Practice Makes Perfect

#1

Many have encounter this issue. What is the problem here?

``````def cube(number):
return number**3
#define cube function
def by_three(number):
return number % 3
#define the by_three function
if by_three(number) == 0:
# Add the condition if the modulo is zero in the by_three
#function is divided evenly
return cube(number)
#outcome of the condition is true
else:
#outcome if the condition is false
return False``````

#2

Hey,can you format code ?

1.Copy and paste you code here
2. Select all of your code
3. use `</>` to format it

#3

@newb528
In this line of code..
Problem1.

``````def by_three(number):
return number % 3``````

You do not have to return number % 3 ?

hint*

Problem2.

If/else statement is inside the function and you're calling `by_three()` function inside function itself as if's condition ?...
Your if/else statement will never execute as your function halts code execution after it sees `return` statement and runs it.

Note..

``````def anyfunc():
return 3
#code
#more code``````

As function sees the return statement ,it halts its execution and takes execution out of the body of function without running any code after return and

``````#code
#more code``````

never actually gets executed!

`if by_three(number) == 0:`

You have to check if number is divisible by 3 or not.

Example..

If I had to check if a number is divisible by 2 or not.
I'd write something like this..
(If condition is inside If -statement)

``````if number % 2 == 0:

hint**
here you have to check if number is divisible by 3?

#4

Right below is what I worte

``````def cube (number):
return number**3

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

#5

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