Category Archives: Java Number Programs

Java Number Programs

ISBN Number Program In Java

In this tutorial, we will learn to create a Java Program to find ISBN Number Program In Java

Prerequisites

Before starting with this tutorial we assume that you are best aware of the following Java programming topics:

Java Operators.
Basic Input and Output function in Java.
Class and Object in Java.
Basic Java programming.
If-else statements in Java.
Function in java.

International Standard Book Number (ISBN) IN JAVA Program

In world every book has a number called ‘International Standard Book Number (ISBN).It is unique number. With the help of this number, we can easily find any book in the world. ISBN is a 10 digit number. The number is legal if sum of (1*digit1 + 2*digit2 + 3*digit3 + 4*digit4 + 5*digit5 + 6*digit6 + 7*digit7 + 8*digit8 + 9*digit9 + 10*digit10) is divisible by sum%11==0 gives remainder Zero.

Let’s take a Number = 0306406152 the find sum
Sum = (0*1) + (3*2) + (0*3) + (6*4) + (4*5) + (0*6) + (6*7) + (1*8) + (5*9) + (2*10)
–“– = 0 + 6 + 0 + 24 + 20 + 0 + 42 + 8 + 45 + 20
Sum = 165, and 165 / 11 = 15. So 165 is divisible by 11 then the number is International Standard Book Number (ISBN).

Java Program to check the number is ISBN Number Program or not.

In this program we would find the given number is tech number first will take the input from the user and check whether it is Tech number or not. If it’s not then the program will show the output as it’s not a tech number.

// java program to find ISBN number
import java.io.*;
class ISBNumber {

    public static void main(String[] args) throws IOException {
        // Delacig variables..
        long isbn_no;
        int c = 0, i, t, d, dNo;
        String str;
        // Input a 10-digit ISBN number from user
        InputStreamReader in = new InputStreamReader(System.in);
        BufferedReader br = new BufferedReader(in);
        System.out.print("enter a 10-digit ISBN number: ");
        isbn_no = Long.parseLong(br.readLine());

        // checking the length is 10 or not of given number else , exit from the program
        str = "" + isbn_no;
        if (str.length() != 10) {
            System.out.println("Illegal ISBN");
            return;
        }
        // : Logic for an ISBN 0201530821
        //S = 0*10 + 2*9 + 0*8 + 1*7 + 5*6 + 3*5 + 0*4 + 8*3 + 2*2 + 1*1 = 99
        // if divisible by 11.
        // compute the  digits
        c = 0;
        for (i = 0; i < str.length(); i++) {
            d = Integer.parseInt(str.substring(i, i + 1));
            dNo = i + 1;
            t = dNo * d;
            c += t;
        }

        // checking  condtion  is divisible by 11 or not
      
        if ((c % 11) != 0) {
            System.out.println("Illegal ISBN");
        } else {
            System.out.println("Legal ISBN");
        }
    }
}

// java program to find ISBN number

import java.io.*;

class ISBNumber {

public static void main(String[] args) throws IOException {

// Delacig variables..

long isbn_no;

int c = 0, i, t, d, dNo;

String str;

// Input a 10-digit ISBN number from user

InputStreamReader in = new InputStreamReader(System.in);

BufferedReader br = new BufferedReader(in);

System.out.print("enter a 10-digit ISBN number: ");

isbn_no = Long.parseLong(br.readLine());

// checking the length is 10 or not of given number else , exit from the program

str = "" + isbn_no;

if (str.length() != 10) {

System.out.println("Illegal ISBN");

return;

}

// : Logic for an ISBN 0201530821

//S = 0*10 + 2*9 + 0*8 + 1*7 + 5*6 + 3*5 + 0*4 + 8*3 + 2*2 + 1*1 = 99

// if divisible by 11.

// compute the digits

c = 0;

for (i = 0; i < str.length(); i++) {

d = Integer.parseInt(str.substring(i, i + 1));

dNo = i + 1;

t = dNo * d;

c += t;

}

// checking condtion is divisible by 11 or not

if ((c % 11) != 0) {

System.out.println("Illegal ISBN");

} else {

System.out.println("Legal ISBN");

}

Output

In the above program, we have first declared and initialized a set variables required in the program.

isbn_no = it will hold entered number.
c = 0 itr will count length.
i= for iteration .
c= calculate the sum of number.

After that we take 10 digit number form user.

Then we check that given number is a 10 digit number or not.

if not we we exit from program

else we proceed the logic for ISbn number as shown in image below.

Find isbn by multiplying each digit from left to right by 1, 2, 3,…….,10.

and at last we will check if number is completely divisible by 11 then it is a leagal isbn number if not it is not a isbn number.

Program to check Twisted Prime Program in Java

In this tutorial, we will learn to create a Java Program to check Twisted Prime Program in Java.

Prerequisites

Before starting with this tutorial we assume that you are best aware of the following Java programming topics:

Java Operators.
Basic Input and Output function in Java.
Class and Object in Java.
Basic Java programming.
If-else statements in Java.
Function in java.

