number):
return "that number"

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

#2

your problem is the return of the funtions:
in the first is something like: return number*number*number
in the second you lost the return after the if

#3

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

what is wrong in mycode??

#4

In the second funtion, the cube(number) is not assigned to the return value, you cant do two things:

return cube(number) #include all in one line. I recomend this one
or
return number

#5

thx bro, but i already pass it.
now i get trouble insection 17/19.

#6

¿Cual es tu problema?

#7

why is it:

if number%3 ==0 ?

I don't understand why it's set equal to 0

#8

++++++Intruction 17/19+++++
First, def a function, shut_down, that takes one argument s. Don't forget the parentheses or the colon!Then, if the shut_down function receives an s equal to "yes", it should return "Shutting down"Alternatively, elif s is equal to "no", then the function should return "Shutdown aborted".Finally, if shut_down gets anything other than those inputs, the function should return "Sorry"
++++++mycode++++++++
def shut_down(s):
yes="Shutting down"
no="Shutdown aborted"
return s
if s==yes:
shut_down(yes)
elif s==no:
shut_down(no)
else:
shut_down("Sorry")

"Oops, try again.
Your function failed on the message yes. It returned 'yes' when it should have returned 'Shutting down'"

#9

jgonzal5
you need to somehow define the sentence "if that number is divisible by 3" which you can do like this: "number % 3 == 0", which means you want only numbers divisible by 3