C# program to find the Least Common Multiple of two numbers

Here, we are going to learn how to find the Least Common Multiple of two numbers in C#?
Submitted by Nidhi, on October 09, 2020

Here we will find the LCM of two numbers. The LCM is the smallest number, which is the multiple of both numbers.

Program:

The source code to find the LCM of two specified numbers is given below. The given program is compiled and executed successfully on Microsoft Visual Studio.

//C# program to find the LCM of two numbers.

using System;

class Demo
{
    static void Main()
    {
        int firstNumber=0;
        int secondNumber=0;

        int temp1=0;
        int temp2=0;

        int lcm=0;

        Console.Write("Enter the value of 1st number: ");
        firstNumber = Convert.ToInt32(Console.ReadLine());

        Console.Write("Enter the value of 2nd number: ");
        secondNumber = Convert.ToInt32(Console.ReadLine());

        temp1 = firstNumber;
        temp2 = secondNumber;

        while (firstNumber != secondNumber)
        {
            if (firstNumber > secondNumber)
            {
                firstNumber = firstNumber - secondNumber;
            }
            else
            {
                secondNumber = secondNumber - firstNumber;
            }
        }
        lcm = (temp1 * temp2) / firstNumber;

        Console.WriteLine("Least Common Multiple is : " + lcm);
    }
}

Output:

Enter the value of 1st number: 9
Enter the value of 2nd number: 15
Least Common Multiple is : 45
Press any key to continue . . .

Explanation:

Here, we created a class Demo that contains a Main() method. Here we found the LCM of two numbers. The LCM is the smallest number, which is the multiple of both numbers.

Console.Write("Enter the value of 1st number: ");
firstNumber = Convert.ToInt32(Console.ReadLine());

Console.Write("Enter the value of 2nd number: ");
secondNumber = Convert.ToInt32(Console.ReadLine());

temp1 = firstNumber;
temp2 = secondNumber;

while (firstNumber != secondNumber)
{
    if (firstNumber > secondNumber)
    {
        firstNumber = firstNumber - secondNumber;
    }
    else
    {
        secondNumber = secondNumber - firstNumber;
    }
}
lcm = (temp1 * temp2) / firstNumber;

In the above code, we read the value of numbers and find the LCM of both numbers and then print the result on the console screen.






Comments and Discussions

Ad: Are you a blogger? Join our Blogging forum.





Languages: » C » C++ » C++ STL » Java » Data Structure » C#.Net » Android » Kotlin » SQL
Web Technologies: » PHP » Python » JavaScript » CSS » Ajax » Node.js » Web programming/HTML
Solved programs: » C » C++ » DS » Java » C#
Aptitude que. & ans.: » C » C++ » Java » DBMS
Interview que. & ans.: » C » Embedded C » Java » SEO » HR
CS Subjects: » CS Basics » O.S. » Networks » DBMS » Embedded Systems » Cloud Computing
» Machine learning » CS Organizations » Linux » DOS
More: » Articles » Puzzles » News/Updates


© https://www.includehelp.com some rights reserved.