# Question

Given a non-negative integer num, repeatedly add all its digits until the result has only one digit.

For example:

Given `num = 38`

, the process is like: `3 + 8 = 11`

, `1 + 1 = 2`

. Since 2 has only one digit, return it.

Follow up: Could you do it without any loop/recursion in O(1) runtime?

# Solution

12,345 = 1 × (9,999 + 1) + 2 × (999 + 1) + 3 × (99 + 1) + 4 × (9 + 1) + 5.

12,345 = (1 × 9,999 + 2 × 999 + 3 × 99 + 4 × 9) + (1 + 2 + 3 + 4 + 5).

Thus, add digits is the same as modulo 9 except when the modulo is 0, the sum is 9.

```
class Solution(object):
def addDigits(self, num):
"""
:type num: int
:rtype: int
"""
if num == 0:
return 0
mod = num % 9
return 9 if mod == 0 else mod
```