# Sum of first N natural numbers which are divisible by 2 and 7

• Last Updated : 13 Jun, 2022

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 using namespace std; // Function to calculate the sum// of numbers divisible by 2 or 7int 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 codeint 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 7public 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 7def 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 codeif __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 7using System; class GFG{// Function to calculate the sum// of numbers divisible by 2 or 7public 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 codepublic static int Main(){    int N = 20;    Console.WriteLine( sum(N));    return 0;}} // This code is contributed// by SoumikMondal



## Javascript



Output:

117

Time Complexity: O(1)

Auxiliary Space: O(1)

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