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

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 (X*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/(X*Y) and the sum will be:

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

Therefore, the result will be:

S = S1 + S2 - S3

Below is the implementation of the above approach:

C++

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// 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;
  
// 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 / (X * Y))) * (2 * (X * Y)
                      + (N / (X * Y) - 1) * (X * Y))/ 2;
  
    return S1 + S2 - S3;
}
  
// Driver code
int main()
{
    int N = 14;
    int X = 3, Y = 5;
  
    cout << sum(N, X, Y);
  
    return 0;
}

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Java

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// Java program to find sum of numbers from
// 1 to N which are divisible by X or Y
  
public class GFG{
      
    // 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 / (X * Y))) * (2 * (X * Y)
                          + (N / (X * Y) - 1) * (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
}

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C#

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// C# program to find sum of numbers from
// 1 to N which are divisible by X or Y
   
using System;
public class GFG{
       
    // 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 / (X * Y))) * (2 * (X * Y)
                          + (N / (X * Y) - 1) * (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));
       
    }
      
}

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PHP

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<?php
// 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 / ($X * $Y))) * (2 * ($X * $Y)
                    + ((int)$N / ($X * $Y) - 1) * (int)($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.
?>

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Output:

45


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Improved By : Ryuga, ChitraNayal