# Sum of first N natural numbers which are divisible by 2 and 7

Given a number N. The task is to find the sum of all those numbers from 1 to N that are divisible by 2 or by 7.

Examples:

```Input : N = 7
Output : 19
sum = 2 + 4 + 6 + 7

Input : N = 14
Output : 63
sum = 2 + 4 + 6 + 7 + 8 + 10 + 12 + 14
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: To solve the problem, follow the below steps:

->Find the sum of numbers that are divisible by 2 upto N. Denote it by S1.
->Find the sum of numbers that are divisible by 7 upto N. Denote it by S2.
->Find the sum of numbers that are divisible by 14(2*7) 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 2 upto N will be N/2 and the series will be 2, 4, 6, 8, ….

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

For S2: The total numbers that will be divisible by 7 up to N will be N/7 and the series will be 7, 14, 21, 28, ……

```Hence,
S2 = ((N/7)/2) * (2 * 7 + (N/7 - 1) * 7)
```

For S3: The total numbers that will be divisible by 14 upto N will be N/14.

```Hence,
S3 = ((N/14)/2) * (2 * 14 + (N/14 - 1) * 14)
```

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 2 or 7 ` `#include ` `using` `namespace` `std; ` ` `  `// Function to calculate the sum ` `// of numbers divisible by 2 or 7 ` `int` `sum(``int` `N) ` `{ ` `    ``int` `S1, S2, S3; ` ` `  `    ``S1 = ((N / 2)) * (2 * 2 + (N / 2 - 1) * 2) / 2; ` `    ``S2 = ((N / 7)) * (2 * 7 + (N / 7 - 1) * 7) / 2; ` `    ``S3 = ((N / 14)) * (2 * 14 + (N / 14 - 1) * 14) / 2; ` ` `  `    ``return` `S1 + S2 - S3; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `N = 20; ` ` `  `    ``cout << sum(N); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java  program to find sum of  ` `// numbers from 1 to N which  ` `// are divisible by 2 or 7  ` ` `  `import` `java.io.*; ` ` `  `class` `GFG { ` `// Function to calculate the sum  ` `// of numbers divisible by 2 or 7  ` `public` `static` `int` `sum(``int` `N)  ` `{  ` `    ``int` `S1, S2, S3;  ` ` `  `    ``S1 = ((N / ``2``)) * (``2` `* ``2` `+  ` `        ``(N / ``2` `- ``1``) * ``2``) / ``2``;  ` `    ``S2 = ((N / ``7``)) * (``2` `* ``7` `+  ` `        ``(N / ``7` `- ``1``) * ``7``) / ``2``;  ` `    ``S3 = ((N / ``14``)) * (``2` `* ``14` `+  ` `        ``(N / ``14` `- ``1``) * ``14``) / ``2``;  ` ` `  `    ``return` `S1 + S2 - S3;  ` `}  ` ` `  `// Driver code  ` `    ``public` `static` `void` `main (String[] args) { ` ` `  `    ``int` `N = ``20``;  ` `    ``System.out.println( sum(N));  ` `    ``}  ` `}  ` ` `  `// This code is contributed by ajit `

## Python3

 `# Python3 implementation of  ` `# above approach ` ` `  `# Function to calculate the sum  ` `# of numbers divisible by 2 or 7  ` `def` `sum``(N):  ` `     `  `    ``S1 ``=` `((N ``/``/` `2``)) ``*` `(``2` `*` `2` `+` `(N ``/``/` `2` `-` `1``) ``*` `2``) ``/``/` `2` `    ``S2 ``=` `((N ``/``/` `7``)) ``*` `(``2` `*` `7` `+` `(N ``/``/` `7` `-` `1``) ``*` `7``) ``/``/` `2` `    ``S3 ``=` `((N ``/``/` `14``)) ``*` `(``2` `*` `14` `+` `(N ``/``/` `14` `-` `1``) ``*` `14``) ``/``/` `2` ` `  `    ``return` `S1 ``+` `S2 ``-` `S3 ` ` `  ` `  `# Driver code  ` `if` `__name__``=``=``'__main__'``: ` `    ``N ``=` `20` ` `  `    ``print``(``sum``(N)) ` ` `  `# This code is written by ` `# Sanjit_Prasad `

## C#

 `// C# program to find sum of  ` `// numbers from 1 to N which ` `// are divisible by 2 or 7 ` `using` `System; ` ` `  `class` `GFG ` `{ ` `// Function to calculate the sum ` `// of numbers divisible by 2 or 7 ` `public` `static` `int` `sum(``int` `N) ` `{ ` `    ``int` `S1, S2, S3; ` ` `  `    ``S1 = ((N / 2)) * (2 * 2 +  ` `          ``(N / 2 - 1) * 2) / 2; ` `    ``S2 = ((N / 7)) * (2 * 7 +  ` `          ``(N / 7 - 1) * 7) / 2; ` `    ``S3 = ((N / 14)) * (2 * 14 + ` `          ``(N / 14 - 1) * 14) / 2; ` ` `  `    ``return` `S1 + S2 - S3; ` `} ` ` `  `// Driver code ` `public` `static` `int` `Main() ` `{ ` `    ``int` `N = 20; ` `    ``Console.WriteLine( sum(N)); ` `    ``return` `0; ` `} ` `} ` ` `  `// This code is contributed  ` `// by SoumikMondal `

## PHP

 ` `

Output:

```117
```

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