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# Sum of multiples of A and B less than N

• Last Updated : 24 Mar, 2023

Given a number N, the task is to find the sum of all the multiples of A and B below N.
Examples:

```Input:N = 11, A= 8, B= 2
Output: Sum = 30
Multiples of 8 less than 11 is 8 only.
Multiples of 2 less than 11 is 2, 4, 6, 8, 10 and their sum is 30.
As 8 is common in both so it is counted only once.

Input: N = 100, A= 5, B= 10
Output: Sum = 950```

Brute Force Approach:

A brute force approach to solve this problem would be to iterate over all the numbers below N and check if each number is a multiple of A or B. If it is a multiple of either A or B, add it to the sum.

Below is the implementation of the above approach:

## C++

 `#include ` `using` `namespace` `std;` `#define ll long long int`   `// Function to find the sum of all` `// multiples of A and B below N` `ll sumMultiples(ll A, ll B, ll n)` `{` `    ``ll sum = 0;` `    ``for``(ll i=1;i

Output

`Sum = 950`

Time Complexity: O(n)

Auxiliary Space: O(1)

Efficient approach: As the multiples of A will form an AP series a, 2a, 3a….
and B forms an AP series b, 2b, 3b …
On adding the sum of these two series we will get the sum of multiples of both the numbers but there might be some common multiples so remove the duplicates from the sum of these two series by subtracting the multiples of lcm(A, B). So, subtract the series of lcm(A, B) .
So the sum of multiples of A and B less than N is Sum(A)+Sum(B)-Sum(lcm(A, B)).
Below is the implementation of the above approach:

## C++

 `// CPP program to find the sum of all` `// multiples of A and B below N` `#include ` `using` `namespace` `std;` `#define ll long long int `   `// Function to find sum of AP series` `ll sumAP(ll n, ll d)` `{` `    ``// Number of terms` `    ``n /= d;`   `    ``return` `(n) * (1 + n) * d / 2;` `}`   `// Function to find the sum of all` `// multiples of A and B below N` `ll sumMultiples(ll A, ll B, ll n)` `{` `    ``// Since, we need the sum of` `    ``// multiples less than N` `    ``n--;`   `    ``// common factors of A and B` `    ``ll common = (A * B) / __gcd(A, B);`   `    ``return` `sumAP(n, A) + sumAP(n, B) - sumAP(n, common);` `}`   `// Driver code` `int` `main()` `{` `    ``ll n = 100, A = 5, B = 10;`   `    ``cout << ``"Sum = "` `<< sumMultiples(A, B, n);`   `    ``return` `0;` `}`

## Java

 `// Java program to find the sum of all` `// multiples of A and B below N`   `class` `GFG{`   `static` `int` `__gcd(``int` `a, ``int` `b) ` `    ``{ ` `      ``if` `(b == ``0``) ` `        ``return` `a; ` `      ``return` `__gcd(b, a % b);  ` `    ``} ` `    `  `// Function to find sum of AP series` `static` `int` `sumAP(``int` `n, ``int` `d)` `{` `    ``// Number of terms` `    ``n /= d;`   `    ``return` `(n) * (``1` `+ n) * d / ``2``;` `}`   `// Function to find the sum of all` `// multiples of A and B below N` `static` `int` `sumMultiples(``int` `A, ``int` `B, ``int` `n)` `{` `    ``// Since, we need the sum of` `    ``// multiples less than N` `    ``n--;`   `    ``// common factors of A and B` `    ``int` `common = (A * B) / __gcd(A,B);`   `    ``return` `sumAP(n, A) + sumAP(n, B) - sumAP(n, common);` `}`   `// Driver code` `public` `static` `void` `main(String[] args)` `{` `    ``int` `n = ``100``, A = ``5``, B = ``10``;`   `    ``System.out.println(``"Sum = "``+sumMultiples(A, B, n));` `}` `}` `// this code is contributed by mits`

## Python3

 `# Python 3 program to find the sum of ` `# all multiples of A and B below N` `from` `math ``import` `gcd,sqrt`   `# 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 A and B below N` `def` `sumMultiples(A, B, n):` `    `  `    ``# Since, we need the sum of` `    ``# multiples less than N` `    ``n ``-``=` `1`   `    ``# common factors of A and B` `    ``common ``=` `int``((A ``*` `B) ``/` `gcd(A, B))`   `    ``return` `(sumAP(n, A) ``+` `sumAP(n, B) ``-` `            ``sumAP(n, common))`   `# Driver code` `if` `__name__ ``=``=` `'__main__'``:` `    ``n ``=` `100` `    ``A ``=` `5` `    ``B ``=` `10`   `    ``print``(``"Sum ="``, ``int``(sumMultiples(A, B, n)))`   `# This code is contributed by` `# Surendra_Gangwar`

## C#

 `// C# program to find the sum of all` `// multiples of A and B below N`   `class` `GFG{`   `static` `int` `__gcd(``int` `a, ``int` `b) ` `    ``{ ` `    ``if` `(b == 0) ` `        ``return` `a; ` `    ``return` `__gcd(b, a % b); ` `    ``} ` `    `  `// Function to find sum of AP series` `static` `int` `sumAP(``int` `n, ``int` `d)` `{` `    ``// Number of terms` `    ``n /= d;`   `    ``return` `(n) * (1 + n) * d / 2;` `}`   `// Function to find the sum of all` `// multiples of A and B below N` `static` `int` `sumMultiples(``int` `A, ``int` `B, ``int` `n)` `{` `    ``// Since, we need the sum of` `    ``// multiples less than N` `    ``n--;`   `    ``// common factors of A and B` `    ``int` `common = (A * B) / __gcd(A,B);`   `    ``return` `sumAP(n, A) + sumAP(n, B) - sumAP(n, common);` `}`   `// Driver code` `public` `static` `void` `Main()` `{` `    ``int` `n = 100, A = 5, B = 10;`   `    ``System.Console.WriteLine(``"Sum = "``+sumMultiples(A, B, n));` `}` `}` `// this code is contributed by mits`

## PHP

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## Javascript

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Output

`Sum = 950`

Time Complexity: O(log(min(a, b))), where a and b are two parameters of gcd.

Auxiliary Space: O(log(min(a, b)))

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