Bluej for ICSE Schools

Happy New Year 2011

2010-12-31T23:13:00.001-08:00

A Very Safe, Happy and Prosperous New Year to All Our Valued readers.

We must take a pledge today to:

1. Stop Ragging in all the Institutions and make Indian Education System a Ragging Free Place.

2. Make Indian Institutions a Innovation Place Where Innovative Ideas are generated.

3. Make Indian Institutions a Leader in Education and Research.

Ragging India Team

Average of Array

2010-06-18T22:03:00.001-07:00

First of all define a class name "ArrayAverage" that has double type array to contain some values. Now take a double type data that calculates the average of array (result/nums.length). And finally it will display the result on the command prompt with message by using the System.out.println().

Program:

public class ArrayAverage{
public static void main(String[] args) {
    double nums[]={1.0,2.3,3.4,4.5,40.5};
    double result=0.0;
    int i=0;
    for(i=0; i < nums.length; i++){
      result=result + nums[i];
    }
    System.out.println("Average is =" + result/nums.length);
}
}

Method In Java - toUpperCase()

2010-06-18T22:00:00.001-07:00

How to use toUpperCase() method of the String class ?

Procedure for converting letters of the string in uppercase letter using toUpperCase() method.

The following program convert the string "india" into "INDIA" by using toUpperCase() method.

toUpperCase(): returns a string value.

Code of this program:

public class ConvertInUpperCase{

public static void main(String args[]){
    String roseindia = "india";
    System.out.println("String is : " + roseindia);
    String upper = roseindia.toUpperCase();
    System.out.println("String in uppercase letter: " + upper);
}
}

Output of program:

C:\java_work>javac ConvertInUpperCase.java

C:\java_work>java ConvertInUpperCase
String is : india
String in uppercase letter: INDIA

C:\java_work>

HOW TO REDUCE GREEN HOUSE GAS AT HOME

2010-06-09T00:33:00.001-07:00

Not every family can afford to fix solar panels to their roofs or erect a wind turbine, but there are still steps you can take to save energy and therefore money.

1. Switch your electricity supply to a green tariff that promises to supply electricity from renewable sources.

2. Save on your heating bill by insulating and draught-proofing your home.

3. Switch to energy efficient light bulbs. They can use one quarter the electricity and last ten times as long.

4. Recycle paper, plastic, glass and metals. You not only save natural resources, but the energy that goes into making and transporting them.

5. Choose products that have less packaging -then you have less to recycle

6. Don't leave electrical equipment, like TVs and computers on standby they can use up to 75% of the electricity they use when turned on.

7. Take your mobile phone charger out of the socket when not being used.

8. Turn your thermostat down 1°C – you could save 10% on your fuel bills.

9. Only use your washing machine when there is a full load. Set your washing machine to warm or cool washes.

10. Fix leaking taps. A hot tap dripping could waste enough water each day for one bath.

11. Don't overheat your water. The recommended level is 60°C.

12. A full freezer uses less electricity than a half empty one – and if that isn't a good reason for buying more ice cream….

13. Kettles – if you don't need a full kettle of water don't boil a full kettle of water!

14. Dry your washing outside on the line when the weather is fine.

15. Hang reflective foil (with the shiny side facing the radiator) behind radiators fitted to outside walls to reduce heat loss.

16. Line your curtains and close them at dusk.

17. When choosing electrical appliances, make sure what you are buying is energy efficient.

18. Wearing t-shirts in winter and turning the heater up is just silly, put on warmer clothes.

19. Remember to turn the lights off when you leave the room ...... unless somebody else is still in the room!

Prime Number in Java

2009-12-24T05:08:00.001-08:00

To get prime number between 1 to given number, first of all we have to define a class "PrimeNumber". Java I/O package has a input stream and a output stream in which input stream is used for reading the stream and memory allocating and the output stream used for writing bytes. As in this program we are going to insert certain instruction by creating buffer reader class. Here we have to create a buffer for the string class that can be used to instantiate a changeable object for storing and processing a string of character. Now use the ParseInt method for converting the parses the string argument and define 'num' as an integer.

Now applying in this program we use two 'for' loop. For loop will start from 1 to entered number. And another loop will start and divide it from 2 to less than those number. If number is divided by any number that means it is not prime otherwise prime number.

