One of the most common programming exercise for beginners is, write a program to check if given number is prime or not? There are many methods to check if a number is prime or not, but most common of them is trial division, which is what we will see in this tutorial. In my opinion, these kind of program is their first steps towards algorithmic understanding. You first come with a solution, which is driven by the fact that

Further, you come to know that if its not divisible by 2 then there is no need to checking for any other even number, and you increment counter by 2, instead of 1. So in a way, you learn how to optimize your solution by taking advantage of facts available.

After this you can try something like Fibonacci series, or may be finding factorial of a number in Java to do some more practice on programming. This will not only teach you language basics e.g. loops, control statements like if-else, use of arithmetic and relational operator but also helps to develop programming logic.

By the way, you can even take this problem of

This question is also asked on written test and interview as how to print prime numbers from 1 to 100 or finding prime factor of a number in Java.

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This is our Java program to see if a given integer number is prime or not. As I said, a number is called a prime number if its only divisible by 1 or itself, which means prime number doesn't have any positive divisor other than itself. There are many ways to check if number is prime or not, or generating list of primes. Easiest of them is known as

That's all in this program about

The Coding Interview Bootcamp: Algorithms + Data Structures

Data Structures and Algorithms: Deep Dive Using Java

Algorithms and Data Structures - Part 1 and 2

*prime numbers*are natural numbers, which are not divisible by any positive number other than 1 and themselves. You write a for loop to check every number, starting from 1 to given number, to see if given number is divisible by any positive number or not. This leads you to the solution. Then you find some more fact that there is no need to check till N-1, where N is the number we are checking for primeness, and checking till square root of N is enough. This reduces a lot of time, especially while*checking a large number is prime or not*.Further, you come to know that if its not divisible by 2 then there is no need to checking for any other even number, and you increment counter by 2, instead of 1. So in a way, you learn how to optimize your solution by taking advantage of facts available.

After this you can try something like Fibonacci series, or may be finding factorial of a number in Java to do some more practice on programming. This will not only teach you language basics e.g. loops, control statements like if-else, use of arithmetic and relational operator but also helps to develop programming logic.

By the way, you can even take this problem of

*checking if a number is prime or not,*to next level by try to implement different algorithms for finding primes e.g.**sieve of Atkin**or**sieve of Eratosthenes**. In fact in programming challenges, you often need to build your prime number cache up-to a certain number to progress further in finding solution.This question is also asked on written test and interview as how to print prime numbers from 1 to 100 or finding prime factor of a number in Java.

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__Java Program to Check if an Integer is Prime or Not__

This is our Java program to see if a given integer number is prime or not. As I said, a number is called a prime number if its only divisible by 1 or itself, which means prime number doesn't have any positive divisor other than itself. There are many ways to check if number is prime or not, or generating list of primes. Easiest of them is known as **trial division**, which is a natural way of finding prime. In trial division, you divide . It consists of testing whether n is a multiple of any integer between 2 and sqrt{n}. In this program I have presented three solution or method to check if number is prime. First solution is implementation of trial division, where we are checking from 2 to sqrt(n), we are using java.lang.Math class for calculating square root. Since this function returns double, we need to cast result back into integer. Our second method of**checking if a number is prime or not**is little bit optimized that this as it doesn't check division by even numbers other than two.Third method is most optimized of all three methods of prime number checking. By the way there is another exercise for you to do after this is checking if a number is Armstrong number or not.import java.util.Scanner; /** * Java Program to check if a number is Prime or Not. This program accepts a * number from command prompt and check if it is prime or not. * * @author http://java67.blogspot.com */ public class Testing { public static void main(String args[]) { Scanner scnr = new Scanner(System.in); int number = Integer.MAX_VALUE; System.out.println("Enter number to check if prime or not "); while (number != 0) { number = scnr.nextInt(); System.out.printf("Does %d is prime? %s %s %s %n", number, isPrime(number), isPrimeOrNot(number), isPrimeNumber(number)); } } /* * Java method to check if an integer number is prime or not. * @return true if number is prime, else false */ public static boolean isPrime(int number) { int sqrt = (int) Math.sqrt(number) + 1; for (int i = 2; i < sqrt; i++) { if (number % i == 0) { // number is perfectly divisible - no prime return false; } } return true; } /* * Second version of isPrimeNumber method, with improvement like not * checking for division by even number, if its not divisible by 2. */ public static boolean isPrimeNumber(int number) { if (number == 2 || number == 3) { return true; } if (number % 2 == 0) { return false; } int sqrt = (int) Math.sqrt(number) + 1; for (int i = 3; i < sqrt; i += 2) { if (number % i == 0) { return false; } } return true; } /* * Third way to check if a number is prime or not. */ public static String isPrimeOrNot(int num) { if (num < 0) { return "not valid"; } if (num == 0 || num == 1) { return "not prime"; } if (num == 2 || num == 3) { return "prime number"; } if ((num * num - 1) % 24 == 0) { return "prime"; } else { return "not prime"; } } }OutputEnter number to check if prime or not 2? Does 2 is prime? true prime number true 3? Does 3 is prime? true prime number true 4? Does 4 is prime? false not prime false 5? Does 5 is prime? true prime true 6? Does 6 is prime? false not prime false 7? Does 7 is prime? true prime true 17? Does 17 is prime? true prime true 21? Does 21 is prime? false not prime false 131? Does 131 is prime? true prime true 139? Does 139 is prime? true prime true

That's all in this program about

**how to check if a number is prime or not**. number must be integer, as concept of prime is only for natural numbers and not for floating point numbers. As I said there are couple of more algorithms for checking if a number is prime or not, and some of the algorithm is optimized for finding prime numbers. It imperative to know at-least one fast way of finding prime number,as this trial division is not fast enough for real world problems. I suggest exploring sieve of Atkin and sieve of Eratosthenes way of finding prime numbers.**Further Learning**The Coding Interview Bootcamp: Algorithms + Data Structures

Data Structures and Algorithms: Deep Dive Using Java

Algorithms and Data Structures - Part 1 and 2

Your isPrimeOrNot() method is not a correct way to detemine if a number is prime. (num * num - 1 ) % 24 == 0 is a property that is true for all primes but it can be true for some non primes as well eg num = 25.

