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: EASY
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):
	binaryArr = list(bin(n))
	binaryArr.reverse()
	return binaryArr.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'))