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++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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PHP

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<?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
?>

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Output:

117


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Improved By : Sanjit_Prasad, SoM15242, jit_t