Sum of all the prime divisors of a number

23 May, 2022
Given a number N. The task is to find the sum of all the prime divisors of N. 
Examples: 
Input: 60
Output: 10
2, 3, 5 are prime divisors of 60

Input: 39
Output: 16
3, 13 are prime divisors of 39
A naive approach will be to iterate for all numbers till N and check if the number divides N. If the number divides N, check if that number is prime or not. Add all the prime numbers till N which divides N. 
Below is the implementation of the above approach:  
C++
C
Java
C#
Python3
PHP
Javascript
// CPP program to find sum of// prime divisors of N#include <bits/stdc++.h>using namespace std;#define N 1000005 // Function to check if the// number is prime or not.bool isPrime(int n){    // Corner cases    if (n <= 1)        return false;    if (n <= 3)        return true;     // This is checked so that we can skip    // middle five numbers in below loop    if (n % 2 == 0 || n % 3 == 0)        return false;     for (int i = 5; i * i <= n; i = i + 6)        if (n % i == 0 || n % (i + 2) == 0)            return false;     return true;} // function to find sum of prime// divisors of Nint SumOfPrimeDivisors(int n){    int sum = 0;    for (int i = 1; i <= n; i++) {        if (n % i == 0) {            if (isPrime(i))                sum += i;        }    }    return sum;}// Driver codeint main(){    int n = 60;    cout << "Sum of prime divisors of 60 is " << SumOfPrimeDivisors(n) << endl;}
// C program to find sum of// prime divisors of N#include <stdio.h>#include <stdbool.h> #define N 1000005 // Function to check if the// number is prime or not.bool isPrime(int n){    // Corner cases    if (n <= 1)        return false;    if (n <= 3)        return true;     // This is checked so that we can skip    // middle five numbers in below loop    if (n % 2 == 0 || n % 3 == 0)        return false;     for (int i = 5; i * i <= n; i = i + 6)        if (n % i == 0 || n % (i + 2) == 0)            return false;     return true;} // function to find sum of prime// divisors of Nint SumOfPrimeDivisors(int n){    int sum = 0;    for (int i = 1; i <= n; i++) {        if (n % i == 0) {            if (isPrime(i))                sum += i;        }    }    return sum;} // Driver codeint main(){    int n = 60;    printf("Sum of prime divisors of 60 is %d\n",SumOfPrimeDivisors(n));} // This code is contributed by kothavvsaakash.
// Java program to find sum// of prime divisors of Nimport java.io.*;import java.util.*; class GFG{// Function to check if the// number is prime or not.static boolean isPrime(int n){    // Corner cases    if (n <= 1)        return false;    if (n <= 3)        return true;     // This is checked so that    // we can skip middle five    // numbers in below loop    if (n % 2 == 0 || n % 3 == 0)        return false;     for (int i = 5;             i * i <= n; i = i + 6)        if (n % i == 0 ||            n % (i + 2) == 0)            return false;     return true;} // function to find// sum of prime// divisors of Nstatic int SumOfPrimeDivisors(int n){    int sum = 0;    for (int i = 1;             i <= n; i++)    {        if (n % i == 0)        {            if (isPrime(i))                sum += i;        }    }    return sum;} // Driver codepublic static void main(String args[]){    int n = 60;    System.out.print("Sum of prime divisors of 60 is " +                          SumOfPrimeDivisors(n) + "\n");}}
// C# program to find sum// of prime divisors of Nusing System;class GFG{     // Function to check if the// number is prime or not.static bool isPrime(int n){    // Corner cases    if (n <= 1)        return false;    if (n <= 3)        return true;     // This is checked so that    // we can skip middle five    // numbers in below loop    if (n % 2 == 0 || n % 3 == 0)        return false;     for (int i = 5;             i * i <= n; i = i + 6)        if (n % i == 0 ||            n % (i + 2) == 0)            return false;     return true;} // function to find// sum of prime// divisors of Nstatic int SumOfPrimeDivisors(int n){    int sum = 0;    for (int i = 1;            i <= n; i++)    {        if (n % i == 0)        {            if (isPrime(i))                sum += i;        }    }    return sum;} // Driver codepublic static void Main(){    int n = 60;    Console.WriteLine("Sum of prime divisors of 60 is " +                        SumOfPrimeDivisors(n) + "\n");}} // This code is contributed// by inder_verma.
