Sum of common divisors of two numbers A and B

Given two number A and B, the task is to find the sum of common factors of two numbers A and B. The numbers A and B is less than 10^8.
Examples:

Input: A = 10, B = 15
Output: Sum = 6
The common factors are 1, 5, so their sum is 6 

Input: A = 100, B = 150
Output: Sum = 93

Naive Approach: Iterate from i = 1 to minimum of A and B and check whether i is a factor of both A and B. If i is a factor of A and B then add it to sum. Display the sum at the end of the loop.
Below is the implementation of the above approach: 

C++

// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;

// print the sum of common factors
int sum(int a, int b)
{
    // sum of common factors
    int sum = 0;

    // iterate from 1 to minimum of a and b
    for (int i = 1; i <= min(a, b); i++)

        // if i is the common factor
        // of both the numbers
        if (a % i == 0 && b % i == 0)
            sum += i;

    return sum;
}

// Driver code
int main()
{
    int A = 10, B = 15;

    // print the sum of common factors
    cout << "Sum = " << sum(A, B) << endl;

    return 0;
}

Java

// Java implementation of above approach
import java.io.*;

class GFG {



// print the sum of common factors
static int sum(int a, int b)
{
    // sum of common factors
    int sum = 0;

    // iterate from 1 to minimum of a and b
    for (int i = 1; i <= Math.min(a, b); i++)

        // if i is the common factor
        // of both the numbers
        if (a % i == 0 && b % i == 0)
            sum += i;

    return sum;
}

// Driver code


    public static void main (String[] args) {
            int A = 10, B = 15;

    // print the sum of common factors
    System.out.print("Sum = " + sum(A, B));
    }
}
// This code is contributed by shs..

Python 3

# Python 3 implementation of
# above approach

# print the sum of common factors
def sum(a, b):

    # sum of common factors
    sum = 0

    # iterate from 1 to minimum of a and b
    for i in range (1, min(a, b)):

        # if i is the common factor
        # of both the numbers
        if (a % i == 0 and b % i == 0):
            sum += i

    return sum

# Driver Code
A = 10
B = 15

# print the sum of common factors
print("Sum =", sum(A, B))

# This code is contributed
# by Akanksha Rai

C#

// C# implementation of above approach


using System;

class GFG {



// print the sum of common factors
static int sum(int a, int b)
{
    // sum of common factors
    int sum = 0;

    // iterate from 1 to minimum of a and b
    for (int i = 1; i <= Math.Min(a, b); i++)

        // if i is the common factor
        // of both the numbers
        if (a % i == 0 && b % i == 0)
            sum += i;

    return sum;
}

// Driver code


    public static void Main () {
            int A = 10, B = 15;

    // print the sum of common factors
    Console.WriteLine("Sum = " + sum(A, B));
    }
}
// This code is contributed by shs..

PHP

<?php
// PHP implementation of above approach

// print the sum of common factors
function sum($a, $b)
{
    // sum of common factors
    $sum = 0;

    // iterate from 1 to minimum of a and b
    for ($i = 1; $i <= min($a, $b); $i++)

        // if i is the common factor
        // of both the numbers
        if ($a %$i == 0 && $b %$i == 0)
            $sum += $i;

    return $sum;
}

// Driver code
$A = 10; $B = 15;

// print the sum of common factors
echo "Sum = " , sum($A, $B);

// This code is contributed by shs.
?>

Javascript

<script>
// Javascript implementation of above approach

// print the sum of common factors
function sum(a, b)
{
    // sum of common factors
    var sum = 0;

    // iterate from 1 to minimum of a and b
    for (var i = 1; i <= Math.min(a, b); i++)

        // if i is the common factor
        // of both the numbers
        if (a % i == 0 && b % i == 0)
            sum += i;

    return sum;
}


var A = 10, B = 15;

// print the sum of common factors
document.write("Sum = " + sum(A, B) + "<br>");

//This code is contributed by SoumikMondal
</script>

Output:

Sum = 6

Time Complexity: O(min(a, b))

Auxiliary Space: O(1)

An efficient approach is to use the same concept used in Common divisors of two numbers. Calculate the greatest common divisor (gcd) of given two numbers, and then find the sum of divisors of that gcd.

