Refactorable number

Given an integer n. Check whether the number is refactorable or not. A refactorable number is an integer n that is divisible by count of all it’s divisors.

Input:  n = 8
Output: yes
8 has 4 divisors: 1, 2, 4, 8
Since 8 is divisible by 4 therefore 8 is 
refactorable number.

Input : n = 4 
Output: no
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This solution is pretty straightforward. The idea is to iterate from 1 to sqrt(n) and count all the divisors of a number. After that we just need to check whether the number n is divisible by it’s total count or not.


// C++ program to check whether number is 
// refactorable or not
#include <bits/stdc++.h>

// Function to count all divisors
bool isRefactorableNumber(int n)
    int divCount = 0;  // Initialize result

    for (int i=1; i<=sqrt(n); ++i)
        if (n%i==0)
            // If divisors are equal, count 
            // only one.
            if (n/i == i)

            else // Otherwise count both

    return n % divCount == 0;

//Driver program to test above function
int main()
    int n = 8;
    if (isRefactorableNumber(n))

    n = 14;
    if (isRefactorableNumber(n))

    return 0;


# Python program to check whether number is
# refactorable or not
import math

def isRefactorableNumber(n):

    # Initialize result
    divCount = 0

    for i in range(1,int(math.sqrt(n))+1):

        if n % i == 0:

            # If divisors are equal, count only one
            if n/i == i:
                divCount += 1

            else:  # Otherwise count both
                divCount += 2

    return n % divCount == 0

# Driver program to test above function
n = 8
if isRefactorableNumber(n):
    print "yes"
    print "no"

n = 14
if (isRefactorableNumber(n)):
    print "yes"
    print "no"

Time complexity: O(sqrt(n))
Auxiliary space: O(1)

Facts about refactorable number

  1. There is no refactorable number which is perfect.
  2. There is no three consecutive integers can all be refactorable.
  3. Refactorable number have natural density zero.


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