Python: digital root of a number: it returns None

Question

Given a number, add up the digits to get a new number. Add up the digits of that to get another new number. Keep doing this until you get a number that has only one digit. That number is the digital root the given number.

For example, if n = 45893, we add up the digits to get 4 + 5 + 8 + 9 + 3 = 29. We then add up the digits of 29 to get 2 + 9 = 11. We then add up the digits of 11 to get 1 + 1 = 2. Since 2 has only one digit, 2 is our digital root.

def sum_digit_root(num):
    n=str(num)
    if len(n)<=1:
        return num

    else:
        sum=0
        for i in range(len(n) ):
            sum+=int(n[i] )

        sum_digit_root(sum)


print('\n---print sum_digit_root(num)   ---------------------------', sum_digit_root(45893))

yes! changed to ' return sum_digit_root(sum) ' , now it works. Thanks a lot. — xin pds, Feb 25 '20 at 11:52

score 2 · Accepted Answer · answered Feb 25 '20 at 12:02

You forget the return statement in the else branch:

def sum_digit_root(num):
    n=str(num)
    if len(n)<=1:
        return num

    else:
        sum=0
        for i in range(len(n) ):
            sum+=int(n[i] )

        return sum_digit_root(sum)
    #   ^^^^^^

Python: digital root of a number: it returns None

1 Answers1