Sum of all the multiples of 3 and 7 below N

Given a number N, the task is to find the sum of all the multiples of 3 and 7 below N.
Note: A number must not repeat itself in the sum.

Examples:

Input: N = 10
Output: 25
3 + 6 + 7 + 9 = 25

Input: N = 24
Output: 105
3 + 6 + 7 + 9 + 12 + 14 + 15 + 18 + 21 = 105

Approach:



The formula for the sum of an AP series is :
n * ( a + l ) / 2
Where n is the number of terms, a is the starting term, and l is the last term.

Below is the implementation of the above approach:

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// CPP program to find the sum of all
// multiples of 3 and 7 below N
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to find sum of AP series
long long sumAP(long long n, long long d)
{
    // Number of terms
    n /= d;
  
    return (n) * (1 + n) * d / 2;
}
  
// Function to find the sum of all
// multiples of 3 and 7 below N
long long sumMultiples(long long n)
{
    // Since, we need the sum of
    // multiples less than N
    n--;
  
    return sumAP(n, 3) + sumAP(n, 7) - sumAP(n, 21);
}
  
// Driver code
int main()
{
    long long n = 24;
  
    cout << sumMultiples(n);
  
    return 0;
}
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// Java program to find the sum of all
// multiples of 3 and 7 below N
import java.util.*;
  
class solution
{
  
// Function to find sum of AP series
static long sumAP(long n, long d)
{
    // Number of terms
    n /= d;
  
    return (n) * (1 + n) * d / 2;
}
  
// Function to find the sum of all
// multiples of 3 and 7 below N
static long sumMultiples(long n)
{
    // Since, we need the sum of
    // multiples less than N
    n--;
  
    return sumAP(n, 3) + sumAP(n, 7) - sumAP(n, 21);
}
  
// Driver code
public static void main(String args[])
{
    long n = 24;
  
    System.out.println(sumMultiples(n));
  
 }
}
  
//This code is contributed by Surendra_Gangwar
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# Python3 program to find the sum of 
# all multiples of 3 and 7 below N
  
# Function to find sum of AP series
def sumAP(n, d):
      
    # Number of terms
    n = int(n / d);
  
    return (n) * (1 + n) * (d / 2);
  
# Function to find the sum of all
# multiples of 3 and 7 below N
def sumMultiples(n):
  
    # Since, we need the sum of
    # multiples less than N
    n -= 1;
  
    return int(sumAP(n, 3) + 
               sumAP(n, 7) - 
               sumAP(n, 21));
  
# Driver code
n = 24;
  
print(sumMultiples(n));
  
# This code is contributed 
# by mits
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// C# program to find the sum of all 
// multiples of 3 and 7 below N 
using System; 
  
class GFG 
  
// Function to find sum of AP series 
static long sumAP(long n, long d) 
    // Number of terms 
    n /= d; 
  
    return (n) * (1 + n) * d / 2; 
  
// Function to find the sum of all 
// multiples of 3 and 7 below N 
static long sumMultiples(long n) 
    // Since, we need the sum of 
    // multiples less than N 
    n--; 
  
    return sumAP(n, 3) + sumAP(n, 7) -
                         sumAP(n, 21); 
  
// Driver code 
static public void Main(String []args) 
    long n = 24; 
  
    Console.WriteLine(sumMultiples(n)); 
  
// This code is contributed 
// by Arnab Kundu
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<?php
// PHP program to find the sum of all
// multiples of 3 and 7 below N
  
// Function to find sum of AP series
function sumAP($n, $d)
{
    // Number of terms
    $n = (int)($n / $d);
  
    return ($n) * (1 + $n) * ($d / 2);
}
  
// Function to find the sum of all
// multiples of 3 and 7 below N
function sumMultiples($n)
{
    // Since, we need the sum of
    // multiples less than N
    $n--;
  
    return sumAP($n, 3) + 
           sumAP($n, 7) - sumAP($n, 21);
}
  
// Driver code
$n = 24;
  
echo sumMultiples($n);
  
// This code is contributed 
// by Akanksha Rai
?>
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Output:
105




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