Program:

import java.io.*;

class PrimeNumber {
public static void main(String[] args) throws Exception{
    int i;
    BufferedReader bf = new BufferedReader(
              new InputStreamReader(System.in));
    System.out.println("Enter number:");
    int num = Integer.parseInt(bf.readLine());
    System.out.println("Prime number: ");
    for (i=1; i < num; i++ ){
      int j;
      for (j=2; j<i; j++){
        int n = i%j;
        if (n==0){
          break;
        }
      }
      if(i == j){
        System.out.print(" "+i);
      }
    }
}
}

Preparing table of a number by using loop

2009-12-24T05:03:00.001-08:00

the methods of preparing the table of a given number by using loop condition. As we know the loop statements are used to repeat a statement or process multiple times according to a specified condition. Loop checks certain condition and if it finds the condition is valuable then all the statements written under loop are executed.

Here we will take a number a=25 of which we have to prepare a table. Define the integer a=25 and b=1 as the initial point. Now apply "while" condition of loop and confine b<=10 as we have to make a table of 25. Again define another integer as c=a*b, this will be the result when we multiply 'a' with 'b'. Here we have to multiply 'a' with 'b' up to 10 times like a*1, a*2....................a*9, a*10. So make define b=b+1 as increment operator.

prorgram:

class PreparingTable{
public static void main(String[] args) {
    int a=25, b=1;
    System.out.println("the table of "+a+"= ");
    while(b<=10){
      int c = a*b;
      System.out.println(c);
      b = b+1;
    }
}
}

Listing out leap years between certain period

2009-10-24T00:32:00.001-07:00

we have to find out the leap years between 1990 and 2006. First define the two years under a class "leapyears". Let i = 2006 and n=1990. Now with the help of for loop method initialize the year as n=1990 and n<=i. Also apply the increment statement in the loop as we have to check one by one.

As we know a leap year is divisible by 4, define an integer l=n%4. So if 'n' is divisible by 4 or l=0, then the particular year can be a leap year. For checking this, apply the if statement and if this satisfies then, the year will be a leap year. For listing out each year write "+n" in the System.out.println.

Code of the program:

class leapyears
{
public static void main(String[] args)
{
    int i=2006;
    int n;
    for (n=1990; n<=i ; n++){
      int l=n%4;
      if (l==0){
        System.out.println("leap year: "+n);
      }
    }
}
}

Checking whether a year is leap or not

2009-10-24T00:30:00.000-07:00

we have taken the year 2000. So define an integer n=2000 in the class "Leapyear" and now apply "if else" condition. As we know leap year is divided by the integer 4 and so applying if condition as n/4=0, then "n" is a leap year. Now in the System.out.println write the message that the year is a leap year. Again applying "else" condition the output will be that the year is not a leap year.

Program Code:

class Leapyear
{
public static void main(String[] args)
{
    int n=2000;
    if (n%4==0){
    System.out.println("The given year is a leap year");
      }
    else{
    System.out.println("This is not a leap year");
}
}
}

WAP to construct a triangle with the ‘*’

2009-10-06T06:24:00.001-07:00

First of all make a class named 'triangle' under the Java I/O package and as we have to use the Buffer class, the application of all try and catch block is important for avoiding any kind of error. After creating BufferedReader object and input stream reader define an integer 'a' and apply parseInt method for the conversion of string into integer.

Now apply the for loop and define an integer 'i' and it should be either less than or equal to the integer "a" (the input number). Again define another integer type variable "j" in another for loop. Here in the second for loop "j" the number of times we have to print *.

Now compile and run the program and insert any number on your command window and surely you will get a triangle shape with star *. You can take any other object instead of *.

Code of the program:

import java.io.*;

class triangle{
public static void main(String[] args) {
    try{
      BufferedReader object = new BufferedReader(new InputStreamReader(System.in));
      System.out.println("enter the number");
      int a= Integer.parseInt(object.readLine());
      for (int i=1; i<a;i++ ){
        for (int j=1; j<=i;j++ ){
          System.out.print("*");
        }
        System.out.println("");
      }
    }
    catch(Exception e){}
}
}

A program for calculating area and perimeter of a rectangle

2009-10-06T06:23:00.001-07:00

First of all create a class named RecArea under Java.io package. Now define two integer variable 'l' and 'w'. As the program will be based on keyboard numerical input, it is important for every programmer to use correct data without any mistake. In this case the exception methods like try/catch mechanism helps in detecting user input errors. So before starting the functional code, enclosed it with try clause so that any error in the statement causes the execution of the catch clauses.