Twisted Prime Number

A given number called twisted prime number if it is a prime number and reverse of it also a prime number.

Some Examples of Twisted Prime number 2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79.

Take a number : “97” is a prime number then
Twisted Prime Number “97” is a prime number
and its reverse “79” is also a prime number.

Program to check Twisted Prime Program in Java

In this program we would find the given number is twisted prime number or not , first will take the input from the user and check whether it is prime number or not. If it’s prime then we will twist the number and also check it is prime or not.

// java program to find twisyed prime number
import java.util.*;  
class T_Prime   
{  
    public static void main(String[] args)  
    {  
    //  decalering variables..
    int num, rev,sum=0 , count;  
    Scanner sc = new Scanner(System.in);      
    System.out.println("Enter a prime number?");  
    num = sc.nextInt();  
    
  // reversing a number
        while(num!=0)  
        {  
        rev = num%10;  
        sum = sum*10 + rev;  
        num= num/10;  
        }  
        
    // condition for twisted prime number
    count=0;
    for (int j = 2; j <= sum / 2; j++)  
    {  
        if ((sum % j) == 0)  
        {  
            count = 1;  
            break;  
        }  
    }  
    // condition for twisted prime number
    if (count == 0)  
            System.out.println("Twisted Prime");  
     else  
            System.out.println("Not Twisted Prime");  
}  
}

// java program to find twisyed prime number

import java.util.*;

class T_Prime

{

public static void main(String[] args)

{

// decalering variables..

int num, rev,sum=0 , count;

Scanner sc = new Scanner(System.in);

System.out.println("Enter a prime number?");

num = sc.nextInt();

// reversing a number

while(num!=0)

{

rev = num%10;

sum = sum*10 + rev;

num= num/10;

}

// condition for twisted prime number

count=0;

for (int j = 2; j <= sum / 2; j++)

{

if ((sum % j) == 0)

{

count = 1;

break;

}

// condition for twisted prime number

if (count == 0)

System.out.println("Twisted Prime");

else

System.out.println("Not Twisted Prime");

}

Output

Twisted Prime number

Not a prime number.

In the above program, we have first declared and initialized a set variables required in the program.

num = it will hold entered number.
rev = it will hold reverse or a number
sum = for adding values
count= counting the divisibility
i= For iteration.

After Declaring variables we take a prime number from user.

First of all we reveres the given number and then check the reverse number is also a prime number or not.

In above image we reversing a number then at last we will check the reverse number is also prime or not .

if the reverse number is also a prime then we can say the entered prime number is twisted prime number.

else it is not a twisted prime number.

Program to check Special Number Program in Java

In this tutorial, we will learn to create a Java Program to check Special Number Program in Java

Prerequisites

Before starting with this tutorial we assume that you are best aware of the following Java programming topics:

Java Operators.
Basic Input and Output function in Java.
Class and Object in Java.
Basic Java programming.
If-else statements in Java.
loops in java

Special Number in Java

Number said to special if the sum of factorial of its digits is equal to the Given number.

Example: 145.
-> factor of 1! + factor of 4! + factor of 5!
-> 1 + 24 + 120.
-> 145.

So the number is a special number.

Program to check Special Number Program in Java

In this program we would find the given number is special number or not , first will take the input from the user and then we calculate sum of factor of number gives number it self it is Special number. Let have a look at the code.

// java program to fing given number is a special or not ..
import java.util.Scanner;
 
class Main {
  public static void main(String[] args) {
    Scanner in=new Scanner(System.in);
    // decalring variables...
    int count, number, sum=0, temp;
    
    // taking value from user
    System.out.println("Enter a number");
    number = in.nextInt();
    temp = number;
    
    // Find the factorial of its digits.
    while(temp != 0){
      count = temp%10;
      sum += factorial(count);
      temp = temp/10;
    }
    
    // Checking the sum is equal to the original number or not
    if(sum == number)
      System.out.println("Special number");
    else
      System.out.println("Not a Special Number");
 
  }
    
  //function to calculate the factorial of the digits
  private static int factorial(int no){
    int fact=1;
    for(int i=2; i<=no; i++){
      fact = fact*i;
    }
    return fact;
  }
}

// java program to fing given number is a special or not ..

import java.util.Scanner;

class Main {

public static void main(String[] args) {

Scanner in=new Scanner(System.in);

// decalring variables...

int count, number, sum=0, temp;

// taking value from user

System.out.println("Enter a number");

number = in.nextInt();

temp = number;

// Find the factorial of its digits.

while(temp != 0){

count = temp%10;

sum += factorial(count);

temp = temp/10;

}

// Checking the sum is equal to the original number or not

if(sum == number)

System.out.println("Special number");

else

System.out.println("Not a Special Number");

}

//function to calculate the factorial of the digits

private static int factorial(int no){

int fact=1;

for(int i=2; i<=no; i++){

fact = fact*i;

}

return fact;

}

Output

Special Number.

Not a Special Number.

