The LCM program in Java outputs the LCM of the given numbers. In Arithmetic, the lowest common multiple, least common multiple or smallest common multiple of the two numbers p and q are represented by LCM(p, q). It is a lowest positive number that is completely divisible by the numbers p and q. Since, the division of any number by zero is undefined, therefore, the definition of LCM only holds true when both the numbers p and q are not equal to zero. Let’s discuss different approaches to find LCM.

Using the Java for Loop

Filename: LCMExample.java

public class LCMExample
 {             
 public static void main(String argvs[])
 {
                int a = 78, b = 117; // Given numbers
                int ans; // contains LCM of the given numbers
                // finding maximum between the given numbers.
                ans = ( a > b) ? a : b;
                // finding LCM
                for( ; ; )
                {
                    if(ans % a == 0 && ans % b == 0)
                    {
                        // We found our LCM.
                        // No need to check further.
                        break;
                    }
                    // Increamenting by 1.
                    ans = ans + 1;
                }
                // Displaying result
                System.out.println("The LCM of the numbers " + a + " and " + b + " is " + ans);
 }
 } 

Output:

The LCM of the numbers 78 and 117 is 234

Explanation: The LCM of any two numbers is always greater than or equal largest of the given numbers. Therefore, we have assigned the largest of the given two numbers in ans variable. Since, there is no condition in the for loop, it is evaluated true. In each iteration, we are incrementing the ans variable by 1 and checking whether the updated ans value is completely divisible by the given numbers or not. If it is divisible, we forcefully terminate the loop using the break statement; otherwise, the iteration continues.

Using Multiples of the Given Numbers

In this approach, we find the multiples of the given numbers and then find the smallest multiple, which is common in both the given numbers. The smallest common multiple is answer. Let’s understand it with the help of an example.

The LCM of 9 and 12 will be:

Multiples of 9: 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99, 108, 117, 126, 135, 144, 153, …

Multiples of 12: 12, 24, 36, 48, 60, 72, 84, 96, 108, 120, 132, 144, 156, 168, 180, 192, …

Common multiples are: 36, 72, 108, 144

Lowest Common Multiple is 36. Hence, LCM(9, 12) = 36.

Filename: LCMExample1.java

import java.util.List;
 import java.util.ArrayList;
 public class LCMExample1
 {
 public static void main(String argvs[])
 {
                int a = 78, b = 117; // Given numbers
                int ans = 0; // contains the LCM of the given numbers
                // List for storing multiples
                List<Integer> al1 =  new ArrayList<Integer>();
                List<Integer> al2 = new ArrayList<Integer>();
                // Finding and storing multiples of the given numbers
                for(int i = 1; i <= b; i++)
                {
                                // generating multiples in increasing order
                                int multiples = i * a;  
                                al1.add(multiples); // storing in the list
                }
                for(int i = 1; i <= a; i++)
                {
                                // generating multiples in increasing order
                                int multiples = i * b;
                                al2.add(multiples); // storing in the list
                }
                // Converting the lists into Integer arrays.
                Integer[] f1 = new Integer[al1.size()];
                f1 = al1.toArray(f1);
                Integer[] f2 = new Integer[al2.size()];
                f2 = al2.toArray(f2);
                // Two pointer approach to find the lowest multiple
                for(int i = f1.length - 1, j = f2.length - 1; i >= 0 && j >= 0;)
                {
                                if((int)f1[i] == (int)f2[j])
                                {
                                               ans = f1[i]; // updating our result
                                               i--;
                                               j--;
                                }
                                else if(f1[i] > f2[j])
                                {
                                               i--;
                                }
                                else
                                {
                                               j--;
                                }
                }
                System.out.println("The LCM of the numbers " + a + " and " + b + " is " + ans);
 }             
 } 

Output:

The LCM of the numbers 78 and 117 is 234

Explanation: In the above approach, we are finding and storing all the multiples of the first number till the second number and vice-versa. This is because LCM(a, b) <= a*b. Then we are converting the lists into integer arrays. The first integer array contains all the multiples of the first input number, and the second integer array contains the multiples of the second input number in ascending order. Finally, using the two-pointers approach, we are finding the LCM of the given numbers. This approach is not widely used because the numbers become large and finding the multiples of a large number takes a lot of time. Also, the two-pointers approach becomes time consuming.

Using GCD of the Given Numbers

In Mathematics, the multiplication of two numbers is always equal to the product of LCM and GCD of the numbers. That is,

a * b = LCM(a, b) * GCD(a, b)

Therefore,

LCM(a, b) =

The following program implements the above-written formula to get the LCM of the given numbers.

