Related Articles

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

• Last Updated : 31 Mar, 2021

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 ``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

 ``

## Javascript

 ``
Output:
`45`

Time Complexity : O(1)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

My Personal Notes arrow_drop_up