In the above program, we have first declared and initialized a set variables required in the program.

number = it will hold entered number.
temp = it will hold temp the value of number.
sum = for adding values
count= count will count size of a number
i= For iteration.

After Declaring values first we will take a number form user

and pass number to temp variable where we break the number

and find factor of its each digit and calculate sum of all digit.

if sum of all digit is equal to the original number the given number is a Special Number

As shown in image if number equal to the sum of digit factor then it is Special if not not a Special number.

Program to check Niven Number Program in Java

In this tutorial, we will learn to create a Java Program to check Niven Number Program in Java

Prerequisites

Before starting with this tutorial we assume that you are best aware of the following Java programming topics:

Java Operators.
Basic Input and Output function in Java.
Class and Object in Java.
Basic Java programming.
If-else statements in Java.
Function in java.

Harshad (Or Niven) Number

A Number is called a Harshad number if sum of the digits is divisible by the sum of its digits

For example:-> 18 is a Harshad number base of 10, because sum of the its digit which is “(1 and 8) = 9″ (1 + 8 = 9), and 18 is divisible by 9 (18 % 9 == 0).

Program to check Niven Number Program in Java

In this program we would find the given number is Niven number first of all we will take the input from the user and check whether it is Niven number or not. If it’s not then the program will show the output as it’s not a Niven number.

// java program to find niven nmber
import java.util.*;
class HarshadNo
{
    public static void main(String args[])
    {
        // decalring variable and taking value from user
        Scanner sc = new Scanner(System.in);
         
        System.out.print("Enter a number : ");
        int num = sc.nextInt();
        int temp = num, d, sum = 0;
         
        //caluclating sum of digits
        while(temp>0)
        {
            d = temp%10;
            sum = sum + d;
            temp = temp/10;
        }
         // printing the result 
        if(num%sum == 0)
            System.out.println(num+"  is  Harshad Number.");
        else
            System.out.println(naum+"  is not a Harshad Number.");      
    }
}

// java program to find niven nmber

import java.util.*;

class HarshadNo

{

public static void main(String args[])

{

// decalring variable and taking value from user

Scanner sc = new Scanner(System.in);

System.out.print("Enter a number : ");

int num = sc.nextInt();

int temp = num, d, sum = 0;

//caluclating sum of digits

while(temp>0)

{

d = temp%10;

sum = sum + d;

temp = temp/10;

}

// printing the result

if(num%sum == 0)

System.out.println(num+" is Harshad Number.");

else

System.out.println(naum+" is not a Harshad Number.");

}

Output

Niven number

Not a Niven number

In the above program, we have first declared and initialized a set variables required in the program.

num= it will hold entered number.
temp = it will hold temp value
sum= for adding digit sum.
d= it will hold digits of a number

After declaring variables we take a number form user.

After that we add digit sum of a number.

and then we divide the sum with original number to find it is a Niven number or not.

Niven number (or harshad number)if it is divisible by the sum of its digits .

Program to check Happy Number Program in Java

In this tutorial, we will learn to create a Java Program to check Happy Number Program in Java

Prerequisites

Before starting with this tutorial we assume that you are best aware of the following Java programming topics:

Java Operators.
Basic Input and Output function in Java.
Class and Object in Java.
Basic Java programming.
If-else statements in Java.
Function in java.

What is a happy number ?

A number is called a Happy Number when we replace the number by the sum of the squares of its digits, and repeat the process until the number reach to =”1″.

3<sup>2</sup> + 2<sup>2</sup> = 13
1<sup>2</sup> + 3<sup>2</sup> = 10
1<sup>2</sup> + 0<sup>2</sup> = 1

32 + 22 = 13

12 + 32 = 10

12 + 02 = 1

Program to check Happy Number Program in Java

In this program we would find the given number is a Happy number or not , first of all we will take the input from the user and check whether it is Happy number or not. If it’s not then the program will show the output as it’s not a happy number.

// java program t find given number is a happy or not
import java.util.Scanner;
public class HappyNum  
{     
    //isHappyNumber() will determine whether a number is happy or not  
    public static int Happy(int number){  
        int rem = 0, sum = 0;  
          
        //Calculates the sum of squares of digits  
        while(number > 0){  
            rem = number%10;  
            sum = sum + (rem*rem);  
            number = number/10;  
        }  
        return sum;  
    }  
      
    public static void main(String[] args) {  
        // declaring variables
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter any number= ");
        int num = sc.nextInt();
        int temp = num;  
          
        while(temp != 1 && temp != 4){  
            temp = Happy(temp);  
        }  
          
        //conditon for happy number  
        if(temp == 1)  
            System.out.println(num + " is a happy number");  
        else if(temp == 4)  
            System.out.println(num + " is not a happy number");     
    }  
}

// java program t find given number is a happy or not

import java.util.Scanner;

public class HappyNum

{

//isHappyNumber() will determine whether a number is happy or not

public static int Happy(int number){

int rem = 0, sum = 0;

