Cube(1) returned 27 instead of 1


#1


I couldn't find anyone else that has the same problem as I did. I've tried using other examples on this forum and everyone else's code looks exactly like mine so I'm not sure what is causing the issue. I appreciate the help!


https://www.codecademy.com/courses/python-beginner-c7VZg/1/5?curriculum_id=4f89dab3d788890003000096#


I'm receiving the following error:
cube(1) returned 27 instead of 1


The coding appears to be correct and should give me a pass but it's not.


def cube(number):
   return 3 ** 3

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


#2

your cube function:

def cube(number):
   return 3 ** 3

why 3 ** 3? if we call the function:

def cube(number):
   return 3 ** 3

print cube(5)

the number 5 we pass as argument doesn't matter, it only calculates cube of 3. Why is that?


#3

2.) Make that function return the cube of that number (i.e. that number multiplied by itself and multiplied by itself once again).

And if you need another hint read your step 2 instructions carefully its tricky but with Stetim's example you should be able to correct yourself! :thumbsup:


#4

Thank you guys! I see what the error is. Even though I was returning, I wasn't defining.


#5

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