Check if count of divisors is even or odd in Python

Suppose we have a number n, we have to find its total number of divisors are even or odd.

So, if the input is like n = 75, then the output will be Even, as the divisors are [1, 3, 5, 15, 25, 75].

To solve this we shall follow one simple and efficient approach. We have observed that when a number is perfect square then only it has odd number of divisors. So if the number is not perfect square then it will have even divisors. So here we will only check whether the number is perfect square or not and based on this we can return "odd" or "even" as output.

To solve this, we will follow these steps −

• if n < 1 is non-zero, then
  • return
• sqrt := square root of n
• if sqrt*sqrt is same as n, then
  • return 'Odd'
• otherwise,
  • return 'Even'

Let us see the following implementation to get better understanding −

Example

 Live Demo

def solve(n):
   if n < 1:
      return
   sqrt = n**0.5
   if sqrt*sqrt == n:
      return 'Odd'
   else:
      return 'Even'
n = 75
print(solve(n))

Input

75

Output

Even