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Sum of all the multiples of 3 and 7 below N
  • Last Updated : 05 Apr, 2021

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:  



  • We know that multiples of 3 form an AP as S3 = 3 + 6 + 9 + 12 + 15 + 18 + 21 + …
  • And the multiples of 7 form an AP as S7 = 7 + 14 + 21 + 28 + …
  • Now, Sum = S3 + S7 i.e. 3 + 6 + 7 + 9 + 12 + 14 + 15 + 18 + 21 + 21 + …
  • From the previous step, 21 is repeated twice. In fact, all of the multiples of 21 (or 3*7) will be repeated as they are counted twice, once in the series S3 and again in the series S7. So, the multiples of 21 need to be discarded from the result.
  • So, the final result will be S3 + S7 – S21

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:  

C++




// C++ 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;
}

Java




// 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

Python3




# 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

C#




// 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

PHP




<?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
?>

Javascript




<script>
 
// JavaScript 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 = parseInt(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
let n = 24;
 
document.write(sumMultiples(n));
 
// This code is contributed by mohan1240760
 
</script>
Output: 
105

 

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