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

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 <bits/stdc++.h>
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

<?php
// PHP program to find sum of numbers
// from 1 to N which are divisible by 2 or 7

// Function to calculate the sum
// of numbers divisible by 2 or 7
function sum($N)
{
    $S1 = (int)(($N / 2)) * (int)(2 * 2 +
           (int)($N / 2 - 1) * 2) / 2;
    $S2 = (int)(($N / 7)) * (int)(2 * 7 +
           (int)($N / 7 - 1) * 7) / 2;
    $S3 = (int)(($N / 14)) * (int)(2 * 14 +
           (int)($N / 14 - 1) * 14) / 2;

    return ($S1 + $S2) - $S3;
}

// Driver code
$N = 20;

echo sum($N);

// This Code is Contributed by akt_mit
?>

Javascript

<script>
// javascript  program to find sum of
// numbers from 1 to N which
// are divisible by 2 or 7

// Function to calculate the sum
// of numbers divisible by 2 or 7
function sum(N)
{
    var S1, S2, S3;

    S1 = (((N / 2)) * parseInt(2 * 2 +
        parseInt(N / 2 - 1) * 2) / 2);
    S2 = (parseInt(parseInt(N / 7)) * (2 * 7 +
        parseInt(N / 7 - 1) * 7) / 2);
    S3 = (parseInt(parseInt(N / 14)) * (2 * 14 +
        parseInt(N / 14 - 1) * 14) / 2);

    return S1 + S2 - S3;
}

// Driver code
var N = 20;
document.write( sum(N));

// This code is contributed by shikhasingrajput
</script>

Output:

117

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

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