Sum of multiples of a number up to N

09 Jun, 2022
Given a number a and limit N. Find the sum of multiple of a upto N.
Examples : 
Input : a = 4, N = 23
Output : sum = 60
[Multiples : 4, 8, 12, 16, 20]

Input :a = 7, N = 49
Output :sum = 196
[Multiples: 7, 14, 21, 28, 35, 42, 49]
The basic idea is to iterate from i = a to i = n, i++ and check whether i % a == 0 or not.If zero then add i to sum(initially sum = 0).Thus we will get the sum.It will take O(n) time.We can modify the loop as i = a, i <= n, i = i + a to reduce the number of iterations.But it will also take O(m) time if there is m multiples of a.To get the result in O(1) time we can use the formula of summation of n natural numbers.For the above example, a = 4 and N = 23, number of multiples of a, m = N/a(integer division). The multiples are 4, 8, 12, 16, 20. We can write it as 4 X [1, 2, 3, 4, 5]. So we can get the sum of multiples as:
 sum = a * (Summation of 1 to m [natural numbers from 1 to m]) 
 sum = 4 * (m*(m+1) / 2)
 sum = 4 * (5*6 / 2) = 4 * 15 = 60 
 
C++
Java
Python3
C#
PHP
Javascript
// C++ program to find sum of multiples of a number// up to N efficiently#include <iostream>using namespace std; // Function for calculating sum of multiples of// a upto Nint calculate_sum(int a, int N){    // Number of multiples    int m = N / a;     // sum of first m natural numbers    int sum = m * (m + 1) / 2;     // sum of multiples    int ans = a * sum;     return ans;} // Driver codeint main(){    int a = 7, N = 49;    cout << "Sum of multiples of "         << a << " up to " << N << " = "         << calculate_sum(a, N) << endl;    return 0;}
// Java program to find sum of multiples// of a number up to N efficiently class GFG {     // Function for calculating sum// of multiples of a upto Nstatic int calculate_sum(int a, int N) {         // Number of multiples    int m = N / a;     // sum of first m natural numbers    int sum = m * (m + 1) / 2;     // sum of multiples    int ans = a * sum;     return ans;} // Driver codepublic static void main(String[] args) {         int a = 7, N = 49;    System.out.println("Sum of multiples of " + a +                       " up to " + N + " = " +                               calculate_sum(a, N));}} // This code is contributed by Anant Agarwal.
"""Python program to find sum ofmultiples of a number up to N""" # Calculates sum of multiples of# a number upto Ndef calculate_sum(a, N):     # Number of multiples    m = N / a     # sum of first m natural numbers    sum = m * (m + 1) / 2     # sum of multiples    ans = a * sum     print("Sum of multiples of ", a,          " up to ", N, " = ", ans) # Driver Codecalculate_sum(7, 49)     # This code is contributed by Abhishek Agrawal.
// C# program to find sum of multiples// of a number up to N efficientlyusing System; class GFG {     // Function for calculating sum    // of multiples of a upto N    static int calculate_sum(int a, int N)    {         // Number of multiples        int m = N / a;         // sum of first m natural numbers        int sum = m * (m + 1) / 2;         // sum of multiples        int ans = a * sum;         return ans;    }     // Driver code    public static void Main()    {         int a = 7, N = 49;        Console.WriteLine("Sum of multiples of " + a +         " up to " + N + " = " + calculate_sum(a, N));    }} // This code is contributed by vt_m.
<?php// PHP program to find sum// of multiples of a number// up to N efficiently // Function for calculating sum// of multiples of a upto Nfunction calculate_sum($a, $N){    // Number of multiples    $m = $N / $a;     // sum of first m    // natural numbers    $sum = $m * ($m + 1) / 2;     // sum of multiples    $ans = $a * $sum;     return $ans;} // Driver code$a = 7;$N = 49;echo "Sum of multiples of ". $a ,         " up to " . $N . " = " .          calculate_sum($a, $N) ; // This code is contributed by Sam007?>
<script> // Javascript program to find sum// of multiples of a number// up to N efficiently // Function for calculating sum// of multiples of a upto Nfunction calculate_sum(a, N){         // Number of multiples    m = N / a;     // Sum of first m    // natural numbers    sum = m * (m + 1) / 2;     // Sum of multiples    ans = a * sum;     return ans;} // Driver codelet a = 7;let N = 49; document.write("Sum of multiples of "+ a +               " up to " + N + " = " +               calculate_sum(a, N)); // This code is contributed by mohan1240760 </script>
Output : 
Sum of multiples of 7 upto 49 = 196
Time complexity: O(1)Auxiliary space: O(1) 
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