//Calculates the sum of squares of digits

while(number > 0){

rem = number%10;

sum = sum + (rem*rem);

number = number/10;

}

return sum;

}

public static void main(String[] args) {

// declaring variables

Scanner sc = new Scanner(System.in);

System.out.print("Enter any number= ");

int num = sc.nextInt();

int temp = num;

while(temp != 1 && temp != 4){

temp = Happy(temp);

}

//conditon for happy number

if(temp == 1)

System.out.println(num + " is a happy number");

else if(temp == 4)

System.out.println(num + " is not a happy number");

}

Output

Happy Number

Not a Happy Number

In the above program, we have first declared and initialized a set variables required in the program.

num,number= it will hold entered number.
temp = it will hold temp value
sum= for adding digit sum.
rem= it will hold remainder of numbers.

After Declaring variables we will take a number form user to find given number is a happy number or not.

after taking values we will check given number is valid number or not.

and the temp value to function Happy() where we iterate the number to its last value.

the function return the square of each digit present in number and finally at last we will check if the sum is equal to 1 it is a Happy number.

if it is not equal to “1” than it is not a happy number.

Program to check CoPrime Numbers Program in Java

In this tutorial, we will learn to create a Java Program to check CoPrime Numbers Program in Java

Prerequisites

Before starting with this tutorial, we assume that you are best aware of the following Java programming concepts:

Java Operators.
Basic Input and Output function in Java.
Class and Object in Java.
Basic Java programming.
If-else statements in Java.
For loop in Java.

What is Co-prime Number.

The numbers are said to be co-prime, if their GCD is “1”
So 13,14 and 15 are co-prime numbers.

Program to check CoPrime Numbers Program in Java

In this program we would find the given two number are co-prime or not first will take the input from the user and check they arr coprime or not. let have a look at the code

//java program to find co -prime
import java.io.*;
import java.util.*;
class Co_prime
{
    public static void main(String args[])
    {
      //declaring variables and taking vlaues from user..
        Scanner in= new Scanner(System.in);
        int num1,num2,a,i,c=0;
        System.out.println("Enter two Numbers");
        num1= in.nextInt();
        num2= in.nextInt();
        a=num1*num2;
      // condition for co-prime number 
        for(i=1;i<num1;i++)
        {
            if(num1%i==0 && num2%i==0)
            {
                c=i;
            }
        }
// if result is 1 they are co-prime number 
        if(c==1)
        {
            System.out.println("Numbers are Co prime : "+num1+" & "+num2);
        }
        else
        {
            System.out.println("Numbers are not Co prime : "+num1+" & "+num2);
        }
    }// end of main method
}// end of class

//java program to find co -prime

import java.io.*;

import java.util.*;

class Co_prime

{

public static void main(String args[])

{

//declaring variables and taking vlaues from user..

Scanner in= new Scanner(System.in);

int num1,num2,a,i,c=0;

System.out.println("Enter two Numbers");

num1= in.nextInt();

num2= in.nextInt();

a=num1*num2;

// condition for co-prime number

for(i=1;i<num1;i++)

{

if(num1%i==0 && num2%i==0)

{

c=i;

}

// if result is 1 they are co-prime number

if(c==1)

{

System.out.println("Numbers are Co prime : "+num1+" & "+num2);

}

else

{

System.out.println("Numbers are not Co prime : "+num1+" & "+num2);

}

}// end of main method

}// end of class

Output

Co-prime

Not a Co-prime

In the above program, we have first declared and initialized a set variables required in the program.

Num1,num2 = it will hold entered numbers
a= for iteration.
c= for count.

After declaring variables we take two values from user.

Co-prime numbers are the numbers whose common factor is 1 lets look at the example.

Number  2 3
The factors of 2 are 1, 2. and factor of 3 is 1,3
They are Co-Prime.

Input : 4 8
The factors of 4 are 1, 2 and 4. and factor of 8 is 1,2,4 and 8
They are not Co-Prime.

Number 2 3

The factors of 2 are 1, 2. and factor of 3 is 1,3

They are Co-Prime.

Input : 4 8

The factors of 4 are 1, 2 and 4. and factor of 8 is 1,2,4 and 8

They are not Co-Prime.

after that we find factors of given number

if number has common factor 1 then they are co-prime

if not they are not co-prime numbers.

Program to check Circular Prime Program in Java

In this tutorial, we will learn to create a Java Program to check Circular Prime Program in Java Java

Prerequisites

Before starting with this tutorial we assume that you are best aware of the following Java programming topics:

Java Operators.
Basic Input and Output function in Java.
Class and Object in Java.
Basic Java programming.
If-else statements in Java.
Function in java.

Circular Prime Number.

Prime number is called a circular prime if all its cyclic permutations digits,remains a prime number.
Examples:

number = 113 
its all cyclic permutations of 113 (311
and 131) are prime.

number = 113

its all cyclic permutations of 113 (311

and 131) are prime.

