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# 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

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 ` `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 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 ` `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

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## Javascript

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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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