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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

A naive approach is to iterate through 1 to and find the multiples of A and B and add them to sum. At the end of the loop display the sum.

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 
?> 

Output:

Sum = 950

My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

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