LCM Program in Java and Python

Explanation:

The LCM is the smallest number that is exactly divisible by both given numbers. This program searches for that number by testing possible multiples.

The search starts from one of the given numbers and continues up to the product of the two numbers. The product is always a common multiple, so the answer must be found by that point.

For each value, the program checks whether it is divisible by both numbers using the modulus operator. If both remainders are 0, the value is a common multiple.

Because the search moves upward, the first common multiple found is the least common multiple. The loop can stop immediately after printing it.

The loop tests candidate multiples in increasing order. The first candidate whose remainders with both numbers are zero is stored as the answer and the search stops immediately.

The candidate variable behaves like a counter through possible answers. Each loop iteration compares the candidate with both numbers through remainder checks, and the stored result is returned only when both checks pass.

The LCM is the smallest number that is exactly divisible by both input numbers. A simple method starts from the larger of the two numbers and tests multiples until one is divisible by both. Another method uses the relationship between product, HCF and LCM. In this program, the loop or formula is designed to find the first common multiple, not merely any common multiple. The stopping condition is important because the first valid value encountered in increasing order is the least common multiple.

Java Program:

/**
* The class LcmMethod_1 takes 2 numbers as Input and finds their LCM.
* This is Method 1
* @author : www.guideforschool.com
* @Program Type : BlueJ Program - Java
*/

import java.util.Scanner;
class LcmMethod_1
{
    public static void main(String args[])
    {
        Scanner sc = new Scanner(System.in);
        int a,b,lcm=1;
        System.out.print("Enter the 1st number : ");
        a=sc.nextInt();
        System.out.print("Enter the 2nd number : ");
        b=sc.nextInt();
        for(int i=a;i<=a*b;i++) //Even if you start the for loop by 1, you will get the answer, but starting it from either the first or the second number reduces the number of times the for loop is executed.
        {
            if(i%a==0 && i%b==0) //Checking the first number which is divisible by both the numbers
            {
                lcm=i;
                break; //exiting from the loop, as we don’t need anymore checking after getting the LCM
            }
        }
        System.out.println("L.C.M. = "+lcm);
    }
}

Equivalent Python Program:

# Read the number and keep any required copy for digit or divisor processing.
# Loops and conditions implement the number-property test step by step.
# Display the result according to the flag/counter/calculated value.

a = int(input("Enter the first number: "))
b = int(input("Enter the second number: "))

lcm = 1

for i in range(a, a * b + 1):
    if i % a == 0 and i % b == 0:
        lcm = i
        break

print("L.C.M. =", lcm)

Output: