# 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 = 5Output: 7 sum = 3 + 4Input: N = 12Output: 42 sum = 3 + 4 + 6 + 8 + 9 + 12

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

- Find the sum of numbers that are divisible by 3 upto N. Denote it by S1.
- Find the sum of numbers that are divisible by 4 upto N. Denote it by S2.
- Find the sum of numbers that are divisible by 12(3*4) 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:

S_{n}= (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 <bits/stdc++.h>` `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

`<?php` `// PHP 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` `function` `sum(` `$N` `)` `{` ` ` `$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` `$N` `= 20;` `echo` `sum(12);` `// This code is contributed` `// by inder_verma` `?>` |

## Javascript

`<script>` `// Javascript 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` `function` `sum(N)` `{` ` ` `var` `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` `var` `N = 20;` `document.write( sum(12));` `</script>` |

**Output:**

42

**Time Complexity: **O(1), since there is no loop or recursion.

**Auxiliary Space: **O(1), since no extra space has been taken.

## Please

Loginto comment...