~~~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)
```

Difficulty: introductory

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 & 0xaaaaaaaa != 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):
    # Write code here...
    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