Now create an abstract buffer class which is the super class of all classes and represents a stream of input bytes. The InputSreamReader reads the character stream and stores it in the buffer class. Now use parseInt for both length and width of the rectangle. This is an instance of class method and is used to convert a string to an integer. Define the area as l*w and perimeter as 2*(l+w) and in the end use the catch exception.

Now compile and run the program and input the value as you see the message and get the ultimate result. If you find any kind of error, then check the whole program again.

Code of the program:

import java.io.*;
class RecArea
{
public static void main(String[] args)
{

int l=0;
int w=0;

    try{

      BufferedReader br1 = new BufferedReader(new InputStreamReader(System.in));
      System.out.println("Enter length of rectangle : ");
      l = Integer.parseInt(br1.readLine());
      System.out.println("Enter width of rectangle : ");
      w = Integer.parseInt(br1.readLine());
      int area = l*w;
      System.out.println("Area of Rectangle : "+area);
      int perimiter = 2*(l+w);
      System.out.println("Perimeter: " + perimiter);

}catch(Exception e){System.out.println("Error : "+e);}

}
}

Palindrome Number

2009-09-14T05:38:00.001-07:00

With the help of this program, we are going to determine whether the given number is palindrome or not. To achieve the desired result, firstly we have to define a class named "Palindrome". After that we will ask the user to enter any integer type number and then we will reverse it. After reversing the number we will check whether the given number is palindrome or not. If the given number is larger, then it will display a message "Out of range!".

import java.io.*;

public class Palindrome  {
      public static void main(String [] args){
      try{
      BufferedReader object = new BufferedReader(
                  new InputStreamReader(System.in));
      System.out.println("Enter number");
      int num= Integer.parseInt(object.readLine());
        int n = num;
        int rev=0;
        System.out.println("Number: ");
        System.out.println(" "+ num);
        for (int i=0; i<=num; i++){
          int r=num%10;
          num=num/10;
          rev=rev*10+r;
          i=0;
        }
        System.out.println("After reversing the number: "+ " ");
        System.out.println(" "+ rev);
        if(n == rev){
        System.out.print("Number is palindrome!");
        }
        else{
        System.out.println("Number is not palindrome!");
        }
      }
      catch(Exception e){
        System.out.println("Out of range!");
      }
  }
}

Calculate factorial of any given number - factorial of number in double

2009-09-14T05:37:00.001-07:00

import java.io.*;

class Factorial1{
  public static void main(String[] args) {
      try{
        BufferedReader object = new BufferedReader(
        new InputStreamReader(System.in));
        System.out.println("enter the number");
        int a= Integer.parseInt(object.readLine());
        double fact= 1;
        System.out.println("Factorial of " +a+ ":");
        for (int i= 1; i<=a; i++){
              fact=fact*i;
        }
        System.out.println(fact);
    }
    catch (Exception e){
      System.out.println("Array out of bounds exaception");
    }
  }
}

Program to calculate factorial of any given number

2009-09-06T23:23:00.001-07:00

This Java programming tutorial will teach you the methods for writing program to calculate factorial of any given number. First of all define a class "Factorial" under the Java I/O package. Java I/O package has a input stream and a output stream in which input stream is used for reading the stream and memory allocating and the output stream is used for writing bytes. As in this program we are going to insert certain instruction by creating buffer reader class, it is necessary to use 'try' and 'catch' block for catching and handling exceptions during execution of the program.

Here, we have to create a buffer for the string class that can be used to instantiate a changeable object for storing and processing a string of character. The strings length and content change as soon as any object is inserted, replaced or removed from the StringBuffer object.

Now create a buffer object that inherits properties from the string object class. Now create an InputStreamReader that reads bytes and decodes them into character by using certain 'charset'. Now use the ParseInt method for converting the parses the string argument to a decimal integer and define 'a' as an integer. Take an integer variable as fact=1 and insert the message in the System method.

Now applying for loop with conditions as integer i=1(intializer), i<=a and i++ as increment operator. So output result will be like fact=fact*i.

Here is the code of the program:

import java.io.*;
class Factorial{
  public static void main(String[] args) {
      try{
        BufferedReader object = new BufferedReader(new InputStreamReader(System.in));
        System.out.println("enter the number");
        int a= Integer.parseInt(object.readLine());
        int fact= 1;
        System.out.println("Factorial of " +a+ ":");
        for (int i= 1; i<=a; i++){
              fact=fact*i;
        }
        System.out.println(fact);
    }
    catch (Exception e){}
  }
}

WAP to calculate area and perimeter of a circle

2009-09-06T23:22:00.001-07:00

The given example will teach you the method for preparing a program to calculate the area and perimeter of a circle. First of all name a class as "CircleArea" under Java I/O package and define and integer r=o, which is the radius of the circle. Now use try exception to handle errors and other exceptional events. As we have to input the value of radius here create a buffered class with an object as 'br1'. This create a buffering character input stream that uses a default sized input buffer. The InputStreamReader here works as a translator that converts byte stream to character stream. Now type message that "Enter radius of circle" in the System.out.println method.

