- Related Questions & Answers
- Count of common multiples of two numbers in a range in C++
- Find the sum of all Truncatable primes below N in Python
- Sum of sum of first n natural numbers in C++
- Print first m multiples of n in C#
- Minimize the sum of squares of sum of N/2 paired formed by N numbers in C++
- Sum of two large numbers in C++
- Sum of all multiples in JavaScript
- Finding sum of multiples in JavaScript
- Sum of the first N Prime numbers
- Minimum numbers needed to express every integer below N as a sum in C++
- Sum of squares of the first n even numbers in C Program
- Sum of two numbers modulo M in C++
- Sum of first n natural numbers in C Program
- Sum of squares of first n natural numbers in C Program?
- Sum of square of first n odd numbers

- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who

In this problem, we have given three integers M1, M2, and N. Our task is to create a program to find the sum of multiples of two numbers below N.

Here, we will add all the elements below N which are multiples of either M1 or M2

**Let’s take an example to understand the problem,**

**Input **

N = 13, M1 = 4, M2 = 6

**Output **

20

**Explanation** − Number that are multiples of 4 and 6 that are less than 13 are 4, 6, 8, 12.

A simple solution to the problem is to be looping from 1 to N and adding all values that can be divided by M1 or M2.

**Step 1** − sum = 0 , i = 0. Loop from i = 1 to N.

**Step 1.1** − if (i%M1 == 0) or (i%M2 == 0), sum + = i

**Step 2** − Return sum.

Program to illustrate the working of our solution,

#include <iostream< using namespace std; int calcMulSum(int N, int M1, int M2){ int sum = 0; for (int i = 0; i < N; i++) if (i%M1 == 0 || i%M2 == 0) sum += i; return sum; } int main(){ int N = 24, M1 = 4, M2 = 7; cout<<"The sum of multiples of "<<M1<<" and "<<M2<<" below "<<N<<" is "<<calcMulSum(N, M1, M2); return 0; }

The sum of multiples of 4 and 7 below 24 is 102

This is not the best solution to our problem as it takes O(n) time complexity.

A better solution will be using mathematical formulas for the sum of series.

Here, we will use the formula for the sum of the series. The final sum will be the **(multiples of M1 + multiples of M2 - multiples of M1*M2)**

Sum of multiple of x upto n terms is given by,

Sum(X) = (n * (1+n) * X)/2

Let’s formulate the sum,

sum = ( ( ((n/M1)*(1+(n/M1))*M1)/2) + ((n/M2)*(1+(n/M2))*M2)/2 ) - ((n/M1*M2)*(1+(n/M1*M2))*M1*M2)/2 ) )

Program to illustrate the solution,

#include <iostream> using namespace std; int calcMulSum(int N, int M1, int M2){ N--; return (((N/M1) * (1 + (N/M1)) * M1 / 2) + ((N/M2) * (1 + (N/M2)) * M2 / 2) - ((N/(M1*M2)) * (1 + (N/(M1*M2))) * (M1*M2) / 2)); } int main(){ int N = 24, M1 = 4, M2 = 7; cout<<"The sum of multiples of "<<M1<<" and "<<M2<<" below "<<N<<" is "<<calcMulSum(N, M1, M2); return 0; }

The sum of multiples of 4 and 7 below 24 is 102

Advertisements