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++

// 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; 
} 

Java

// 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 
} 

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 
  
# 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 / (X * Y))) * floor (2 * (X * Y) + 
               floor(N / (X * Y) - 1) * (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{ 
       
    // 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)); 
       
    } 
      
} 

PHP

<?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. 
?> 

Output:

Time Complexity : O(1)

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C++

Java

Python3

C#

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