Given an integer `n` return `"odd"` if the number of its divisors is odd. Otherwise return `"even"`.

**Note**: big inputs will be tested.

## Examples:

All prime numbers have exactly two divisors (hence `"even"`).

For `n = 12` the divisors are `[1, 2, 3, 4, 6, 12]` – `"even"`.

For `n = 4` the divisors are `[1, 2, 4]` – `"odd"`.

Source: codewars | Difficulty: EASY
Tags: Mathematics

Starter code: def oddity(n):
	

Solutions:
def oddity(n):
	return 'odd' if n ** 0.5 == int(n ** 0.5) else 'even'


oddity = lambda n: ['odd', 'even'][n ** 0.5 % 1 != 0]


import math

def oddity(n):
	return math.sqrt(n) % 1 == 0 and 'odd' or 'even'


import math

def oddity(n):
	return 'even' if math.sqrt(n) % 1 else 'odd'


import math

def oddity(n):
	if int(math.sqrt(n)) ** 2 == n:
		return 'odd'
	return 'even'


def oddity(n):
	counter = 0
	for x in range(1, int(n ** 0.5) + 1):
		(a, b) = divmod(n, x)
		if not b:
			if a != x:
				counter += 2
			else:
				counter += 1
	return ('even', 'odd')[counter % 2]


def oddity(n):
	return 'odd' if (n ** 0.5).is_integer() else 'even'


def oddity(n):
	return 'even' if n ** 0.5 % 1 else 'odd'


def oddity(n):
	return 'odd' if round(n ** 0.5) ** 2 == n else 'even'