Sum of first N natural numbers which are divisible by X or Y

0 0 6 minutes read

Given a number N. Given two numbers X and Y, the task is to find the sum of all those numbers from 1 to N that are divisible by X or by Y.
Examples:

Input : N = 20
Output : 98

Input : N = 14
Output : 45

Approach: To solve the problem, follow the below steps:
->Find the sum of numbers that are divisible by X upto N. Denote it by S1.
->Find the sum of numbers that are divisible by Y upto N. Denote it by S2.
->Find the sum of numbers that are divisible by both X and Y upto N. Denote it by S3.
->The final answer will be S1 + S2 – S3.
In order to find the sum, we can use the general formula of A.P. which is:

Sn = (n/2) * {2*a + (n-1)*d}

For S1: The total numbers that will be divisible by X upto N will be N/X and the sum will be:

Hence, 
S1 = ((N/X)/2) * (2 * X + (N/X - 1) * X)

For S2: The total numbers that will be divisible by Y upto N will be N/Y and the sum will be:

Hence, 
S2 = ((N/Y)/2) * (2 * Y + (N/Y - 1) * Y)

For S3: The total numbers that will be divisible by both X and Y upto N will be N/lcm(X, Y) and the sum will be:

Hence, 
S2 = ((N/lcm(X, Y))/2) * ((2 * lcm(X,Y)) + (N/lcm(X,Y) - 1) * lcm(X,Y))/2

Therefore, the result will be:

S = S1 + S2 - S3

Below is the implementation of the above approach:

C++

// C++ program to find sum of numbers from
// 1 to N which are divisible by X or Y
#include <bits/stdc++.h>
using namespace std;
 
int gcd(int a, int b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}
  
// Function to return LCM of two numbers
int lcm(int a, int b)
{
    return (a / gcd(a, b)) * b;
}
 
// Function to calculate the sum
// of numbers divisible by X or Y
int sum(int N, int X, int Y)
{
    int S1, S2, S3;
 
    S1 = ((N / X)) * (2 * X + (N / X - 1) * X) / 2;
    S2 = ((N / Y)) * (2 * Y + (N / Y - 1) * Y) / 2;
    S3 = ((N / lcm(X,Y))) * (2 * (lcm(X,Y))
                      + (N / (lcm(X,Y)) - 1) * (lcm(X,Y)))/ 2;
 
    return S1 + S2 - S3;
}
 
// Driver code
int main()
{
    int N = 24;
    int X = 4, Y = 6;
 
    cout << sum(N, X, Y);
 
    return 0;
}

Java

// Java program to find sum of numbers from
// 1 to N which are divisible by X or Y
 
public class GFG{
     
   static int gcd(int a, int b)
    {
        if (a == 0)
            return b;
        return gcd(b % a, a);
    }
      
    // method to return LCM of two numbers
    static int lcm(int a, int b)
    {
        return (a / gcd(a, b)) * b;
    }
      
   
    // Function to calculate the sum
    // of numbers divisible by X or Y
    static int sum(int N, int X, int Y)
    {
        int S1, S2, S3;
     
        S1 = ((N / X)) * (2 * X + (N / X - 1) * X) / 2;
        S2 = ((N / Y)) * (2 * Y + (N / Y - 1) * Y) / 2;
        S3 = ((N / ( lcm(X, Y)))) * (2 * ( lcm(X, Y))
                          + (N / ( lcm(X, Y)) - 1) * ( lcm(X, Y)))/ 2;
     
        return S1 + S2 - S3;
    }
     
    // Driver code
    public static void main(String []args)
    {
        int N = 14;
        int X = 3, Y = 5;
     
        System.out.println(sum(N, X, Y));
     
    }
    // This code is contributed by Ryuga
}

Python3

# Python 3 program to find sum of numbers from
# 1 to N which are divisible by X or Y
from math import ceil, floor
 
def gcd(a,b):
    if a == 0:
        return b
    return gcd(b % a, a)
   
def lcm(a,b):
    return (a / gcd(a,b))* b
   
# Function to calculate the sum
# of numbers divisible by X or Y
def sum(N, X, Y):
    S1 = floor(floor(N / X) * floor(2 * X +
               floor(N / X - 1) * X) / 2)
    S2 = floor(floor(N / Y)) * floor(2 * Y +
               floor(N / Y - 1) * Y) / 2
    S3 = floor(floor(N / (lcm(X,Y)))) * floor (2 * (lcm(X,Y)) +
               floor(N / (lcm(X,Y)) - 1) * (lcm(X,Y)))/ 2
 
    return S1 + S2 - S3
 
# Driver code
if __name__ == '__main__':
    N = 14
    X = 3
    Y = 5
 
    print(int(sum(N, X, Y)))
 