Program to check Circular Prime Program in Java

In this program we would find the given number is a Circular Prime number or not , first of all we will take the input from the user and check whether it is Circular Prime number or not. Lets have a look at the code.

import java.util.*;
class CPrime
{
    static boolean prime(int num)
    {
        boolean count=true;
        for(int i=2;i<=num/2;i++)
        {
            if(num%i==0)
            {
                count=false;
                break;
            }
        }
        return(count);
    }
    public static void main(String arr[])
    {
        Scanner sc=new Scanner(System.in);
        int no,ct=0,temp,base;
        System.out.println("Enter any Prime Number");
        no=sc.nextInt();
        temp=no;
        while(temp>0)
        {
            ct++;
            temp=temp/10;
        }
        base=(int)Math.pow(10,ct-1);
        boolean count=true;
        for(int j=1;j<=ct;j++)
        {
            no=(no%base)*10+(no/base);
            if(CPrime.prime(no)==false)
            {
                count=false;
                break;
            }
            System.out.println(no);
        }
        if(count==true)
        {
            System.out.println("Number is Circular Prime");
        }
        else
        {
            System.out.println("Number is Not Circular Prime");
        }
    }
}

import java.util.*;

class CPrime

{

static boolean prime(int num)

{

boolean count=true;

for(int i=2;i<=num/2;i++)

{

if(num%i==0)

{

count=false;

break;

}

return(count);

}

public static void main(String arr[])

{

Scanner sc=new Scanner(System.in);

int no,ct=0,temp,base;

System.out.println("Enter any Prime Number");

no=sc.nextInt();

temp=no;

while(temp>0)

{

ct++;

temp=temp/10;

}

base=(int)Math.pow(10,ct-1);

boolean count=true;

for(int j=1;j<=ct;j++)

{

no=(no%base)*10+(no/base);

if(CPrime.prime(no)==false)

{

count=false;

break;

}

System.out.println(no);

}

if(count==true)

{

System.out.println("Number is Circular Prime");

}

else

{

System.out.println("Number is Not Circular Prime");

}

Output

Circular prime number

Not a Circular prime number

In the above program, we have first declared and initialized a set variables required in the program.

no,num= it will hold entered number.
temp = it will hold temp value
sum= for adding digit sum.
count,ct= it will hold flag value.
i,j= for iteration.

After declaring variables first we take a prime number from user.

then we check all the possible combination of a given number and all are prime number or not.

Here we created a function prime with in the class Cprime where we checked each and every possible combination of given number and all are prime or not and if they are prime.

A Circular Prime number is a prime number that remains prime under each every cyclic shifts of its digits.

Prime Number Up to N Terms Program in Java

In this tutorial, we will learn to create a Java Prime Number Up to N Terms Program in Java.

Prerequisites

Before starting with this tutorial we assume that you are best aware of the following java programming topics:

Java Operators.
Basic Input and Output
Class and Object.
Basic Java programming.
String.
User define Function.

What is a Prime Number?

A number is called a prime number if it fully divisible by the number itself or 1 only.

For example :-> 7 is a prime number, it is fully divisible by 7 only between 2-7 and “1” only.

Prime Number Up to N Terms Program in Java

In this program we will find prime n umber between 1-N. We would first declared and initialized the required variables. Next, we would ask user to enter value of N. let have a look at Code.

// java program to find n prime numbers
import java.util.Scanner;

public class Prime_Number {
// declaring variables and taking value from user..
    private static Scanner scanner = new Scanner( System.in );

    public static void main(String[] args) {
        System.out.println("Enter Size of N: ");
        String input = scanner.nextLine();
        int n = Integer.parseInt( input );
        System.out.println("List of  prime number between 1 - " + n);
// condttion for prime number..
        for (int number = 2; number <= n; number++)
        {
            boolean p = true;
            for (int i=2; i <= number/2; i++)
            {
                if ( number % i == 0)
                {
                    p = false;
                    break;
                }
            }
			
            if ( p == true )
                System.out.println(number);
        }
    }
}

// java program to find n prime numbers

import java.util.Scanner;

public class Prime_Number {

// declaring variables and taking value from user..

private static Scanner scanner = new Scanner( System.in );

public static void main(String[] args) {

System.out.println("Enter Size of N: ");

String input = scanner.nextLine();

int n = Integer.parseInt( input );

System.out.println("List of prime number between 1 - " + n);

// condttion for prime number..

for (int number = 2; number <= n; number++)

{

boolean p = true;

for (int i=2; i <= number/2; i++)

{

if ( number % i == 0)

{

p = false;

break;

}

if ( p == true )

System.out.println(number);

}

Output

N Prime Number

In the above program, we have first declared and initialized a set variables required in the program.

n = it will hold entered N size.
number = For iteration till N.
p= for condition of prime.

After that we declare variables and taking value from user.

Then we check condition for prime between 2 to N.

Here we iterate loop form two to given number and check the condition for every number that is prime or not and last we will print list of all prime number between 1 to N as shown in above image.

Program to check Amicable Numbers in Java

In this tutorial, we will learn to create a Java Program to check Amicable Numbers in Java using Java programming.

