Program to find the LCM of 2 numbers in C Sharp

LCM of 2 Numbers- The least common multiple (LCM) of two or more numbers is the smallest number (not counting 0) which is a multiple of all of the numbers.

Let’s take an Example for calculating the LCM of 12 and 8.

break down 12 and 8 into prime factors:
prime factors of 12= 2*2*3
prime factors of 8= 2*2*2

the highest factor for number 12 is 3
the highest factor for number 8 is 2

now take the highest power of each prime factor in the prime factorization and multiply them:
=3 *power(2,3)
=3*8
=24

C# program to find the LCM of 2 numbers.

Method-1

Code

class Program
    {
        static void Main(string[] args)
        {
            Console.WriteLine("Please Enter First Number");

            bool flag=false;

            int n1,n2;
            flag = int.TryParse(Console.ReadLine(), out n1);

            if(n1<0)
            {
                Console.WriteLine("Number should not be negative");
                return;
            }
            flag = false;

            Console.WriteLine("Please Enter Second Number");
            flag = int.TryParse(Console.ReadLine(), out n2);

            if (n2 < 0)
            {
                Console.WriteLine("Number should not be negative");
                return;
            }
          // // maximum number between n1 and n2 is stored in minMultiple
            int minvalue = Math.Min(n1, n2);

            while(true)
            {
                if(minvalue%n1==0 && minvalue%n2==0)
                {
                    break;
                }
                minvalue++;
            }

            Console.WriteLine("LCM of {0} and {1} is {2}" ,n1,n2,minvalue);
        }
    }

Output

Please Enter First Number
11
Please Enter Second Number
22
LCM of 11 and 22 is 22

Method-2

Code

class Program
    {
        static void Main(string[] args)
        {
            Console.WriteLine("Please Enter First Number");

            bool flag = false;

            int n1, n2;
            flag = int.TryParse(Console.ReadLine(), out n1);

            if (n1 < 0)
            {
                Console.WriteLine("Number should not be negative");
                return;
            }
            flag = false;

            Console.WriteLine("Please Enter Second Number");
            flag = int.TryParse(Console.ReadLine(), out n2);

            if (n2 < 0)
            {
                Console.WriteLine("Number should not be negative");
                return;
            }

            int minvalue = Math.Min(n1, n2);

            int gcd=1;
            for (int i = 1; i <= n1 && i<=n2; i++)
            {
                // // Checks if i is factor of both integers
                if(n1%i==0 && n2%i==0)
                {
                    gcd = i;

                }
            }
            int lcm = (n1 * n2) / gcd;

            Console.WriteLine("LCM of {0} and {1} is {2}", n1, n2, lcm);
        }
    }

Output

Please Enter First Number
11
Please Enter Second Number
22
LCM of 11 and 22 is 22

Method-3

Code:

class Program
    {
        static void Main(string[] args)
        {
            Console.WriteLine("Please Enter First Number");

            bool flag = false;

            int n1, n2;
            flag = int.TryParse(Console.ReadLine(), out n1);

            if (n1 < 0)
            {
                Console.WriteLine("Number should not be negative");
                return;
            }
            flag = false;

            Console.WriteLine("Please Enter Second Number");
            flag = int.TryParse(Console.ReadLine(), out n2);

            if (n2 < 0)
            {
                Console.WriteLine("Number should not be negative");
                return;
            }

            int numerator = Math.Max(n1, n2);
            int denominator = Math.Min(n1, n2);

            int remainder = numerator % denominator;

             //check the remainder value is zero  after divsion operation
            while(remainder!=0)
            {
                numerator = denominator;
                denominator = remainder;
                remainder = numerator % denominator;

            }

            int gcd = denominator;
            int lcm = (n1 * n2) / gcd;
            

            Console.WriteLine("LCM of {0} and {1} is {2}", n1, n2, lcm);
        }
    }