Sum of multiples of A and B less than N - GeeksforGeeks

07 Apr, 2021
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
 
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++
Java
Python3
C#
PHP
Javascript
// 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 seriesll 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 Nll 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 codeint main(){    ll n = 100, A = 5, B = 10;     cout << "Sum = " << sumMultiples(A, B, n);     return 0;}
// 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 seriesstatic 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 Nstatic 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 codepublic 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
# Python 3 program to find the sum of# all multiples of A and B below Nfrom math import gcd,sqrt # Function to find sum of AP seriesdef 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 Ndef 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 codeif __name__ == '__main__':    n = 100    A = 5    B = 10     print("Sum =", int(sumMultiples(A, B, n))) # This code is contributed by# Surendra_Gangwar
// 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 seriesstatic 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 Nstatic 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 codepublic 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 program to find the sum of all// multiples of A and B below Nfunction __gcd($a,$b){    if ($b == 0)        return $a;    return __gcd($b, $a % $b);} // Function to find sum of AP seriesfunction sumAP($n, $d){    // Number of terms    $n = (int)($n / $d);     return ($n) * (1 + $n) * $d / 2;} // Function to find the sum of all// multiples of A and B below Nfunction sumMultiples($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?>
<script> // JavaScript  program to find the sum of all// multiples of A and B below Nfunction __gcd(a,b){    if (b == 0)        return a;    return __gcd(b, a % b);} // Function to find sum of AP seriesfunction 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 Nfunction 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 codelet n = 100;let A = 5;let B = 10; document.write( "Sum = " + sumMultiples(A, B, n));  // This code is contributed by bobby </script>
Sum = 950
 
SURENDRA_GANGWAR
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