Prerequisites

Before starting with this tutorial we assume that you are best aware of the following Java programming topics:

Java Operators.
Basic Input and Output function in Java.
Class and Object in Java.
Basic Java programming.
If-else statements in Java.
For loop in Java.

What is Amicable Number?

Pairs of number said to be Amicable Numbers, if sum of divisors of first number equals to the second number.
Some of the Amicable Numbers are: (220, 284), (1184, 1210), (2620, 2924), (5020, 5564), (6232, 6368) etc…

let’s have an example of one Amicable Number.
Divisors of 220 are = “1, 2, 4, 5, 10, 11, 20, 22, 44, 55, 110” on adding them we get
(1+2+4+5+10+11+20+22+44+55+110)=284
Now we will check if the Divisors of 284 = 1, 2, 4, 5, 10, 11, 20, 22, 44, 55, 110
1+2+7+71+142=220.
So we can see clearly 220 and 284 are Amicable Numbers.

Program to check Amicable Numbers in Java

In this program , we will learn to create a Program to check Amicable Numbers.First we would declared and initialized the required variables. Next, we would prompt user to input the values. Later we will find that they are Amicable Number or not.

// Java program to find Amicable number.
import java.util.Scanner;

class Amicable
    {
      public static void main(String args[])
      {
            // taking values from user..
            int no1,no2,sum1=0,sum2=0,i;
            Scanner in = new Scanner(System.in);
            System.out.print("Enter  first number: ");
            no1 = in.nextInt();
            System.out.print("Enter second number: ");
            no2 = in.nextInt();
            
            for (i = 1; i <= no1; i++) {
                if (no1 % i == 0)
                    sum1 += i;
            }
            for (int i = 1; i <= no2; i++) {
                if (no2 % i == 0)
                    sum2 += i;
            }
            // Checking condition of Amicable ..
            if (sum1 == sum2)
                System.out.println("Numbers are amicable.");
            else
                System.out.println("Numbers are not amicable.");
				System.out.println("\n");			
    }

}

// Java program to find Amicable number.

import java.util.Scanner;

class Amicable

{

public static void main(String args[])

{

// taking values from user..

int no1,no2,sum1=0,sum2=0,i;

Scanner in = new Scanner(System.in);

System.out.print("Enter first number: ");

no1 = in.nextInt();

System.out.print("Enter second number: ");

no2 = in.nextInt();

for (i = 1; i <= no1; i++) {

if (no1 % i == 0)

sum1 += i;

}

for (int i = 1; i <= no2; i++) {

if (no2 % i == 0)

sum2 += i;

}

// Checking condition of Amicable ..

if (sum1 == sum2)

System.out.println("Numbers are amicable.");

else

System.out.println("Numbers are not amicable.");

System.out.println("\n");

}

Output

In the above program, we have first declared and initialized a set variables required in the program.

no1= it will hold first entered number.
no2= it will hold second entered number.
sum1= it will calculate first divisor sum.
sum2= it will calculate second divisor sum.
i = for iteration.

After that taking values from user we will assigned the values in variable “no1” and “no2” respectively.

Now, we have two different values in variable “no1” and in “no2” and now with the modulus (%) operator we find all the divisor of numbers.

and for second number.

Steps for finding Amicable Number are as follow.

Read and initialize numbers in variables (no1, no2) from the user.
Find the divisor of both the number (no1, no2) using for loop.
Find the sum of divisors in variables sum1 and in sum2.
Checking if the sum of divisors of number no1 and no2 are equal to each other or not.

If equal, the numbers are Amicable.

Else, not Amicable.

So in that manner we will find the entered number are Amicable or Not.

Program to check Ugly Number in Java

In this tutorial, we will learn to create a Java Program to check Ugly Number in Java using Java programming.

Prerequisites

Before starting with this tutorial we assume that you are best aware of the following Java programming topics:

Java Operators.
Basic Input and Output function in Java.
Class and Object in Java.
Basic Java programming.
If-else statements in Java.
For loop in Java.
User define functions.

What is Ugly Numbers?

Ugly numbers is a number whose prime factors are “2, 3 or 5“. Some of the Ugly number are as “1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15”. So the number after dividing gives factors either 2,3 or 5, are called Ugly number.

Program to check Ugly Number in Java

In this program , we will learn to create a Program to check Ugly Number.First we would declared and initialized the required variables. Next, we would prompt user to input the values. Later we will find that they the number is Ugly or not.

