A trick I learned in elementary school to determine whether or not a number was divisible by three is to add all of the integers in the number together and to divide the resulting sum by three. If there is no remainder from dividing the sum by three, then the original number is divisible by three as well.

Given a series of numbers as a string, determine if the number represented by the string is divisible by three.

You can expect all test case arguments to be strings representing values greater than 0. 

Example:

```
"123"      -> true
"8409"     -> true
"100853"   -> false
"33333333" -> true
"7"        -> false
```

Difficulty: introductory

Starter code:

def divisible_by_three(st):
	

Solutions:
def divisible_by_three(s): 
    return int(s) % 3 == 0



def divisible_by_three(st):
    return (sum([int(i) for i in st]) % 3) == 0


def divisible_by_three(st): 
    return sum([int(c) for c in st]) % 3 == 0

import re
def divisible_by_three(st):
    return re.match("(((((0|3)|6)|9)|((2|5)|8)((((0|3)|6)|9))*((1|4)|7))|(((1|4)|7)|((2|5)|8)((((0|3)|6)|9))*((2|5)|8))(((((0|3)|6)|9)|((1|4)|7)((((0|3)|6)|9))*((2|5)|8)))*(((2|5)|8)|((1|4)|7)((((0|3)|6)|9))*((1|4)|7)))+$", st) != None

def divisible_by_three(s):
    return not int(s) % 3

# You can just do "int(st) % 3 == 0"

def divisible_by_three(st): 
    while len(st) > 1:
        st = str(sum(map(int, st)))
    return st in "369"

def divisible_by_three(st): 
    num_str = str(st)
    sum = 0
    # iterate and add the numbers up
    for i in range(0, len(num_str)):
        sum += int(num_str[i])
    # check if numbers are divisible by 3
    if sum % 3 == 0:
        return True
    else:
        return False
    pass

def divisible_by_three(st): 
    sum = 0
    while int(st) > 0:
       digit = int(st)%10
       sum = sum + digit
       st = int(st)//10
    if sum%3 == 0:
        return True
    else:
        return False 

def divisible_by_three(string):
    return not (sum(int(n) for n in string) % 3)

def divisible_by_three(stg): 
    return sum(int(d) for d in stg) % 3 == 0
    
# `int(stg) % 3 == 0` is straight but not what was asked