ReplyDeleteAlso I think the code would clearer if you didn't add one to sqrt but instead used i <= sqrt in your for() loop.

The code is not valid to check the number is prime or not, which is written in the method isPrimeOrNot(int num).

ReplyDeleteThis method will also print some non-prime number like: 25,35,49...

You are correct the code has bug, it prints 25 has prime number.

DeleteCheck this out.

ReplyDeleteI think its the best way to do without any errors or bugs.

beginners may also like it as it lucid and easy.

int limit = 100;

System.out.println("Prime numbers between 1 and " + limit);

//loop through the numbers one by one

for(int i=1; i < 100; i++){

boolean isPrime = true;

//check to see if the number is prime

for(int j=2; j < i ; j++){

if(i % j == 0){

isPrime = false;

break;

}

}

// print the number

if(isPrime)

System.out.print(i + " ");

}

Ashim Kumar Halder

Android Developer

ashimkr.halder@gmail.com

You should add line if(number < 2) return false; to method isPrimeNumber

ReplyDeleteHi, actually if(number %2==0) will also check number<2 condition.

Delete1%2 equals 0 or -4%2 equals 0.

Thanks

ReplyDeletepublic static boolean isPrime(int number){

ReplyDeleteif(number%2==0){

return false;

}

for(int i=3;i<number;i=i+2){

if(number%i==0){

return false;

}

}

return true;

}

hi I hope this works

ReplyDeleteimport java.util.Scanner;

public class Prime {

public static void main(String[] args) {

Prime p= new Prime();

Scanner sc= new Scanner(System.in);

System.out.println("Enter the number to check its prime or not: ");

int num=sc.nextInt();

System.out.println(p.primenum(num));

}

private boolean primenum(int num)

{

boolean isprime=true;

if(num==2||num==3)

{

return isprime;

}

if(num%2==0)

{

return isprime=false;

}

for(int i=3;i<num;i+=2)

{

if(num%i==0)

{

return isprime=false;

}

}

return isprime;

}

}

Hi, this program checks if a number (+ve integer) entered is prime or not. It prompts the user to enter a +ve integer, checks it and displays if it's a prime number or not.

ReplyDeleteTry and run the program. If there is any question or comment on the program, please drop it here. Thanks

..............................................................................

import java.util.Scanner ;

public class PrimeNumbers

{

public static void main(String[] args)

{

Scanner input = new Scanner(System.in) ;

int counter = 1;

System.out.printf(" \n Please enter a +ve Integer :") ;

int num = input.nextInt() ;

if (num == 1 || num % num == 0)

{

++counter ;

if (num == 1)

{

--counter ;

}

}

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

{

if (( num == i ) ^ (num % i == 0))

{

counter ++;

}

}

if ( counter == 2)

{

System.out.printf("\n %d is a prime number", num) ;

}

else

{

System.out.printf("\n %d is not a prime number", num) ;

}

}

}

Can you explain the logic? Why you are checking counter and what is the use of bitwise ^ operator here?

Deleteimport java.util.Scanner ;

ReplyDeletepublic class PrimeNumbers

{

public static void main(String[] args)

{

int i,count=1;

Scanner input = new Scanner(System.in) ;

System.out.printf(" \n Please enter an Integer :") ;

int num = input.nextInt() ;

for(i=2;i0)

{

System.out.printf("\n %d is a prime number", num) ;

}

else

{

System.out.printf("\n %d is not a prime number", num) ;

}

}

}

public static boolean isPrime(int num) {

ReplyDeleteboolean isPrime = true;

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

if(num%i==0) {

isPrime = false;

break;

}

}

return isPrime;

}

public static boolean isPrime(int num) {

ReplyDeleteboolean isPrime = true;

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

if(num%i==0) {

isPrime = false;

break;

}

}

return isPrime;

}

public class PrimeExample{

ReplyDeletepublic static void main(String args[]){

int i,m=0,flag=0;

int n=3;//it is the number to be checked

m=n/2;

if(n==0||n==1){

System.out.println(n+" is not prime number");

}else{

for(i=2;i<=m;i++){

if(n%i==0){

System.out.println(n+" is not prime number");

flag=1;

break;

}

}

if(flag==0) { System.out.println(n+" is prime number"); }

}//end of else

}

}

public class PrimeExample{

ReplyDeletepublic static void main(String args[]){

int i,m=0,flag=0;

int n=3;//it is the number to be checked

m=n/2;

if(n==0||n==1){

System.out.println(n+" is not prime number");

}else{

for(i=2;i<=m;i++){

if(n%i==0){

System.out.println(n+" is not prime number");

flag=1;

break;

}

}

if(flag==0) { System.out.println(n+" is prime number"); }

}//end of else

}

}