Skip to content
Related Articles

Related Articles

Improve Article

Find sum of divisors of all the divisors of a natural number

  • Difficulty Level : Hard
  • Last Updated : 30 Jul, 2021

Given a natural number n, the task is to find sum of divisors of all the divisors of n.

Examples: 

Input : n = 54
Output : 232
Divisors of 54 = 1, 2, 3, 6, 9, 18, 27, 54.
Sum of divisors of 1, 2, 3, 6, 9, 18, 27, 54 
are 1, 3, 4, 12, 13, 39, 40, 120 respectively.
Sum of divisors of all the divisors of 54 = 
1 + 3 + 4 + 12 + 13 + 39 + 40 + 120 = 232.

Input : n = 10
Output : 28
Divisors of 10 are 1, 2, 5, 10
Sums of divisors of divisors are 
1, 3, 6, 18.
Overall sum = 1 + 3 + 6 + 18 = 28

Using the fact that any number n can be expressed as product of prime factors, n = p1k1 x p2k2 x … where p1, p2, … are prime numbers. 
All the divisors of n can be expressed as p1a x p2b x …, where 0 <= a <= k1 and 0 <= b <= k2. 
Now sum of divisors will be sum of all power of p1 – p10, p11,…., p1k1 multiplied by all power of p2 – p20, p21,…., p2k1 
Sum of Divisor of n 
= (p10 x p20) + (p11 x p20) +…..+ (p1k1 x p20) +….+ (p10 x p21) + (p11 x p21) +…..+ (p1k1 x p21) +……..+ 
   (p10 x p2k2) + (p11 x p2k2) +……+ (p1k1 x p2k2). 
= (p10 + p11 +…+ p1k1) x p20 + (p10 + p11 +…+ p1k1) x p21 +…….+ (p10 + p11 +…+ p1k1) x p2k2
= (p10 + p11 +…+ p1k1) x (p20 + p21 +…+ p2k2).

Now, the divisors of any pa, for p as prime, are p0, p1,……, pa. And sum of divisors will be (p(a+1) – 1)/(p -1), let it define by f(p). 
So, sum of divisors of all divisor will be, 
= (f(p10) + f(p11) +…+ f(p1k1)) x (f(p20) + f(p21) +…+ f(p2k2)).

So, given a number n, by prime factorization we can find the sum of divisors of all the divisors. But in this problem we are given that n is product of element of array. So, find prime factorization of each element and by using the fact ab x ac = ab+c.



Below is the implementation of this approach:  

C++




// C++ program to find sum of divisors of all
// the divisors of a natural number.
#include<bits/stdc++.h>
using namespace std;
 
// Returns sum of divisors of all the divisors
// of n
int sumDivisorsOfDivisors(int n)
{
    // Calculating powers of prime factors and
    // storing them in a map mp[].
    map<int, int> mp;
    for (int j=2; j<=sqrt(n); j++)
    {
        int count = 0;
        while (n%j == 0)
        {
            n /= j;
            count++;
        }
 
        if (count)
            mp[j] = count;
    }
 
    // If n is a prime number
    if (n != 1)
        mp[n] = 1;
 
    // For each prime factor, calculating (p^(a+1)-1)/(p-1)
    // and adding it to answer.
    int ans = 1;
    for (auto it : mp)
    {
        int pw = 1;
        int sum = 0;
 
        for (int i=it.second+1; i>=1; i--)
        {
            sum += (i*pw);
            pw *= it.first;
        }
        ans *= sum;
    }
 
    return ans;
}
 
// Driven Program
int main()
{
    int n = 10;
    cout << sumDivisorsOfDivisors(n);
    return 0;
}

Java




// Java program to find sum of divisors of all
// the divisors of a natural number.
import java.util.HashMap;
 
class GFG
{
 
    // Returns sum of divisors of all the divisors
    // of n
    public static int sumDivisorsOfDivisors(int n)
    {
 
        // Calculating powers of prime factors and
        // storing them in a map mp[].
        HashMap<Integer, Integer> mp = new HashMap<>();
        for (int j = 2; j <= Math.sqrt(n); j++)
        {
            int count = 0;
            while (n % j == 0)
            {
                n /= j;
                count++;
            }
            if (count != 0)
                mp.put(j, count);
        }
 
        // If n is a prime number
        if (n != 1)
            mp.put(n, 1);
 