Now use the parseInt() method of the Integer class in order to convert from external numeric format to internal format. Now create the Math class in which all the mathematical functions are defined. This Math class can be imported from the java.lang.* package. Write the program for both the cases: radius and perimeter.

Before ending the program use the Catch mechanism that detects and catch user input errors. In the end compile and run the program and enter your desired value as radius for calculating the radius and perimeter of the circle.

Code of the program:

import java.io.*;
class CircleArea{
  public static void main(String[] args){
    int r=0;
           try{
        BufferedReader br1 = new BufferedReader(new InputStreamReader(System.in));
        System.out.println("Enter Radius of Circle  : ");
        r = Integer.parseInt(br1.readLine());
        double area = java.lang.Math.PI*r*r;
        System.out.println("Area of Circle : "+area);
        double  perimeter =2*java.lang.Math.PI*r ;
        System.out.println("Perimeter of Circle : "+perimeter);
      }
      catch(Exception e){
        System.out.println("Error : "+e);
      }
  }
}

Program to list all even numbers between two numbers

2009-08-07T23:34:00.000-07:00

First create a class named AllEvenNum under the java.io package. Now use the try/catch exception to avoid any kind of input error. After this create a buffer class in which all the input data are stored and modified. Then give message as to "Enter number" in the System method.

As we have to find out all the even numbers between 1 and the input number, define an integer variable 'num'. Now apply ParseInt method that parses the string character into decimal integer. Again apply for loop in which define an integer i=1 and i<= num also with an increment operator. Then apply the if condition that i/2=0 i.e. to find even numbers which are divided by the integer 2. In the end apply the catch exception.

      import java.io.*;

class AllEvenNum{
  public static void main(String[] args) {
    try{
      BufferedReader br1 = new BufferedReader(new InputStreamReader(System.in));
      System.out.println("Enter number : ");
      int num = Integer.parseInt(br1.readLine());
      System.out.println("Even Numbers:");
      for (int i=1;i <=num ; i++){
        if(i%2==0 ){
          System.out.print(i+",");
        }
      }
    }
    catch(Exception e){}

  }
}

Determining the largest number

2009-08-07T23:33:00.001-07:00

This example of Java programming will teach you the coding for determining the largest number amongst three. Here we have taken three integers as x = 500, y = 70 and z = 3000. After defining these three integers under the class "largernumber" apply "if" and "else" conditions that can help you in finding the largest value one by one.

First check if "x>y". If this satisfies then check whether x>z or not. Again if this satisfies then write in the system class that "x is greater". Again the term "else" comes when "x" is not greater than "z". So check again, if "z" is greater than "y" or not. If this satisfies then type in the system class as "z is greater" otherwise (in the else condition) "y" is greater. Now check whether "y" is greater than "z" or not.

If "x" is not greater than "y" as per the first condition, then the condition "else" comes and now you have to check if "y>z" or not. If this satisfies then the output comes as "y is greater".

class largernumber{
  public static void main(String[] args) {
    int x=500, y=70, z=3000;
    if (x>y){
      if (x>z){
        System.out.println("x is greater");
      }
      else{
        if(z>y){
          System.out.println("z is greater");
        }
        else{
          System.out.println("y is greater");
        }
      }
    }
    else{
      if (y>z){
        System.out.println("y is greater");
      }
    }
  }
}

Comparing Two Numbers

2009-07-30T06:41:00.001-07:00

This is a very simple example of Java that teaches you the method of comparing two numbers and finding out the greater one. First of all, name a class "Comparing" and take two numbers in this class. Here we have taken a=24 and b=25, now we have to find out whether a=b, a>b or b>a. To find out this apply if and else condition one by one. Now apply the condition "if (a=b)", if this satisfies then type that both are equal in the system class. If this doesn't satisfy, then check whether a>b by applying the "else if" condition and type the message "a is greater than b" in the system class. Again this doesn't satisfy then 'else' condition as shown in the example will show that b is greater than a.

