# Sum of numbers from 1 to N which are divisible by 3 or 4

Given a number N. The task is to find the sum of all those numbers from 1 to N that are divisible by 3 or by 4.

Examples:

```Input : N = 5
Output : 7
sum = 3 + 4

Input : N = 12
Output : 42
sum = 3 + 4 + 6 + 8 + 9 + 12
```

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

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

1. Find the sum of numbers that are divisible by 3 upto N. Denote it by S1.
2. Find the sum of numbers that are divisible by 4 upto N. Denote it by S2.
3. Find the sum of numbers that are divisible by 12(3*4) upto N. Denote it by S3.
4. 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}

Where,
n -> total number of terms
a -> first term
d -> common difference
```

For S1: The total numbers that will be divisible by 3 upto N will be N/3 and the series will be 3, 6, 9, 12, ….

```Hence,
S1 = ((N/3)/2) * (2 * 3 + (N/3 - 1) * 3)
```

For S2: The total numbers that will be divisible by 4 up to N will be N/4 and the series will be 4, 8, 12, 16, …...

```Hence,
S2 = ((N/4)/2) * (2 * 4 + (N/4 - 1) * 4)
```

For S3: The total numbers that will be divisible by 12 upto N will be N/12.

```Hence,
S3 = ((N/12)/2) * (2 * 12 + (N/12 - 1) * 12)
```

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 3 or 4 ` `#include ` `using` `namespace` `std; ` ` `  `// Function to calculate the sum ` `// of numbers divisible by 3 or 4 ` `int` `sum(``int` `N) ` `{ ` `    ``int` `S1, S2, S3; ` ` `  `    ``S1 = ((N / 3)) * (2 * 3 + (N / 3 - 1) * 3) / 2; ` `    ``S2 = ((N / 4)) * (2 * 4 + (N / 4 - 1) * 4) / 2; ` `    ``S3 = ((N / 12)) * (2 * 12 + (N / 12 - 1) * 12) / 2; ` ` `  `    ``return` `S1 + S2 - S3; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `N = 20; ` ` `  `    ``cout << sum(12); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to find sum of numbers from 1 to N  ` `// which are divisible by 3 or 4  ` `class` `GFG{ ` ` `  `// Function to calculate the sum  ` `// of numbers divisible by 3 or 4  ` `static` `int` `sum(``int` `N)  ` `{  ` `    ``int` `S1, S2, S3;  ` ` `  `    ``S1 = ((N / ``3``)) * (``2` `* ``3` `+ (N / ``3` `- ``1``) * ``3``) / ``2``;  ` `    ``S2 = ((N / ``4``)) * (``2` `* ``4` `+ (N / ``4` `- ``1``) * ``4``) / ``2``;  ` `    ``S3 = ((N / ``12``)) * (``2` `* ``12` `+ (N / ``12` `- ``1``) * ``12``) / ``2``;  ` ` `  `    ``return` `S1 + S2 - S3;  ` `}  ` ` `  `// Driver code  ` ` ``public` `static` `void` `main (String[] args) { ` `    ``int` `N = ``20``;  ` ` `  `    ``System.out.print(sum(``12``));  ` `} ` ` `  `}  `

## Python3

 `# Python3 program to find sum of numbers  ` `# from 1 to N ` `# which are divisible by 3 or 4 ` ` `  `# Function to calculate the sum  ` `# of numbers divisible by 3 or 4  ` `def` `sum``(N): ` ` `  `    ``global` `S1,S2,S3 ` ` `  `    ``S1 ``=` `(((N ``/``/` `3``)) ``*`  `         ``(``2` `*` `3` `+` `(N ``/``/``3` `-` `1``) ``*` `3``) ``/``/``2``) ` `    ``S2 ``=` `(((N ``/``/` `4``)) ``*`  `         ``(``2` `*` `4` `+` `(N ``/``/` `4` `-` `1``) ``*` `4``) ``/``/` `2``) ` `    ``S3 ``=` `(((N ``/``/` `12``)) ``*`  `         ``(``2` `*` `12` `+` `(N ``/``/` `12` `-` `1``) ``*` `12``) ``/``/` `2``) ` ` `  `    ``return` `int``(S1 ``+` `S2 ``-` `S3) ` ` `  `if` `__name__``=``=``'__main__'``: ` `    ``N ``=` `12` `    ``print``(``sum``(N)) ` ` `  `# This code is contributed by Shrikant13 `

## C#

 `// C# program to find sum of  ` `// numbers from 1 to N which  ` `// are divisible by 3 or 4  ` `using` `System; ` ` `  `class` `GFG ` `{ ` ` `  `// Function to calculate the sum  ` `// of numbers divisible by 3 or 4  ` `static` `int` `sum(``int` `N)  ` `{  ` `    ``int` `S1, S2, S3;  ` ` `  `    ``S1 = ((N / 3)) * (2 * 3 +  ` `          ``(N / 3 - 1) * 3) / 2;  ` `    ``S2 = ((N / 4)) * (2 * 4 +  ` `          ``(N / 4 - 1) * 4) / 2;  ` `    ``S3 = ((N / 12)) * (2 * 12 +  ` `          ``(N / 12 - 1) * 12) / 2;  ` ` `  `    ``return` `S1 + S2 - S3;  ` `}  ` ` `  `// Driver code  ` `public` `static` `void` `Main ()  ` `{ ` `    ``int` `N = 20;  ` ` `  `    ``Console.WriteLine(sum(12));  ` `} ` `}  ` ` `  `// This code is contributed ` `// by inder_verma `

## PHP

 ` `

Output:

```42
```

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