        // For each prime factor, calculating (p^(a+1)-1)/(p-1)
        // and adding it to answer.
        int ans = 1;
 
        for (HashMap.Entry<Integer, Integer> entry : mp.entrySet())
        {
            int pw = 1;
            int sum = 0;
            for (int i = entry.getValue() + 1; i >= 1; i--)
            {
                sum += (i * pw);
                pw = entry.getKey();
            }
            ans *= sum;
        }
 
        return ans;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int n = 10;
        System.out.println(sumDivisorsOfDivisors(n));
    }
}
 
// This code is contributed by
// sanjeev2552

Python3




# Python3 program to find sum of divisors
# of all the divisors of a natural number.
import math as mt
 
# Returns sum of divisors of all
# the divisors of n
def sumDivisorsOfDivisors(n):
 
    # Calculating powers of prime factors
    # and storing them in a map mp[].
    mp = dict()
    for j in range(2, mt.ceil(mt.sqrt(n))):
 
        count = 0
        while (n % j == 0):
            n //= j
            count += 1
 
        if (count):
            mp[j] = count
 
    # If n is a prime number
    if (n != 1):
        mp[n] = 1
 
    # For each prime factor, calculating
    # (p^(a+1)-1)/(p-1) and adding it to answer.
    ans = 1
    for it in mp:
        pw = 1
        summ = 0
 
        for i in range(mp[it] + 1, 0, -1):
            summ += (i * pw)
            pw *= it
     
        ans *= summ
 
    return ans
 
# Driver Code
n = 10
print(sumDivisorsOfDivisors(n))
     
# This code is contributed
# by mohit kumar 29

C#




// C# program to find sum of divisors of all
// the divisors of a natural number.
using System;
using System.Collections.Generic;
     
class GFG
{
 
    // Returns sum of divisors of
    // all the divisors of n
    public static int sumDivisorsOfDivisors(int n)
    {
 
        // Calculating powers of prime factors and
        // storing them in a map mp[].
        Dictionary<int,
                   int> mp = new Dictionary<int,
                                            int>();
        for (int j = 2; j <= Math.Sqrt(n); j++)
        {
            int count = 0;
            while (n % j == 0)
            {
                n /= j;
                count++;
            }
            if (count != 0)
                mp.Add(j, count);
        }
 
        // If n is a prime number
        if (n != 1)
            mp.Add(n, 1);
 
        // For each prime factor,
        // calculating (p^(a+1)-1)/(p-1)
        // and adding it to answer.
        int ans = 1;
 
        foreach(KeyValuePair<int, int> entry in mp)
        {
            int pw = 1;
            int sum = 0;
            for (int i = entry.Value + 1;
                     i >= 1; i--)
            {
                sum += (i * pw);
                pw = entry.Key;
            }
            ans *= sum;
        }
 
        return ans;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        int n = 10;
        Console.WriteLine(sumDivisorsOfDivisors(n));
    }
}
 
// This code is contributed
// by Princi Singh

Javascript




<script>
// Javascript program to find sum of divisors of all
// the divisors of a natural number.
     
     // Returns sum of divisors of all the divisors
    // of n
    function sumDivisorsOfDivisors(n)
    {
        // Calculating powers of prime factors and
        // storing them in a map mp[].
        let mp = new Map();
        for (let j = 2; j <= Math.sqrt(n); j++)
        {
            let count = 0;
            while (n % j == 0)
            {
                n = Math.floor(n/j);
                count++;
            }
            if (count != 0)
                mp.set(j, count);
        }
   
        // If n is a prime number
        if (n != 1)
            mp.set(n, 1);
   
        // For each prime factor, calculating (p^(a+1)-1)/(p-1)
        // and adding it to answer.
        let ans = 1;
   
        for (let [key, value] of mp.entries())
        {
            let pw = 1;
            let sum = 0;
            for (let i = value + 1; i >= 1; i--)
            {
                sum += (i * pw);
                pw = key;
            }
            ans *= sum;
        }
   
        return ans;
    }
     
    // Driver code
    let n = 10;
    document.write(sumDivisorsOfDivisors(n));
     
     
 
// This code is contributed by patel2127
</script>

Output: 

28

Optimizations : 
For the cases when there are multiple inputs for which we need find the value, we can use Sieve of Eratosthenes as discussed in this post.

This article is contributed by Anuj Chauhan. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 

Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.




My Personal Notes arrow_drop_up
Recommended Articles
Page :