C++

// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;

// Function to calculate gcd of two numbers
int gcd(int a, int b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}

// Function to calculate all common divisors
// of two given numbers
// a, b --> input integer numbers
int sumcommDiv(int a, int b)
{
    // find gcd of a, b
    int n = gcd(a, b);

    // Find the sum of divisors of n.
    int sum = 0;
    for (int i = 1; i <= sqrt(n); i++) {

        // if 'i' is factor of n
        if (n % i == 0) {

            // check if divisors are equal
            if (n / i == i)
                sum += i;
            else
                sum += (n / i) + i;
        }
    }

    return sum;
}

// Driver program to run the case
int main()
{
    int a = 10, b = 15;
    cout << "Sum = " << sumcommDiv(a, b);

    return 0;
}

Java

//Java implementation of above approach

import java.io.*;

class GFG {

// Function to calculate gcd of two numbers
static int gcd(int a, int b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}

// Function to calculate all common divisors
// of two given numbers
// a, b --> input integer numbers
static int sumcommDiv(int a, int b)
{
    // find gcd of a, b
    int n = gcd(a, b);

    // Find the sum of divisors of n.
    int sum = 0;
    for (int i = 1; i <= Math.sqrt(n); i++) {

        // if 'i' is factor of n
        if (n % i == 0) {

            // check if divisors are equal
            if (n / i == i)
                sum += i;
            else
                sum += (n / i) + i;
        }
    }

    return sum;
}

// Driver program to run the case
    public static void main (String[] args) {

    int a = 10, b = 15;
    System.out.println("Sum = " + sumcommDiv(a, b));
    }
}

Python3

# Python 3 implementation of above approach
from math import gcd,sqrt

# Function to calculate all common divisors
# of two given numbers
# a, b --> input integer numbers
def sumcommDiv(a, b):
    # find gcd of a, b
    n = gcd(a, b)

    # Find the sum of divisors of n.
    sum = 0
    N = int(sqrt(n))+1
    for i in range(1,N,1):
        # if 'i' is factor of n
        if (n % i == 0):
            # check if divisors are equal
            if (n / i == i):
                sum += i
            else:
                sum += (n / i) + i

    return sum

# Driver program to run the case
if __name__ == '__main__':
    a = 10
    b = 15
    print("Sum =",int(sumcommDiv(a, b)))

# This code is contributed by
# Surendra_Gangwar

C#

// C# implementation of above approach

using System;

public class GFG{

// Function to calculate gcd of two numbers
static int gcd(int a, int b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}

// Function to calculate all common divisors
// of two given numbers
// a, b --> input integer numbers
static int sumcommDiv(int a, int b)
{
    // find gcd of a, b
    int n = gcd(a, b);

    // Find the sum of divisors of n.
    int sum = 0;
    for (int i = 1; i <= Math.Sqrt(n); i++) {

        // if 'i' is factor of n
        if (n % i == 0) {

            // check if divisors are equal
            if (n / i == i)
                sum += i;
            else
                sum += (n / i) + i;
        }
    }

    return sum;
}

// Driver program to run the case
    static public void Main (){
        int a = 10, b = 15;
        Console.WriteLine("Sum = " + sumcommDiv(a, b));
    }
}

PHP

<?php
// PHP implementation of above approach

// Function to calculate gcd of two numbers
function gcd($a, $b)
{
    if ($a == 0)
        return $b;
    return gcd($b % $a, $a);
}

// Function to calculate all common divisors
// of two given numbers
// a, b --> input integer numbers
function  sumcommDiv($a, $b)
{
    // find gcd of a, b
$n = gcd($a, $b);

    // Find the sum of divisors of n.
    $sum = 0;
    for ($i = 1; $i <= sqrt($n); $i++) {

        // if 'i' is factor of n
        if ($n % $i == 0) {

            // check if divisors are equal
            if ($n / $i == $i)
                $sum += $i;
            else
                $sum += ($n / $i) + $i;
        }
    }

    return $sum;
}

// Driver program to run the case
    $a = 10;
    $b = 15;
    echo "Sum = " , sumcommDiv($a, $b);


?>

Javascript

<script>

// Javascript implementation of above approach

// Function to calculate gcd of two numbers
function gcd(a, b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}

// Function to calculate all common divisors
// of two given numbers
// a, b --> input integer numbers
function sumcommDiv(a, b)
{
    // find gcd of a, b
    var n = gcd(a, b);

    // Find the sum of divisors of n.
    var sum = 0;
    for (var i = 1; i <= Math.sqrt(n); i++) {

        // if 'i' is factor of n
        if (n % i == 0) {

            // check if divisors are equal
            if (n / i == i)
                sum += i;
            else
                sum += (n / i) + i;
        }
    }

    return sum;
}

// Driver program to run the case
var a = 10, b = 15;
document.write( "Sum = " + sumcommDiv(a, b));

// This code is contributed by rutvik_56.
</script>

Output:

Sum = 6

Time complexity: O(sqrt(n))

Auxiliary Space: O(logn)