# Python 3 program to find# sum of prime divisors of NN = 1000005 # Function to check if the# number is prime or not.def isPrime(n):         # Corner cases    if n <= 1:        return False    if n <= 3:        return True     # This is checked so that     # we can skip middle five    # numbers in below loop    if n % 2 == 0 or n % 3 == 0:        return False     i = 5    while i * i <= n:        if (n % i == 0 or            n % (i + 2) == 0):            return False        i = i + 6     return True # function to find sum# of prime divisors of Ndef SumOfPrimeDivisors(n):    sum = 0    for i in range(1, n + 1) :        if n % i == 0 :            if isPrime(i):                sum += i         return sum # Driver coden = 60print("Sum of prime divisors of 60 is " +              str(SumOfPrimeDivisors(n))) # This code is contributed# by ChitraNayal
<?php// PHP program to find sum// of prime divisors of N$N = 1000005; // Function to check if the// number is prime or not.function isPrime($n){    global $N;    // Corner cases    if ($n <= 1)        return false;    if ($n <= 3)        return true;     // This is checked so that    // we can skip middle five    // numbers in below loop    if ($n % 2 == 0 || $n % 3 == 0)        return false;     for ($i = 5; $i * $i <= $n;                 $i = $i + 6)        if ($n % $i == 0 ||            $n % ($i + 2) == 0)            return false;     return true;} // function to find sum// of prime divisors of Nfunction SumOfPrimeDivisors($n){    $sum = 0;    for ($i = 1; $i <= $n; $i++)    {        if ($n % $i == 0)        {            if (isPrime($i))                $sum += $i;        }    }    return $sum;} // Driver code$n = 60;echo "Sum of prime divisors of 60 is " .                 SumOfPrimeDivisors($n); // This code is contributed// by ChitraNayal?>
<script>// Javascript program to find sum of// prime divisors of N    let N = 1000005;         // Function to check if the    // number is prime or not.    function isPrime(n)    {        // Corner cases    if (n <= 1)        return false;    if (n <= 3)        return true;       // This is checked so that we can skip    // middle five numbers in below loop    if (n % 2 == 0 || n % 3 == 0)        return false;       for (let i = 5; i * i <= n; i = i + 6)        if (n % i == 0 || n % (i + 2) == 0)            return false;       return true;    }         // function to find sum of prime    // divisors of N    function SumOfPrimeDivisors(n)    {        let sum = 0;    for (let i = 1; i <= n; i++) {        if (n % i == 0) {            if (isPrime(i))                sum += i;        }    }    return sum;    }         // Driver code    let n = 60;    document.write("Sum of prime divisors of 60 is "+SumOfPrimeDivisors(n));              // This code is contributed by avanitrachhadiya2155</script>
Sum of prime divisors of 60 is 10
Time Complexity: O(N * sqrt(N)) 
Efficient Approach: The complexity can be reduced using Sieve of Eratosthenes with some modifications. The modifications are as follows:  
Take an array of size N and substitute zero in all the indexes(initially consider all the numbers are prime).
Iterate for all the numbers whose indexes have zero(i.e., it is prime numbers).
Add this number to all it’s multiples less than N
Return the array[N] value which has the sum stored in it.
Below is the implementation of the above approach.  