Filename: LCMExample2.java

public class LCMExample2
 {
 // Method for finding GCD of two numbers
 public static int findGCD(int a, int b)
 {
                // Base cases
                if(a == 0)
                {
                                return b;
                }
                if(b == 0)
                {
                                return a;
                }
                // recursively finding GCD
                return findGCD(b, a % b);
 }
 public static void main(String argvs[])
 {
                int a = 78, b = 117; // Given numbers
                int ans; // contains LCM of the given numbers
                int product = a * b; // calculating product
                int gcd = findGCD(a, b); // invoking the method findGCD() and storing the result                    
                ans = product / gcd; // calculating LCM
                // Displaying result
                System.out.println("The LCM of the numbers " + a + " and " + b + " is " + ans);
 }
 } 

Output:

The LCM of the numbers 78 and 117 is 234

Explanation: In the above approach, first we calculate the product of the given numbers. Then using Euler’s algorithm, we are finding GCD. Finally, we are doing the division of the product of numbers and GCD of the given numbers to get the required LCM.

Finding LCM of more than two numbers

So far, we have only discussed the LCM of the two given numbers. However, it is also possible to find the LCM of more than two numbers. The following program illustrates the same.

Filename: LCMExample3.java

public class LCMExample3
 {             
 //Method implementing Euclid's algorithm
 public static int findGCD(int n1, int n2)
 {
                // base cases
                if(n1 == 0) return n2;
                if(n2 == 0) return n1;
                // recursively finding the GCD
                return findGCD(n2, n1 % n2);
 }
 public static void main(String argvs[])
 {
                // input array
                int arr[] = {6, 10, 16, 28, 78, 90, 112, 188, 200, 290, 310, 444};
                int size = arr.length; // Calculating size of the input array
                // ans contains the LCM of the entire array
                // Initializing ans with the first element of the input array
                int ans = arr[0];
                for(int i = 1; i < size; i++)
                {
                               int product = ans * arr[i]; // calculating product
                               // invoking method fingGCD() and storing the result
                               int gcd = findGCD(ans, arr[i]);
                               ans = product / gcd; // updating our ans
                }
                // Displaying result
                System.out.println("The LCM of the given numbers is " + ans);
 }
 } 

Output:

The LCM of the given numbers is 161387600

Explanation: The above approach is similar to the previous one. The only difference is instead of two numbers, we have taken multiple numbers in an array. Therefore, we have used the Java for-loop to iterate over each element of the array to calculate the LCM of the whole array.

Finding LCM of Fractions

Not only numbers, but LCM of fractions can also be found. In Mathematics, the LCM of fractions is equal to the LCM of all the numerator numbers divided by the GCD of all the numbers of denominators. That is,

LCM( ,  =

The following Java program demonstrates the same.

Filename: LCMExample3.java

public class LCMExample3
 {             
 //Method implementing Euclid's algorithm
 public static int findGCD(int n1, int n2)
 {
                // base cases
                if(n1 == 0) return n2;
                if(n2 == 0) return n1;
                // recursively finding the GCD
                return findGCD(n2, n1 % n2);
 }
 public static void main(String argvs[])
 {
                // input arrays
                // Array for numerators
                int numerators[] = {6, 10, 16, 28, 78, 90, 112, 188, 200, 290, 310, 444};
                // Array for denominators
                int denominators[] = {8, 4, 10, 36, 98, 112, 180, 198, 222, 788, 888, 990};
                int size = numerators.length; // Calculating size of the input arrays
                // lcmOfNum contains the LCM of all the numerators
                // Initializing lcmOfNum with the first element of the <em>numerators</em> array
                int lcmOfNum = numerators[0];
                // gcdOfDnm contains the GCD of all the denominators
                // Initializing gcdOfDnm with the first element of the <em>denominators</em> array
                int gcdOfDnm = denominators[0];
                // Loop for calculating LCM of numerators and GCD of denominators
                for(int i = 1; i < size; i++)
                {
                               int product = lcmOfNum * numerators[i]; // calculating product
                               // invoking method fingGCD() and
                                // storing the result
                               int gcdOfNum = findGCD(lcmOfNum, numerators[i]);
                               lcmOfNum = product / gcdOfNum; // updating our LCM of numerators
                                // calculating GCD of all the denominators
                               gcdOfDnm = findGCD(gcdOfDnm, denominators[i]);
                }
                int ans = lcmOfNum / gcdOfDnm; // calculating the final result
                // Displaying result
                System.out.println("The LCM of the given numbers is " + ans);
 }
 } 

Output:

The LCM of the given fractions is 80693800

Explanation: The two input arrays contain the numerators and denominators of the fractions. numerators[0] and denominators[0] constitute the first fraction ( . numerators[1] and denominators[1] constitute the second fraction ( , and so on. In a single Java for-loop, we are calculating the LCM for the numerators and GCD for the denominators. Finally, we are dividing the calculated LCM and GCD to get our result.

Comparison of The Above Approaches

Those approaches that do not include Euclid’s algorithm to find the LCM should be avoided if given liberty. We have already discussed that Euclid’s algorithm is the best algorithm to find the GCD. Also, we can easily find LCM if we know the GCD of the given numbers. Therefore, the use of Euler’s algorithm should be encouraged to find LCM. However, it is important to know all the mentioned approaches present in the section from the interview point of view.

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