// java program to find ugly number..
public class UglyNo {
   public static boolean UglyNumber(int no) {
      boolean x = true;
      while (no != 1) {
         if (no % 5 == 0) {
            no /= 5;
         }
         else if (no % 3 == 0) {
            no /= 3;
         }
         // To check if number is divisible by 2 or not
         else if (no % 2 == 0) {
            no /= 2;
         }
         else {
            x = false;
            break;
         }
      }
      return x;
   }
   public static int UglyNum(int n) {
      int i = 1;
      int count = 1;
      while (n > count) {
         i++;
         if (UglyNumber(i)) {
            count++;
         }
      }
      return i;
   }
   public static void main(String[] args) {
      int number = 50;
      int no = UglyNum(number);
      System.out.println("The Ugly no. at position " + number + " is " + no);
   }
}

// java program to find ugly number..

public class UglyNo {

public static boolean UglyNumber(int no) {

boolean x = true;

while (no != 1) {

if (no % 5 == 0) {

no /= 5;

}

else if (no % 3 == 0) {

no /= 3;

}

// To check if number is divisible by 2 or not

else if (no % 2 == 0) {

no /= 2;

}

else {

x = false;

break;

}

return x;

}

public static int UglyNum(int n) {

int i = 1;

int count = 1;

while (n > count) {

i++;

if (UglyNumber(i)) {

count++;

}

return i;

}

public static void main(String[] args) {

int number = 50;

int no = UglyNum(number);

System.out.println("The Ugly no. at position " + number + " is " + no);

}

output

In the above program, we have first declared and initialized a set variables required in the program.

number= A number N to find the Nth Ugly Number.
no2= it will hold second entered number.
sum1= it will calculate first divisor’s sum.
sum2= it will calculate second divisor’s sum.
i = for iteration.

After that taking values from user we will assigned these values in to variable “no1” and “no2” respectively.

So what we are do here is that if any number has factor 2,3, or 5 if it has then the number is ugly and if number not have factors 2,3 or 5 then it is not ugly.Let’s take an example for that
Number is : 10
So Prime Factors of 10 is : 2 and 5.
Prime Factors contains only 2,3 or 5 : Here 10 is an Ugly number.

Number is : 11
So Prime Factors of 11 is : 1 and 11.
Prime Factors contains only 2,3 or 5 : Here 11 is not an Ugly number.

so here we have done same using java programming in image below .

So finding that the number is Ugly or not, we divide the number by 2, 3, and 5. When the number gives remainder equivalent to Zero,then it is an Ugly number. Otherwise, it is not an Ugly number.

Java Program to convert Number to Word

In this tutorial, we will learn to create a Java Program to convert Number to Word.

Prerequisites

Before starting with this tutorial we assume that you are best aware of the following java programming topics:

Java Operators.
Basic Input and Output
Class and Object.
Basic Java programming.
String.
User define Function.

Java Program to convert Number to Word

In this program we will convert Number to Word. We would first declared and initialized the required variables. Next, we would ask user to enter a number. let have a look at Code.

// java program to covert number to string...
import java.util.Scanner;
import java.util.Scanner; 
public class Number_to_word 
{ 
public static void main(String[] args)
{
 int num = 0;
 Scanner scanner = new Scanner(System.in);
 System.out.println("Please enter a number(max upto 9 digits)");
 try {
 num = scanner.nextInt();
 if (num == 0) {
 System.out.print("Number in words: Zero");
 } else {
 System.out.print("Number in words: " + number_Word(num));
 }
 } catch (Exception e) {
 System.out.println("Please enter a valid number");
 }
 // close the reader
 scanner.close();
 }  private static String number_Word(int num) {
 
 String words = "";
 String unitarr[] = { "zero", "one", "two", "three", "four", "five", "six", 
 "seven", "eight", "nine", "ten", "eleven", "twelve",
 "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", 
 "eighteen", "nineteen" };
 String tensarr[] = { "zero", "ten", "twenty", "thirty", "forty", "fifty",
 "sixty", "seventy", "eighty", "ninety" };  if (num == 0) {
 return "zero";
 }
 
 if (num < 0) { 
 // converting the number to a string
 String numberStr = "" + num; 
 // removing minus before the number 
 numberStr = numberStr.substring(1); 
 // add minus before the number and convert the rest of number 
 return "minus " + number_Word(Integer.parseInt(numberStr)); 
 } 
 // cconditon for divisible by 1 million
 if ((num / 1000000) > 0) {
 words += number_Word(num / 1000000) + " million ";
 num %= 1000000;
 }
 // cconditon for divisible by 1 thousand
 if ((num / 1000) > 0) {
 words += number_Word(num / 1000) + " thousand ";
 num %= 1000;
 }
 // cconditon for divisible by 1 hundred
 if ((num / 100) > 0) {
 words += number_Word(num / 100) + " hundred ";
 num %= 100;
 }  if (num > 0) {
 if (num < 20) { 
  words += unitarr[num];
 } else { 
 words += tensarr[num / 10]; 
 if ((num % 10) > 0) {
 words += "-" + unitarr[num % 10];
 } 
 }
 }
 return words;
 }
}

// java program to covert number to string...

import java.util.Scanner;

public class Number_to_word

{

public static void main(String[] args)

{

int num = 0;

Scanner scanner = new Scanner(System.in);

System.out.println("Please enter a number(max upto 9 digits)");

try {

num = scanner.nextInt();

if (num == 0) {

System.out.print("Number in words: Zero");

} else {

System.out.print("Number in words: " + number_Word(num));

