Check If Given Number is Ore Number in Python

Suppose we have a number n. We have to check whether n is an ore number or not. As we know an ore number is a number whose divisors have an integer harmonic value.

So, if the input is like 28, then the output will be True as there are six divisors of 28: [1, 2, 4, 7, 14, 28], so

As 3 is an integer so 28 is an ore number.

To solve this, we will follow these steps −

• Define a function get_all_div() . This will take n
• div := a new list
• for i in range 1 to integer part of (square root of n), do
  • if n is divisible by i, then
    • if quotient of (n / i) is i, then
      • insert i at the end of div
    • otherwise,
      • insert i at the end of div
      • insert quotient of (n / i) at the end of div
• return div
• Define a function get_harmonic_mean() . This will take n
• div := get_all_div(n)
• total := 0
• length := size of div
• for i in range 0 to length - 1, do
  • total := total + (n / div[i])
• total := total / n
• return length / total
• From the main method do the following:
• mean := get_harmonic_mean(n)
• if mean is an integer, then
  • return True
• return False

Let us see the following implementation to get better understanding −

Example Code

Live Demo

def get_all_div(n):  
   div = []
   for i in range(1, int(n**(0.5)) + 1):
      if n % i == 0:
         if n // i == i:
           div.append(i)
         else:
           div.append(i)
           div.append(n // i)
   return div

def get_harmonic_mean(n):
   div = get_all_div(n)

   total = 0
   length = len(div)
 
   for i in range(0, length):
      total += (n / div[i])
 
   total /= n
   return length / total
 
def solve(n):
   mean = get_harmonic_mean(n)
   if mean - int(mean) == 0:
      return True
   return False

n = 28
print(solve(n))

Input

28

Output

True