Check if the given number is Ore number or not in Python

PythonServer Side ProgrammingProgramming

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
raja
Published on 15-Jan-2021 06:50:43
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