Description:

#Task:

Write a function that returns true if the number is a "Very Even" number.

If a number is a single digit, then it is simply "Very Even" if it itself is even.

If it has 2 or more digits, it is "Very Even" if the sum of it's digits is "Very Even".


#Examples:
```
input(88) => returns false -> 8 + 8 = 16 -> 1 + 6 = 7 => 7 is odd 

input(222) => returns true

input(5) => returns false

input(841) => returns true -> 8 + 4 + 1 = 13 -> 1 + 3 => 4 is even
```

Note: The numbers will always be 0 or positive integers!

Difficulty: introductory

Starter code:

def is_very_even_number(n):
	

Solutions:
def is_very_even_number(n):
    while len(str(n)) > 1:
        n = sum(int(x) for x in str(n))
    return True if n % 2 == 0 else False

def is_very_even_number(n):
    return n == 0 or (n - 1) % 9 % 2

def is_very_even_number(n):
    if n < 10:
        return n % 2 == 0
    
    return is_very_even_number(sum(int(d) for d in str(n)))

def get_sum(x):
    return sum(int(t) for t in list(str(x)))
 
def is_very_even_number(n):
  result = get_sum(n)
  while len(str(result)) >= 2:
    result = get_sum(result)
  return (result %2 == 0)


def is_very_even_number(n):
    return is_very_even_number(sum(int(i) for i in str(n))) if n > 9 else not n% 2

def is_very_even_number(n):
    while n>=10:
        p, n = n, 0
        while p>0:
            re = p%10
            n += re
            p //= 10
    return n%2==0

def is_very_even_number(n):
    if len(str(n)) == 1:
        return n % 2 == 0
    return is_very_even_number(sum(map(int, list(str(n)))))

def is_very_even_number(n):
    while n >= 10:
        n = sum(map(int, str(n)))
    return n % 2 == 0

def is_very_even_number(n):
    if n < 10:
        return n % 2 == 0
    return is_very_even_number(sum(map(int, str(n))))

def is_very_even_number(n):
    return True if not n else all((not (n%9)%2,  (n%9) ))