C++
Java
Python3
C#
PHP
Javascript
// CPP program to find prime divisors of// all numbers from 1 to n#include <bits/stdc++.h>using namespace std; // function to find prime divisors of// all numbers from 1 to nint Sum(int N){    int SumOfPrimeDivisors[N+1] = { 0 };     for (int i = 2; i <= N; ++i) {         // if the number is prime        if (!SumOfPrimeDivisors[i]) {             // add this prime to all it's multiples            for (int j = i; j <= N; j += i) {                 SumOfPrimeDivisors[j] += i;            }        }    }    return SumOfPrimeDivisors[N];} // Driver codeint main(){    int N = 60;    cout << "Sum of prime divisors of 60 is "         << Sum(N) << endl;}
// Java program to find// prime divisors of// all numbers from 1 to nimport java.io.*;import java.util.*; class GFG{     // function to find prime// divisors of all numbers// from 1 to nstatic int Sum(int N){    int SumOfPrimeDivisors[] = new int[N + 1];          for (int i = 2; i <= N; ++i)    {         // if the number is prime        if (SumOfPrimeDivisors[i] == 0)        {             // add this prime to            // all it's multiples            for (int j = i; j <= N; j += i)            {                 SumOfPrimeDivisors[j] += i;            }        }    }    return SumOfPrimeDivisors[N];} // Driver codepublic static void main(String args[]){    int N = 60;    System.out.print("Sum of prime " +                "divisors of 60 is " +                       Sum(N) + "\n");}}
# Python 3 program to find# prime divisors of# all numbers from 1 to n # function to find prime# divisors of all numbers# from 1 to ndef Sum(N):      SumOfPrimeDivisors = [0] * (N + 1)          for i in range(2, N + 1) :              # if the number is prime        if (SumOfPrimeDivisors[i] == 0) :                      # add this prime to            # all it's multiples            for j in range(i, N + 1, i) :                              SumOfPrimeDivisors[j] += i                  return SumOfPrimeDivisors[N]  # Driver codeN = 60print("Sum of prime" ,      "divisors of 60 is",                  Sum(N));                   # This code is contributed# by Smitha
// C# program to find// prime divisors of// all numbers from 1 to nusing System; class GFG{     // function to find prime// divisors of all numbers// from 1 to nstatic int Sum(int N){    int []SumOfPrimeDivisors = new int[N + 1];         for (int i = 2; i <= N; ++i)    {         // if the number is prime        if (SumOfPrimeDivisors[i] == 0)        {             // add this prime to            // all it's multiples            for (int j = i;                     j <= N; j += i)            {                 SumOfPrimeDivisors[j] += i;            }        }    }         return SumOfPrimeDivisors[N];} // Driver codepublic static void Main(){    int N = 60;    Console.Write("Sum of prime " +                    "divisors of 60 is " +                         Sum(N) + "\n");}} // This code is contributed// by Smitha
<?php// PHP program to find prime// divisors of all numbers// from 1 to n // function to find prime// divisors of all numbers// from 1 to nfunction Sum($N){    for($i = 0; $i <= $N; $i++)        $SumOfPrimeDivisors[$i] = 0;     for ($i = 2; $i <= $N; ++$i)    {         // if the number is prime        if (!$SumOfPrimeDivisors[$i])        {             // add this prime to            // all it's multiples            for ($j = $i; $j <= $N; $j += $i)            {                 $SumOfPrimeDivisors[$j] += $i;            }        }    }    return $SumOfPrimeDivisors[$N];} // Driver code$N = 60;echo "Sum of prime divisors of 60 is " . Sum($N); // This code is contributed by Mahadev99?>
<script> // Javascript program to find// prime divisors of// all numbers from 1 to n         // function to find prime    // divisors of all numbers    // from 1 to n    function Sum(N)    {        let SumOfPrimeDivisors = new Array(N+1);        for(let i=0;i<SumOfPrimeDivisors.length;i++)        {            SumOfPrimeDivisors[i]=0;        }        for (let i = 2; i <= N; ++i)        {              // if the number is prime            if (SumOfPrimeDivisors[i] == 0)            {                  // add this prime to                // all it's multiples                for (let j = i; j <= N; j += i)                {                      SumOfPrimeDivisors[j] += i;                }            }        }        return SumOfPrimeDivisors[N];    }         // Driver code    let N = 60;    document.write("Sum of prime " +                "divisors of 60 is " +                       Sum(N) + "<br>");         // This code is contributed by rag2127     </script>
Sum of prime divisors of 60 is 10
Time Complexity: O(N * log N) 
Time complexity can be reduced by finding all the factors efficiently.
Below approach describe how to find all the factors efficiently. 
