~~~if-not:ruby,python
Return `1` when *any* odd bit of `x` equals 1; `0` otherwise.
~~~
~~~if:ruby,python
Return `true` when *any* odd bit of `x` equals 1; `false` otherwise.
~~~

Assume that:
* `x` is an unsigned, 32-bit integer;
* the bits are zero-indexed (the least significant bit is position 0)


## Examples

```
  2  -->  1 (true) because at least one odd bit is 1 (2 = 0b10)
  5  -->  0 (false) because none of the odd bits are 1 (5 = 0b101)
170  -->  1 (true) because all of the odd bits are 1 (170 = 0b10101010)
```

Source: codewars | Difficulty: EASY
Tags: Bit manipulation

Fundamentals

Starter code: def any_odd(x):
	

Solutions:
MATCH = int('10' * 16, 2)

def any_odd(x):
	return bool(MATCH & x)


def any_odd(n):
	return 1 if '1' in bin(n)[2:][-2::-2] else 0


def any_odd(x):
	return x & 2863311530 != 0


def any_odd(x):
	return int('1' in f'{x:032b}'[::2])


def any_odd(x):
	bin = '{0:b}'.format(x)
	bin = bin[::-1]
	odd = 0
	print(bin)
	for letter in bin:
		if odd % 2 == 1 and letter == '1':
			return 1
		odd += 1
	return 0


def any_odd(x):
	a = bin(x)[2:]
	b = a[-2::-2]
	return 1 if '1' in b else 0


def any_odd(x):
	return '1' in f'0{x:b}'[-2::-2]


def any_odd(x):
	binary = bin(x)
	answer = 0
	binary_reversed = str(binary[::-1])
	for k in range(0, len(binary_reversed)):
		if k % 2 != 0 and binary_reversed[k] == '1':
			answer = 1
	return answer


def any_odd(x):
	return '1' in list(bin(x)[-2::-2])


def any_odd(x):
	bina = bin(x).split('b')[1]
	for (a, b) in enumerate(bina, start=len(bina) - 1):
		if a % 2:
			if '1' in b:
				return 1
	return 0