Now compile and run the program and you will find the desired output. If you are getting any error then check the whole program thoroughly and surely you will get correct result. By compiling and running this exact program, you will find that b is greater than a.

class  Comparing{
  public static void main(String[] args) {
    int a=24, b=25;
    if (a == b){
      System.out.println("Both are equal");
    }
    else if(a>b){
      System.out.println("a is greater than b");
    }
    else{
      System.out.println("b is greater than a");
    }
  }
}

Java prime number generator

2009-07-30T06:39:00.001-07:00

This program find and displays prime numbers between 1 and a given number. * It doesn't include the number entered.

public static void main(String[] args) {

    if(args.length==0)

        System.out.print("Please ennter a Number.");

        System.exit(0);

    int count=0;

    int number= Integer.parseInt(args[0]);

    int[] numColl= new int[number];

     for (int i=1;i<number;i++)

        if(i==1)

            continue;

        else if(i==2||i==3||i==5||i==7)

            count++;

            numColl[i]= i;

             continue;

        else

            if(i%2==0||i%3==0||i%5==0||i%7==0)

                continue;

            else

                numColl[i]= i;

                count++;

                continue;

    Arrays.sort(numColl);

    System.out.println("Prime numbers between 1 and "+number+" are "+ count);

    int N=1;

    for(int i=0;i<numColl.length;i++)

        if(numColl[i]==0)

            continue;

        else

            System.out.println("Prime number " +N+ " is: "+numColl[i]);

            N++;

}

Prime number generator Program 2

2009-07-23T05:28:00.001-07:00

1.           package com.destiney.Prime;

2.           class Prime

3.           {

4.             public static void main( String args[] )

5.             {

6.               int x, y, c = 0;

7.               for( x = 2; x < 1000; x++ )

8.               {

9.                 if( x % 2 != 0 || x == 2 )

10.             {

11.               for( y = 2; y <= x / 2; y++ )

12.               {

13.                 if( x % y == 0 )

14.                 {

15.                   break;

16.                 }

17.               }

18.

19.               if( y > x / 2 )

20.               {

21.                 System.out.println( x );

22.                 c++;

23.               }

24.             }

25.           }

26.           System.out.println( "\nTotal: " + c );

27.         }

28.       }

29.

Prime number generator Program 1

2009-07-23T05:27:00.001-07:00

`class` `Prime_number {`
  `public static void` `main(String[] args) {`
    `int` `num =` `11;`
    `int` `i;`
    `for` `(i=2; i < num ;i++ ){`
      `int` `n = num%i;`
      `if` `(n==0){`
        `System.out.println("not Prime!");`
        `break;`
      `}`
    `}`
    `if(i == num){`
      `System.out.println("Prime number!");`
    `}`
  `}`
`}`

Ragging helpline tips - How to Use it.

2009-06-20T04:46:00.001-07:00

The helpline aims to help victims of ragging within minutes of their complaints. It will be operational at all hours. A four-digit, toll-free number will become operational once the helpline is launched. The steps being planned:

Step 1: A ragged student or a friend calls the helpline.

Step 2: The distress message is automatically stored in a central database with time, location and details of caller.

Step 3: The message is instantly replayed from the central control room, simultaneously alerting the head of the institution, the warden of the hostel and an appointed nodal officer of the affiliating university.

Step 4: An officer at the central control room will decide, immediately after the distress call, whether the case needs immediate intervention from district officials and police. If so, the message will be relayed to the district magistrate and the superintendent of police.

Step 5: The complaint will be placed on a website to enable the media and citizens to keep track of follow-up action.

Step 6: At the institution, its head will be required to act "immediately" on receiving the complaint. The head and the hostel warden will be held responsible for follow-up action.

Step 7: The head and the warden will have to alert an anti-ragging squad they are required to constitute in the institution. This squad is to consist of students and representatives of the staff. Its mandate is to be prepared for physical intervention in ragging cases 24x7.

Step 8: The squad is expected to intervene at the offence site, stop or prevent any ragging, and collect any evidence, including witness testimonies. If the victim called after the incident, or the anti-ragging squad could not arrive on time, the squad will initiate action against the accused.

Step 9: The institute head has to decide, within 24 hours of receiving the call, whether to register an FIR. Independently of any police action, the institute must complete its inquiry and action must be completed in seven days.

Step 10: The central database will also store the status of the action taken. It will also be made available to an NGO nominated by the Centre.

