How to check if the number is divisible by another one Python

Question

I got a problem, when I need to check ff the number is divisible by 3, it should return “Divi”. If it is divisible by 3, it should return “Sible”. If it is divisible by both 2 and 3, it should return “DiviSible”.

I tried this code, however if my x = 6, I want to only show "Divisible" and not "divi" and "sible". However, this code returns the three values to me. Any idea on how to make it work? THANKS!

def fun_divi():
  if(x%2==0):
    print("Divi")
  if(x%3==0):
    print("Sible")
    if(x%2==0) and (x%3==0):
      print("Divisible")
  else:
    print("Not divisible by any")
fizz_buzz()

Answer 1

if(x%2==0) and (x%3==0):
  print("Divisible")
elif(x%2==0):
  print("Divi")
elif(x%3==0):
  print("Sible") 
else:
  print("Not divisible by any")

You should check both diversity first.

Answer 2

You have to use elif , fix your indentation, and bring the multicase if statement to the top. And no need for the brackets in the statements.

def fun_divi(x):
    if x%2==0 and x%3==0:
        print("Divisible")
    elif x%2==0:
        print("Divi")
    elif x%3==0:
        print("Sible")  
    else:
        print("Not divisible by any")

If you would like a simpler way:

def fun_divi(x):

    if not x % 2 or not x % 3:
        if not x%2: print("Divi", end = "")
        if not x%3: print("S" if x%2 else "s", "ible", end = "", sep = "")
        print("")
    else:print("Not divisible by any")

Testing:

 >>> fun_divi(6) Divisible >>> fun_divi(5) Not divisible by any >>> fun_divi(3) Sible >>> fun_divi(2) Divi >>>

---

Answer 3

Use a single resulting variable:

def fun_divi(x):
    res = ''
    if (x % 2 == 0):
        res += "divi"
    if (x % 3 == 0):
        res += "sible"

    print("Not divisible by any" if not res else res.capitalize())

fun_divi(6)   # Divisible
fun_divi(9)   # Sible
fun_divi(5)   # Not divisible by any

Answer 4

thats because you should use the conditions in descending order like

def fun_divi():
    if(x%2==0) and (x%3==0):
        print("Divisible")
    elif(x%3==0):
        print("Divi")
    elif(x%2==0):
        print("Sible")
    else:
        print("Not divisible by any")

Answer 5

Your question is the typical FizzBuzz excersise, which by the way is famous and you should have researched before. Anyway, your code has a couple of things you should fix first:

1. Indentation - Try using an editor previous to write your code and paste it then.
2. If case handling - You are evaluating a case with two conditions at the end, but you're evaluating the same cases individually first, this will left your multicase unused because one of the conditions was already met.
3. Brackets - If cases don't use brackets, so remove them.
4. Variables - Your code shows a call to a function than doesn't exist, and it evaluates variables that you haven't created - I'm referring to "x" in all your evaluations.

After fixing all that, your code should look something like this:

def fun_divi(x):
    if x%2==0 and x%3==0:
        print("DiviSible")

    elif x%2==0:
        print("Divi")

    elif x%3==0:
        print("Sible")

    else:
        print("Not divisible by any")

Answer 6

Following is the code

def func(x): return "DiviSible" if (x%2==0 and x%3==0) else "Sible" if x%3==0 else "Divi" if x%2==0 else "Not Divisible by any"

func(12) 'DiviSible' func(3) 'Divi' func(6) 'DiviSible' func(10) 'Sible' func(11) 'Not Divisible by any'