# This code is contributed by 
# Surendra_Gangwar

C#

// C# program to find sum of numbers from
// 1 to N which are divisible by X or Y
  
using System;
public class GFG{
   
    static int gcd(int a, int b)
    {
        if (a == 0)
            return b;
        return gcd(b % a, a);
    }
      
    // method to return
    // LCM of two numbers
    static int lcm(int a, int b)
    {
        return (a / gcd(a, b)) * b;
    }
      
    // Function to calculate the sum
    // of numbers divisible by X or Y
    static int sum(int N, int X, int Y)
    {
        int S1, S2, S3;
      
        S1 = ((N / X)) * (2 * X + (N / X - 1) * X) / 2;
        S2 = ((N / Y)) * (2 * Y + (N / Y - 1) * Y) / 2;
        S3 = ((N / (lcm(X, Y)))) * (2 * (lcm(X, Y))
                          + (N / (lcm(X, Y)) - 1) * (lcm(X, Y)))/ 2;
      
        return S1 + S2 - S3;
    }
      
    // Driver code
    public static void Main()
    {
        int N = 14;
        int X = 3, Y = 5;
      
        Console.Write(sum(N, X, Y));
      
    }
     
}

PHP

<?php
   
function gcd( $a, $b)
{
   if ($a == 0)
        return $b;
    return gcd($b % $a, $a);
}
  
// Function to return LCM
// of two numbers
function lcm( $a, $b)
{
    return ($a / gcd($a, $b)) * $b;
}  
// PHP program to find sum of numbers from
// 1 to N which are divisible by X or Y
// Function to calculate the sum
// of numbers divisible by X or Y
function sum($N, $X, $Y)
{
    $S1; $S2; $S3;
 
    $S1 = floor(((int)$N / $X)) * (2 * $X + (int)((int)$N / $X - 1) * $X) / 2;
    $S2 = floor(((int)$N / $Y)) * (2 * $Y + (int)((int)$N / $Y - 1) * $Y) / 2;
    $S3 = floor(((int)$N / (lcm( $X, $Y)))) * (2 * (lcm( $X, $Y))
                    + ((int)$N / (lcm( $X, $Y)) - 1) * (int)(lcm( $X, $Y)))/ 2;
 
    return ceil($S1 + ($S2 - $S3));
}
 
// Driver code
    $N = 14;
    $X = 3;
    $Y = 5;
 
    echo  sum($N, $X, $Y);
 
#This code is contributed by ajit.
?>

Javascript

<script>
// javascript program to find sum of numbers from
// 1 to N which are divisible by X or Y
 
function gcd(a, b)
{
if (b == 0)
    return a;
return gcd(b, a % b);
}
  
// Function to return LCM of two numbers
function lcm(a, b)
{
    return (a / gcd(a, b)) * b;
}
 
// Function to calculate the sum
// of numbers divisible by X or Y
function sum(N , X , Y)
{
    var S1, S2, S3;
 
    S1 = (parseInt(N / X)) * (2 * X + parseInt(N / X - 1) * X) / 2;
    S2 = (parseInt(N / Y)) * (2 * Y + parseInt(N / Y - 1) * Y) / 2;
    S3 = (parseInt(N / (lcm(X, Y)))) * (2 * (lcm(X, Y))
                      + parseInt(N / (lcm(X, Y)) - 1) * (lcm(X, Y)))/ 2;
 
    return S1 + S2 - S3;
}
 
// Driver code
var N = 14;
var X = 3, Y = 5;
 
document.write(sum(N, X, Y));
 
// This code is contributed by Princi Singh 
</script>

C

// C program to find sum of numbers from
// 1 to N which are divisible by X or Y
#include <stdio.h>
 
int gcd(int a, int b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}
  
// Function to return LCM of two numbers
int lcm(int a, int b)
{
    return (a / gcd(a, b)) * b;
}
 
// Function to calculate the sum
// of numbers divisible by X or Y
int sum(int N, int X, int Y)
{
    int S1, S2, S3;
 
    S1 = ((N / X)) * (2 * X + (N / X - 1) * X) / 2;
    S2 = ((N / Y)) * (2 * Y + (N / Y - 1) * Y) / 2;
    S3 = ((N / (lcm(X, Y)))) * (2 * (lcm(X, Y)) + (N / (lcm(X, Y)) - 1) * (lcm(X, Y)))/ 2;
    return S1 + S2 - S3;
}
 
// Driver code
int main()
{
    int N = 14;
    int X = 3, Y = 5;
    printf("%d ",sum(N, X, Y));
    return 0;
}

Output

Time Complexity: O(log(min(X,Y)), for calculating gcd
Auxiliary Space: O(log(min(X,Y))

Recommended

Solve DSA problems on GfG Practice.

Solve Problems

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!