How does this prime number test in Java work?

Question

The code snippet below checks whether a given number is a prime number. Can someone explain to me why this works? This code was on a study guide given to us for a Java exam.

public static void main(String[] args)
{    
    int j = 2;
    int result = 0;
    int number = 0;
    Scanner reader = new Scanner(System.in);
    System.out.println("Please enter a number: ");
    number = reader.nextInt();
    while (j <= number / 2)
    {
        if (number % j == 0)
        {
           result = 1;
        }
        j++;
    }
    if (result == 1)
    {
        System.out.println("Number: " + number + " is Not Prime.");
    }
    else
    {
        System.out.println("Number: " + number + " is Prime. ");
    }
}

Answer 1

Overall theory

The condition if (number % j == 0) asks if number is exactly divisible by j

The definition of a prime is

a number divisible by only itself and 1

so if you test all numbers between 2 and number, and none of them are exactly divisible then it is a prime, otherwise it is not.

Of course you don't actually have to go all way to the number , because number cannot be exactly divisible by anything above half number .

Specific sections

While loop

This section runs through values of increasing j, if we pretend that number = 12 then it will run through j = 2,3,4,5,6

  int j = 2;
  .....
  while (j <= number / 2)
  {
      ........
      j++;
  }

If statement

This section sets result to 1, if at any point number is exactly divisible by j . result is never reset to 0 once it has been set to 1.

  ......
  if (number % j == 0)
  {
     result = 1;
  }
  .....

Further improvements

Of course you can improve that even more because you actually need go no higher than sqrt(number) but this snippet has decided not to do that; the reason you need go no higher is because if (for example) 40 is exactly divisible by 4 it is 4*10, you don't need to test for both 4 and 10. And of those pairs one will always be below sqrt(number) .

It's also worth noting that they appear to have intended to use result as a boolean, but actually used integers 0 and 1 to represent true and false instead. This is not good practice.

Answer 2

I've tried to comment each line to explain the processes going on, hope it helps!

int j = 2;   //variable
int result = 0; //variable
int number = 0; //variable
Scanner reader = new Scanner(System.in); //Scanner object
System.out.println("Please enter a number: "); //Instruction
number = reader.nextInt(); //Get the number entered
while (j <= number / 2) //start loop, during loop j will become each number between 2 and 
{                             //the entered number divided by 2
    if (number % j == 0) //If their is no remainder from your number divided by j...
    {
        result = 1;  //Then result is set to 1 as the number divides equally by another number, hergo
    }                //it is not a prime number
    j++;  //Increment j to the next number to test against the number you entered
}
if (result == 1)  //check the result from the loop
{
    System.out.println("Number: " + number + " is Not Prime."); //If result 1 then a prime   
}
else
{
    System.out.println("Number: " + number + " is Prime. "); //If result is not 1 it's not a prime
}    

Answer 3

It works by iterating over all number between 2 and half of the number entered (since any number greater than the input/2 (but less than the input) would yield a fraction). If the number input divided by j yields a 0 remainder ( if (number % j == 0) ) then the number input is divisible by a number other than 1 or itself. In this case result is set to 1 and the number is not a prime number.

Answer 4

Java java.math.BigInteger class contains a method isProbablePrime(int certainty) to check the primality of a number.

isProbablePrime(int certainty) : A method in BigInteger class to check if a given number is prime. For certainty = 1 , it return true if BigInteger is prime and false if BigInteger is composite.

Miller–Rabin primality algorithm is used to check primality in this method.

import java.math.BigInteger;

public class TestPrime {

    public static void main(String[] args) {
        int number = 83;
        boolean isPrime = testPrime(number);
        System.out.println(number + " is prime : " + isPrime);

    }

    /**
     * method to test primality
     * @param number
     * @return boolean
     */
    private static boolean testPrime(int number) {
        BigInteger bValue = BigInteger.valueOf(number);

        /**
         * isProbablePrime method used to check primality. 
         * */
        boolean result = bValue.isProbablePrime(1);

        return result;
    }
}

Output: 83 is prime : true

For more information, see my blog .

Answer 5

Do try

public class PalindromePrime   {
     private static int g ,k ,n =0,i,m ; 

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

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

        System.out.print(" Please Inter Data : "); 
        g = scanner.nextInt();  

        System.out.print(" Please Inter Data 2  : "); 
        m = scanner.nextInt();

        count(g,m);


        }

//      
        //********************************************************************************    


    private static    int count(int L, int R) 

        for( i= L ; i<= R ;i++){
            int count = 0 ;
            for( n = i ; n >=1 ;n -- ){

                if(i%n==0){

                    count = count + 1 ;
                }           
            }
            if(count == 2)
            {       
                b = b +i + "" ; 
            }   

        }

        System.out.print("  Data  : "); 
        System.out.print("  Data : \n "  +b );  

        return R;

        }
}