Sum of GCD of all numbers upto N with N itself (original) (raw)

Last Updated : 15 Jul, 2025

Given an integer N, the task is to find the sum of Greatest Common Divisors of all numbers up to N with N itself.
Examples:

Input: N = 12
Output: 40
Explanation:
GCD of [1, 12] = 1, [2, 12] = 2, [3, 12] = 3, [4, 12] = 4, [5, 12] = 1, [6, 12] = 6, [7, 12] = 1, [8, 12] = 4, [9, 12] = 3, [10, 12] = 2, [11, 12] = 1, [12, 12] = 12. The sum is (1 + 2 + 3 + 4 + 1 + 6 + 1 + 4 + 3 + 2 + 1 + 12) = 40.
Input: N = 2
Output: 3
Explanation:
GCD of [1, 2] = 1, [2, 2] = 2 and their sum is 3.

Try It Yourself

Naive Approach: A simple solution is to iterate over all numbers from 1 to N and find their gcd with N itself and keep on adding them.
Time Complexity: O(N * log N)

Efficient Approach:
To optimize the above-mentioned approach, we need to observe that GCD(i, N) gives one of the divisors of N. So, instead of running a loop from 1 to N, we can check for each divisor of N how many numbers are there with GCD(i, N) same as that divisor.

Illustration:
For example N = 12, its divisors are 1, 2, 3, 4, 6, 12.
Numbers in range [1, 12] whose GCD with 12 is:

1 are {1, 5, 7, 11}

2 are {2, 10}

3 are {3, 9}

4 are {4, 8}

6 is {6}

12 is {12}

So answer is; 1*4 + 2*2 + 3*2 + 4*2 + 6*1 + 12*1 = 40.

So we have to find the number of integers from 1 to N with GCD d, where d is a divisor of N. Let us consider x1, x2, x3, .... xn as the different integers from 1 to N such that their GCD with N is d.
Since, GCD(xi, N) = d then GCD(xi/d, N/d) = 1
So, the count of integers from 1 to N whose GCD with N is d is Euler Totient Function of (N/d).

Below is the implementation of the above approach:

C++ `

// C++ Program to find the Sum // of GCD of all integers up to N // with N itself

#include <bits/stdc++.h> using namespace std;

// Function to Find Sum of // GCD of each numbers int getCount(int d, int n) {

int no = n / d;

int result = no;

// Consider all prime factors
// of no. and subtract their
// multiples from result
for (int p = 2; p * p <= no; ++p) {

    // Check if p is a prime factor
    if (no % p == 0) {

        // If yes, then update no
        // and result
        while (no % p == 0)
            no /= p;
        result -= result / p;
    }
}

// If no has a prime factor greater
// than sqrt(n) then at-most one such
// prime factor exists
if (no > 1)
    result -= result / no;

// Return the result
return result;

}

// Finding GCD of pairs int sumOfGCDofPairs(int n) { int res = 0;

for (int i = 1; i * i <= n; i++) {
    if (n % i == 0) {
        // Calculate the divisors
        int d1 = i;
        int d2 = n / i;

        // Return count of numbers
        // from 1 to N with GCD d with N
        res += d1 * getCount(d1, n);

        // Check if d1 and d2 are
        // equal then skip this
        if (d1 != d2)
            res += d2 * getCount(d2, n);
    }
}

return res;

}

// Driver code int main() { int n = 12;

cout << sumOfGCDofPairs(n);

return 0;

}

Java

// Java program to find the Sum // of GCD of all integers up to N // with N itself class GFG{

// Function to Find Sum of // GCD of each numbers static int getCount(int d, int n) { int no = n / d; int result = no;

// Consider all prime factors
// of no. and subtract their
// multiples from result
for(int p = 2; p * p <= no; ++p)
{

    // Check if p is a prime factor
    if (no % p == 0) 
    {

        // If yes, then update no
        // and result
        while (no % p == 0)
            no /= p;
        result -= result / p;
    }
}

// If no has a prime factor greater
// than Math.sqrt(n) then at-most one such
// prime factor exists
if (no > 1)
    result -= result / no;

// Return the result
return result;

}

// Finding GCD of pairs static int sumOfGCDofPairs(int n) { int res = 0;

for(int i = 1; i * i <= n; i++)
{
    if (n % i == 0)
    {
        
        // Calculate the divisors
        int d1 = i;
        int d2 = n / i;

        // Return count of numbers
        // from 1 to N with GCD d with N
        res += d1 * getCount(d1, n);

        // Check if d1 and d2 are
        // equal then skip this
        if (d1 != d2)
            res += d2 * getCount(d2, n);
    }
}
return res;

}

// Driver code public static void main(String[] args) { int n = 12;

System.out.print(sumOfGCDofPairs(n));

} }

// This code is contributed by Amit Katiyar

Python3

Python3 program to find the sum

of GCD of all integers up to N

with N itself

Function to Find Sum of

GCD of each numbers

def getCount(d, n):

no = n // d;
result = no;

# Consider all prime factors
# of no. and subtract their
# multiples from result
for p in range(2, int(pow(no, 1 / 2)) + 1):

    # Check if p is a prime factor
    if (no % p == 0):

        # If yes, then update no
        # and result
        while (no % p == 0):
            no //= p;
            
        result -= result // p;

# If no has a prime factor greater
# than Math.sqrt(n) then at-most one such
# prime factor exists
if (no > 1):
    result -= result // no;

# Return the result
return result;

Finding GCD of pairs

def sumOfGCDofPairs(n):

res = 0;

for i in range(1, int(pow(n, 1 / 2)) + 1):
    if (n % i == 0):

        # Calculate the divisors
        d1 = i;
        d2 = n // i;

        # Return count of numbers
        # from 1 to N with GCD d with N
        res += d1 * getCount(d1, n);

        # Check if d1 and d2 are
        # equal then skip this
        if (d1 != d2):
            res += d2 * getCount(d2, n);

return res;

Driver code

if name == 'main':

n = 12;

print(sumOfGCDofPairs(n));

This code is contributed by Amit Katiyar

C#

// C# program to find the sum // of GCD of all integers up to N // with N itself using System;

class GFG{

// Function to find sum of // GCD of each numbers static int getCount(int d, int n) { int no = n / d; int result = no;

// Consider all prime factors
// of no. and subtract their
// multiples from result
for(int p = 2; p * p <= no; ++p)
{

    // Check if p is a prime factor
    if (no % p == 0) 
    {

        // If yes, then update no
        // and result
        while (no % p == 0)
            no /= p;
            
        result -= result / p;
    }
}

// If no has a prime factor greater
// than Math.Sqrt(n) then at-most 
// one such prime factor exists
if (no > 1)
    result -= result / no;

// Return the result
return result;

}

// Finding GCD of pairs static int sumOfGCDofPairs(int n) { int res = 0;

for(int i = 1; i * i <= n; i++)
{
    if (n % i == 0)
    {
        
        // Calculate the divisors
        int d1 = i;
        int d2 = n / i;

        // Return count of numbers
        // from 1 to N with GCD d with N
        res += d1 * getCount(d1, n);

        // Check if d1 and d2 are
        // equal then skip this
        if (d1 != d2)
            res += d2 * getCount(d2, n);
    }
}
return res;

}

// Driver code public static void Main(String[] args) { int n = 12;

Console.Write(sumOfGCDofPairs(n));

} }

// This code is contributed by Amit Katiyar

JavaScript

Time Complexity: O(N)
Auxiliary Space: O(1)