Write a function that returns the number of '2's in the factorization of a number.  
For example,
```python
two_count(24)
```
should return 3, since the factorization of 24 is 2^3 x 3
```python
two_count(17280)
```
should return 7, since the factorization of 17280 is 2^7 x 5 x 3^3  
The number passed to two_count (twoCount)  will always be a positive integer greater than or equal to 1.

Difficulty: introductory

Starter code:

def two_count(n):
	

Solutions:
def two_count(n):
    res = 0
    while not n & 1:
        res += 1
        n >>= 1
    return res


def two_count(n):
  return bin(n)[::-1].index('1')

def two_count(n):
    return n.bit_length() - len(bin(n).rstrip('0')) + 2

def two_count(n):
  i = 0
  while True:
      n, r = divmod(n, 2)
      if r:
          break
      i += 1
  return i


def two_count(n):
    
    # store the brinary digits of n as an array
    binaryArr = list(bin(n))
    
    # reverse the array
    binaryArr.reverse()
    
    # return the index of the first location of element '1'
    return binaryArr.index('1')



def two_count(n):
    return bin(n)[::-1].index('1')


def two_count(n):
    if n%2 != 0:
        return 0
    else:
        return 1 + two_count(n//2)


def two_count(n):
  n = bin(n)[2:]
  
  return len(n) - len(n.strip('0'))