}

} catch (Exception e) {

System.out.println("Please enter a valid number");

}

// close the reader

scanner.close();

} private static String number_Word(int num) {

String words = "";

String unitarr[] = { "zero", "one", "two", "three", "four", "five", "six",

"seven", "eight", "nine", "ten", "eleven", "twelve",

"thirteen", "fourteen", "fifteen", "sixteen", "seventeen",

"eighteen", "nineteen" };

String tensarr[] = { "zero", "ten", "twenty", "thirty", "forty", "fifty",

"sixty", "seventy", "eighty", "ninety" }; if (num == 0) {

return "zero";

}

if (num < 0) {

// converting the number to a string

String numberStr = "" + num;

// removing minus before the number

numberStr = numberStr.substring(1);

// add minus before the number and convert the rest of number

return "minus " + number_Word(Integer.parseInt(numberStr));

}

// cconditon for divisible by 1 million

if ((num / 1000000) > 0) {

words += number_Word(num / 1000000) + " million ";

num %= 1000000;

}

// cconditon for divisible by 1 thousand

if ((num / 1000) > 0) {

words += number_Word(num / 1000) + " thousand ";

num %= 1000;

}

// cconditon for divisible by 1 hundred

if ((num / 100) > 0) {

words += number_Word(num / 100) + " hundred ";

num %= 100;

} if (num > 0) {

if (num < 20) {

words += unitarr[num];

} else {

words += tensarr[num / 10];

if ((num % 10) > 0) {

words += "-" + unitarr[num % 10];

}

return words;

}

Output

In the above program, we have first declared and initialized a set variables required in the program.

num = it will hold entered Number
word= it will hold word values.
unitarr[]= it will hold unit value of word.
tensarr[]=it will hold tense value of word.

After that we declare variables and taking value from user.

the following code supports numbers up-to 9999 digits, i.e., numbers from 0 to 9999.

here we create arrays that break individual parts of a strings. One array is used for single digits (unitarr[]), one for numbers from 10 to 19,. etc (tensarr[]). and one for powers of 10. and multiple condition that number is >1000 ,>100 ans so on..as shoen in image below.

And finally we will print the characters or given numbers.

Program to check nth Prime Number In Java

In this tutorial, we will learn to create a Java Program to check nth Prime Number In Java.

Prerequisites

Before starting with this tutorial, we assume that you are best aware of the following Java programming concepts:

Java Operators.
Basic Input and Output function in Java.
Class and Object in Java.
Basic Java programming.
If-else statements in Java.
For loop in Java.

Prime Number.

A number is called a prime if it is divisible by 1 or itself only.

Example:=> 2, 3, 5, 7, 11 and etc…

except 2 all are odd numbers.

Program to check nth Prime Number In Java

In this program we would find the nth number position prime number.first of we would take a value from user and find the nth position prime number.let have a look at the code.

// java program to fing nth postion prime number..
import java.util.Scanner;  
public class Prime_num  
{  
public static void main(String[] args)   
{  
//decalring variables and taking value from user..  
Scanner sc = new Scanner(System.in);  
System.out.print("Enter the postion of n to compute the nth prime number: ");  
int number = sc.nextInt();   
int prime=1, ct=0, i;  
while (ct < number)  
{  
prime=prime+1;  
for (i = 2; i <= prime; i++)  
{   
if (prime % i == 0)   
{  
break;  
}  
}  
if (i == prime)  
{  
//increments the counter variable by 1 if the number is prime number 
ct = ct+1;  
}  
}  
//prints the nth prime number  
System.out.println("The " +number +"th prime number is: " + prime);  
}  
}

// java program to fing nth postion prime number..

import java.util.Scanner;

public class Prime_num

{

public static void main(String[] args)

{

//decalring variables and taking value from user..

Scanner sc = new Scanner(System.in);

System.out.print("Enter the postion of n to compute the nth prime number: ");

int number = sc.nextInt();

int prime=1, ct=0, i;

while (ct < number)

{

prime=prime+1;

for (i = 2; i <= prime; i++)

{

if (prime % i == 0)

{

break;

}

if (i == prime)

{

//increments the counter variable by 1 if the number is prime number

ct = ct+1;

}

//prints the nth prime number

System.out.println("The " +number +"th prime number is: " + prime);

}

Output

In the above program, we have first declared and initialized a set variables required in the program.

number = it will hold entered nth value number
prime= it wil hold value of prime number
i= for iteration .
ct= calculate the count value

After declaring variables we take a nth value of from user.

and we will find the nth position prime number as shown in image below.

Approach to find nth Prime number.

we will used the same approach to find the prime number as w use to do always.

Steps for finding Prime nth Prime number.

take an integer value (number) from the user.
In loop,we execute the condition (ct<number). At Initially, set value “ct=0”.
Increment the value “i” by 1.
Check if the variable “i” is prime number or not.
If true, increment the variable “ct” by “1”.
follow this approach till nth number.
and finally at last print the last prime number value.

Hope this help you.