If we look carefully, all the divisors are present in pairs. For example if n = 100, then the various pairs of divisors are: (1,100), (2,50), (4,25), (5,20), (10,10)
Using this fact we could speed up our program significantly.  
We, however, have to be careful if there are two equal divisors as in the case of (10, 10). In such case, we’d take only one of them. 
Below is the implementation of the above approach. 
C++
C
Java
Python3
C#
Javascript
// C++ program to find sum of// prime divisors of N#include <bits/stdc++.h>using namespace std; // Function to check if the// number is prime or not.bool isPrime(int n){    // Corner cases    if (n <= 1)        return false;    if (n <= 3)        return true;     // This is checked so that we can skip    // middle five numbers in below loop    if (n % 2 == 0 || n % 3 == 0)        return false;     for (int i = 5; i * i <= n; i = i + 6)        if (n % i == 0 || n % (i + 2) == 0)            return false;     return true;} // function to find sum of prime// divisors of Nint SumOfPrimeDivisors(int n){    int sum = 0;    // return type of sqrt function    // if float    int root_n = (int)sqrt(n);    for (int i = 1; i <= root_n; i++) {        if (n % i == 0) {            // both factors are same            if (i == n / i && isPrime(i)) {                sum += i;            }            else {                // both factors are                // not same ( i and n/i )                if (isPrime(i)) {                    sum += i;                }                if (isPrime(n / i)) {                    sum += (n / i);                }            }        }    }    return sum;}// Driver codeint main(){    int n = 60;    cout << "Sum of prime divisors of 60 is "         << SumOfPrimeDivisors(n) << endl;}// This code is contributed by hemantraj712
// C program to find sum of// prime divisors of N#include <stdio.h>#include <stdbool.h>#include <math.h> // Function to check if the// number is prime or not.bool isPrime(int n){    // Corner cases    if (n <= 1)        return false;    if (n <= 3)        return true;     // This is checked so that we can skip    // middle five numbers in below loop    if (n % 2 == 0 || n % 3 == 0)        return false;     for (int i = 5; i * i <= n; i = i + 6)        if (n % i == 0 || n % (i + 2) == 0)            return false;     return true;} // function to find sum of prime// divisors of Nint SumOfPrimeDivisors(int n){    int sum = 0;    // return type of sqrt function    // if float    int root_n = (int)sqrt(n);    for (int i = 1; i <= root_n; i++) {        if (n % i == 0) {            // both factors are same            if (i == n / i && isPrime(i)) {                sum += i;            }            else {                // both factors are                // not same ( i and n/i )                if (isPrime(i)) {                    sum += i;                }                if (isPrime(n / i)) {                    sum += (n / i);                }            }        }    }    return sum;}// Driver codeint main(){    int n = 60;    printf("Sum of prime divisors of 60 is %d\n",SumOfPrimeDivisors(n));}// This code is contributed by hemantraj712
// Java program to find sum of// prime divisors of Nclass GFG{ // Function to check if the// number is prime or not.static boolean isPrime(int n){         // Corner cases    if (n <= 1)        return false;    if (n <= 3)        return true;       // This is checked so that we can skip    // middle five numbers in below loop    if (n % 2 == 0 || n % 3 == 0)        return false;       for(int i = 5; i * i <= n; i = i + 6)        if (n % i == 0 || n % (i + 2) == 0)            return false;       return true;}   // Function to find sum of prime// divisors of Nstatic int SumOfPrimeDivisors(int n){    int sum = 0;         // Return type of sqrt function    // if float    int root_n = (int)Math.sqrt(n);    for(int i = 1; i <= root_n; i++)    {        if (n % i == 0)        {                         // Both factors are same            if (i == n / i && isPrime(i))            {                sum += i;            }            else            {                                 // Both factors are                // not same ( i and n/i )                if (isPrime(i))                {                    sum += i;                }                if (isPrime(n / i))                {                    sum += (n / i);                }            }        }    }    return sum;} // Driver codepublic static void main(String[] args){    int n = 60;    System.out.println("Sum of prime divisors of 60 is " +                       SumOfPrimeDivisors(n));}} // This code is contributed by divyeshrabadiya07
