def cube(number):

return "that number"

def by_three(number):

if cube(number) % 3 == 0:

else :

return False

def cube(number):

return "that number"

def by_three(number):

if cube(number) % 3 == 0:

else :

return False

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

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??

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

number=cube(number) #compose your code

return number

thx bro, but i already pass it.

now i get trouble insection 17/19.

sorry with my bad english

++++++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'"

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