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

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 <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.