Step 11: Failure by the institute to act can lead to a range of punishments — from disciplinary action against the head, warden or members of the staff to declaring the institute doesn't meet academic standards.Source:.telegraphindia

TOLL FREE ANTI-RAGGING HELPLINE- Started on 20th June 2009.

2009-06-20T04:39:00.001-07:00

TOLL FREE ANTI-RAGGING HELPLINE (24x7)

1800-180-5522 AND

155222 - Under Implementation

Make Use of it. Don't misuse it.

New Delhi: With an aim to curb the increasing incidents of ragging, the University Grants Commission (UGC) is all set to start a national call centre to enable ragging victims from across the country to register their complaints by dialing a toll free number from today.

The helpline is equipped with software that will pass on the complaints directly to the head of the institution within 15 minutes to help the administration protect the victims.

The functioning of the helpline will be monitored by the UGC and it will be operational around the clock.

The Education Consultancy India Ltd (ECIL) has been entrusted with the job of setting up of the national call centre.

A strict guideline was issued by the UGC to set up the national call centre to enable ragging victims from across the country to register their complaints by dialing a toll free number.

The decision has been widely hailed by teachers and students, who feel this, would check the growing instances of ragging.

Further, under the Anti-Ragging Law, offenders can be fined up to Rs.2.5 lakh or rusticated for one to four semesters if found guilty.

The new provisions laid down by the UGC also requires that the parents and students give a written undertaking, at the time of admission, that they are aware of the ragging guidelines and will report any such incident.

"This is a welcome step taken by the UGC. It will boost the confidence of new students," said a fresher.source:indiaedunews.net

Bluej Program to check a String for palindrome

2009-05-27T05:16:00.001-07:00

import java.io.*;
class palindrome
{
public static void main(String arg[])throws IOException
{
DataInputStream in=new DataInputStream(System.in);
int l,i,j,flag=1;
String s;
System.out.println("Enter the string:");
s=in.readLine();
l=s.length();
for(i=0,j=l-1;i<=j;i++,j--)
{
if(s.charAt(i)==s.charAt(j))
{}
else
{
flag=0;
break;
}
}
if(flag==1)
System.out.print(s+" is palindrome");
else
System.out.print(s+" is not palindrome");
}
}

Bluej program to display the product of two given matrices

2009-05-27T05:15:00.001-07:00

public class Matrix
{
public static void main(String args[])
{
int a[][] ={{ 1, 2, 5 },{ 3, 4, 4 },{ 5, 6, 2 },};
int b[][] ={{ 7, 8, 9 },{ 1, 2, 3 },{ 4, 9, 5 },};
int c[][] = mult(a, b);
System.out.println("Result: ");
for (int i = 0; i < c.length; ++i)
{
System.out.print("{");
for (int j = 0; j < c.length; ++j)
{
System.out.print(" "+c[j]);
}
System.out.println("}");
}
}

public static int mult(int a[][], int b[][])[][]
{
int rows = a.length;
int cols = b[0].length;
int c[][] = new int[rows][cols];
for (int i = 0; i <>
{
for (int j = 0; j <>
{
c[j] = 0;
for (int k = 0; k <>
{
c[j]= a[k] * b[k][j]+c[j];
}
}
}
return c;
}
}

Bluej Programm to Checks whether a given NUMBER is palindrome

2009-05-27T05:14:00.001-07:00

import java.io.*;
public class Palindromenumber
{
public static void main(String args[])throws IOException
{
DataInputStream din=new DataInputStream(System.in);
int num,remainder,result=0,n;
System.out.println("Enter the number:");
String s=din.readLine();num = Integer.parseInt(s);
n=num;
while(num>0)
{
remainder = num%10;result = result * 10 + remainder;num = num/10;
}
if(n==result)
{
System.out.println(n+"is palindrome" );
}else
{
System.out.println(n+" is not palindrome" );
}
}
}

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class Prime_number { public static void main(String[] args) { int num = 11; int i; for (i=2; i < num ;i++ ){ int n = num%i; if (n==0){ System.out.println("not Prime!"); break; } } if(i == num){ System.out.println("Prime number!"); } }}

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Bluej Programm to Checks whether a given NUMBER is palindrome

`class` `Prime_number {`
`public static void` `main(String[] args) {`
`int` `num =` `11;`
`int` `i;`
`for` `(i=2; i < num ;i++ ){`
`int` `n = num%i;`
`if` `(n==0){`
`System.out.println("not Prime!");`
`break;`
`}`
`}`
`if(i == num){`
`System.out.println("Prime number!");`
`}`
`}`
`}`