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Sum of all the multiples of 3 and 7 below N

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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 

Brute Force Approach:

A brute force approach to solve this problem would be to iterate through all the numbers from 1 to N-1, and check if each number is a multiple of 3 or 7. If it is, add it to the sum. However, we need to make sure that a number is not added to the sum multiple times. To do this, we can use a set to keep track of the numbers that have already been added.

Below is the implementation of the above approach:  

C++




#include <bits/stdc++.h>
using namespace std;
 
// Function to find the sum of all
// multiples of 3 and 7 below N
long long sumMultiples(long long n)
{
    long long sum = 0;
    set<long long> multiples;
 
    for (long long i = 1; i < n; i++) {
        if (i % 3 == 0 || i % 7 == 0) {
            if (multiples.find(i) == multiples.end()) {
                sum += i;
                multiples.insert(i);
            }
        }
    }
 
    return sum;
}
 
// Driver code
int main()
{
    long long n = 24;
 
    cout << sumMultiples(n);
 
    return 0;
}


Java




import java.util.HashSet;
import java.util.Set;
 
public class Main {
 
  // Function to find the sum of all
  // multiples of 3 and 7 below N
  static long sumMultiples(long n) {
    long sum = 0;
    Set<Long> multiples = new HashSet<>();
 
    for (long i = 1; i < n; i++) {
      if (i % 3 == 0 || i % 7 == 0) {
        if (!multiples.contains(i)) {
          sum += i;
          multiples.add(i);
        }
      }
    }
 
    return sum;
  }
 
  // Driver code
  public static void main(String[] args) {
    long n = 24;
 
    System.out.println(sumMultiples(n));
  }
}
 
// This code is contributed by Prajwal Kandekar


Python3




# Function to find the sum of all
# multiples of 3 and 7 below N
def sumMultiples(n):
    sum = 0
    multiples = set()
    for i in range(1, n):
        if i % 3 == 0 or i % 7 == 0:
            if i not in multiples:
                sum += i
                multiples.add(i)
    return sum
 
n = 24
print(sumMultiples(n))


C#




using System;
using System.Collections.Generic;
 
class MainClass {
    // Function to find the sum of all multiples of 3 and 7
    // below N
    static long SumMultiples(long n)
    {
        long sum = 0;
        HashSet<long> multiples = new HashSet<long>();
 
        for (long i = 1; i < n; i++) {
            if (i % 3 == 0 || i % 7 == 0) {
                if (!multiples.Contains(i)) {
                    sum += i;
                    multiples.Add(i);
                }
            }
        }
 
        return sum;
    }
 
    // Driver code
    static void Main()
    {
        long n = 24;
        Console.WriteLine(SumMultiples(n));
    }
}
// This code is contributed by user_dtewbxkn77n


Javascript




// Function to find the sum of all
// multiples of 3 and 7 below N
function sumMultiples(n) {
    let sum = 0;
    let multiples = new Set();
 
    for (let i = 1; i < n; i++) {
        if (i % 3 === 0 || i % 7 === 0) {
            if (!multiples.has(i)) {
                sum += i;
                multiples.add(i);
            }
        }
    }
 
    return sum;
}
 
// Driver code
let n = 24;
 
console.log(sumMultiples(n));


Output

105

Time Complexity: O(N), since we need to iterate through all the numbers from 1 to N-1.

Auxiliary Space: O(N), since the set can potentially store all the multiples.

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

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

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


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Last Updated : 10 Apr, 2023
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