# Python3 program to find sum of# prime divisors of Nimport math # Function to check if the# number is prime or not.def isPrime(n) :     # Corner cases    if (n <= 1) :        return False    if (n <= 3) :        return True      # This is checked so that we can skip    # middle five numbers in below loop    if (n % 2 == 0 or n % 3 == 0) :        return False         i = 5    while i * i <= n :        if (n % i == 0 or n % (i + 2) == 0) :            return False        i = i + 6      return True  # function to find sum of prime# divisors of Ndef SumOfPrimeDivisors(n) :     Sum = 0         # return type of sqrt function    # if float    root_n = (int)(math.sqrt(n))    for i in range(1, root_n + 1) :        if (n % i == 0) :                       # both factors are same            if (i == (int)(n / i) and isPrime(i)) :                Sum += i                         else :                # both factors are                # not same ( i and n/i )                if (isPrime(i)) :                    Sum += i                                 if (isPrime((int)(n / i))) :                    Sum += (int)(n / i)                        return Sum     n = 60print("Sum of prime divisors of 60 is", SumOfPrimeDivisors(n)) # This code is contributed by rameshtravel07
// C# program to find sum of// prime divisors of Nusing System;class GFG {         // Function to check if the    // number is prime or not.    static bool isPrime(int n)    {        // Corner cases        if (n <= 1)            return false;        if (n <= 3)            return true;              // This is checked so that we can skip        // middle five numbers in below loop        if (n % 2 == 0 || n % 3 == 0)            return false;              for (int i = 5; i * i <= n; i = i + 6)            if (n % i == 0 || n % (i + 2) == 0)                return false;              return true;    }          // function to find sum of prime    // divisors of N    static int SumOfPrimeDivisors(int n)    {        int sum = 0;        // return type of sqrt function        // if float        int root_n = (int)Math.Sqrt(n);        for (int i = 1; i <= root_n; i++) {            if (n % i == 0) {                // both factors are same                if (i == n / i && isPrime(i)) {                    sum += i;                }                else {                    // both factors are                    // not same ( i and n/i )                    if (isPrime(i)) {                        sum += i;                    }                    if (isPrime(n / i)) {                        sum += (n / i);                    }                }            }        }        return sum;    }   static void Main() {    int n = 60;    Console.WriteLine("Sum of prime divisors of 60 is " + SumOfPrimeDivisors(n));  }} // This code is contributed by suresh07.
<script>    // Javascript program to find sum of    // prime divisors of N         // Function to check if the    // number is prime or not.    function isPrime(n)    {        // Corner cases        if (n <= 1)            return false;        if (n <= 3)            return true;         // This is checked so that we can skip        // middle five numbers in below loop        if (n % 2 == 0 || n % 3 == 0)            return false;         for (let i = 5; i * i <= n; i = i + 6)            if (n % i == 0 || n % (i + 2) == 0)                return false;         return true;    }     // function to find sum of prime    // divisors of N    function SumOfPrimeDivisors(n)    {        let sum = 0;        // return type of sqrt function        // if float        let root_n = parseInt(Math.sqrt(n), 10);        for (let i = 1; i <= root_n; i++) {            if (n % i == 0) {                // both factors are same                if (i == parseInt(n / i, 10) && isPrime(i)) {                    sum += i;                }                else {                    // both factors are                    // not same ( i and n/i )                    if (isPrime(i)) {                        sum += i;                    }                    if (isPrime(parseInt(n / i, 10))) {                        sum += (parseInt(n / i, 10));                    }                }            }        }        return sum;    }         let n = 60;    document.write("Sum of prime divisors of 60 is "         + SumOfPrimeDivisors(n) + "</br>");   // This code is contributed by muksh07.</script>
Sum of prime divisors of 60 is 10
Time Complexity: O(sqrt(N) * sqrt(N))  
inderDuMCA
Smitha Dinesh Semwal
Mahadev99
ukasp
avanitrachhadiya2155
rag2127
hemantraj712
mukesh07
suresh07
divyeshrabadiya07
rameshtravel07
kothavvsaakash
divisors
Prime Number
sieve
Mathematical
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Prime Number
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