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; ` `} ` |

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## 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` `)); ` `} ` ` ` `} ` |

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## 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 ` |

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## 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 ` |

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## 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 ` `?> ` |

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**Output:**

42

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