Sum of multiples of A and B less than N

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 <bits/stdc++.h>
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<n;i++){
        if(i%A==0 || i%B==0)
            sum+=i;
    }
    return sum;
}

// Driver code
int main()
{
    ll n = 100, A = 5, B = 10;
    cout << "Sum = " << sumMultiples(A, B, n);
    return 0;
}

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 <bits/stdc++.h>
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

<?php
// PHP program to find the sum of all
// multiples of A and B below N
function __gcd($a,$b)
{
    if ($b == 0)
        return $a;
    return __gcd($b, $a % $b);
}

// 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 A and B below N
function sumMultiples($A, $B, $n)
{
    // Since, we need the sum of
    // multiples less than N
    $n--;

    // 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
$n = 100;
$A = 5;
$B = 10;

echo "Sum = " . sumMultiples($A, $B, $n);

// This code is contributed by mits
?>

Javascript

<script>

// JavaScript  program to find the sum of all
// multiples of A and B below N
function __gcd(a,b)
{
    if (b == 0)
        return a;
    return __gcd(b, a % b);
}

// 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 A and B below N
function sumMultiples(A, B, n)
{
    // Since, we need the sum of
    // multiples less than N
    n--;

    // common factors of A and B
    common = parseInt((A * B) /
               __gcd(A, B));

    return sumAP(n, A) +
           sumAP(n, B) -
           sumAP(n, common);
}

// Driver code
let n = 100;
let A = 5;
let B = 10;

document.write( "Sum = " + sumMultiples(A, B, n));


// This code is contributed by bobby

</script>

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)))