Prime number search java

Java program to check prime number

The number which is only divisible by itself and 1 is known as prime number, for example 7 is a prime number because it is only divisible by itself and 1.
This program takes the number (entered by user) and then checks whether the input number is prime or not. The program then displays the result. If you are looking for a program that displays the prime number between two intervals then see: Java program to display prime numbers between 1 to n.

Example: Program to check whether input number is prime or not

To understand this program you should have the knowledge of for loop, if-else statements and break statement.

import java.util.Scanner; class PrimeCheck < public static void main(String args[]) < int temp; boolean isPrime=true; Scanner scan= new Scanner(System.in); System.out.println("Enter any number:"); //capture the input in an integer int num=scan.nextInt(); scan.close(); for(int i=2;i> //If isPrime is true then the number is prime else not if(isPrime) System.out.println(num + " is a Prime Number"); else System.out.println(num + " is not a Prime Number"); > >

Enter any number: 19 19 is a Prime Number

Enter any number: 6 6 is not a Prime Number

You can also use while loop to check the prime number:
Just replace this part of the code in above program:

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Comments

The program fails, when i provided the number ‘ 893 ‘ . It is a prime number but the above program shows that it is not a prime. The following program gives the correct answer. However, The loop runs 8 times, irrespective of prime or not prime. In case of the above program, It exit the loop as soon as it is an even number but for odd numbers, the loop will run more number of times. Please provide me your feedback(pros/cons) on this program. public class PrimeNumber <
public static void main(String args[]) <
PrimeNumber p = new PrimeNumber();
Scanner scan = new Scanner(System.in);
System.out.println(“Enter a number for check in main method:”);
int n = scan.nextInt();
int count = 0;
int number = 0;
for (int i = 2; i if (n % i == 0) <
System.out.println(» count called when » + i);
count = count + 1;
>
number = number + 1;
System.out.println(» The loop called «+number+» times»);
>
if (count == 0) <
System.out.println(n + » is a prime number «);
>
System.out.println(» ================================ «);
p.checkPrime();
>
>

Hi friend try this program and give a feedback please, public class PrimeNumber <
public static void main(String args[]) <
int num=0;
int prime=0;
Scanner sc=new Scanner(System.in);
System.out.println(“Enter the numbe to check”);
num=sc.nextInt();
if(num%2==0 && num>=2)
<
prime=0; >
else
<
prime=1;
>
if(prime==0)
<
System.out.println(” prime number”);
>
else
<
System.out.println(” not prime number”);
>
> >

Umm… actually, 893 is not a prime number mate. It is divisible by 19 at 47 times. So maybe that’s why you’re getting an error.

your example is to complex lets simplify the example of prime public static void main(String args[]) <
int num = new Scanner(System.in).nextInt();
if (num % 2 == 0) <
System.out.println(“not prime”);
> else <
System.out.println(“Prime”);
>
>

This only checks if the number is even, if you enter 2 which is a prime number, it will consider it as not a prime number..

No, I think it does not work if the input number is 1. OP, you need to hardcode to fix the code. if the input is 1.

it is so the program doesn’t allow numbers bellow 4 to enter the for loop. testing the numbers 2 and 3 wouldn’t make sense, because they are prime. However this program allows 1 to be a prime number which is demonstrably false as per the definition of a prime number, so a change to the logic would benefit the people reading this. For example you can get around this by adding the following code:
if(isPrime == 1)
isPrime=false;
>
this code must be placed before the last if structure and after the for loop.

How to identify the prime number count from 1 to 1000 without using the below code. (Required alternative method – should not go one by one and count it) int Count=0;
for(int i=1;i <=1000;i++)
temp=num%i;
if(temp==0)
isPrime=false;
break;
>
>
//If isPrime is true then the number is prime else not
if(isPrime) System.out.println(num + » is Prime Number»);
Count++;
> Thanks in Advance
Degaleesan

for (i=2;i <=num/2;i++)
i just want yo know why this has been used? as per my understanding i=2 means prime has two factors that is 1 and itself.
i dont know about next part..sir can you please explain

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Check If a Number Is Prime in Java

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We’re looking for a new Java technical editor to help review new articles for the site.

1. Introduction

First, let’s go over some basic theory.

Simply put, a number is prime if it’s only divisible by one and by the number itself. The non-prime numbers are called composite numbers. And number one is neither prime nor composite.

In this article, we’ll have a look at different ways to check the primality of a number in Java.

2. A Custom Implementation

With this approach, we can check if a number between 2 and (square root of the number) can accurately divide the number.

The following logic will return true if the number is prime:

public boolean isPrime(int number) < return number >1 && IntStream.rangeClosed(2, (int) Math.sqrt(number)) .noneMatch(n -> (number % n == 0)); >

3. Using BigInteger

BigInteger class is generally used for storing large sized integers, i.e., those greater than 64bits. It provides a few useful APIs for working with int and long values.

One of those APIs is the isProbablePrime. This API returns false if the number is definitely a composite and returns true if there is some probability of it being prime. It is useful when dealing with large integers because it can be quite an intensive computation to verify these fully.

A quick side-note – the isProbablePrime API uses what’s known as “Miller – Rabin and Lucas – Lehmer” primality tests to check if the number is probably prime. In cases where the number is less than 100 bits, only the “Miller – Rabin” test is used, otherwise, both tests are used for checking the primality of a number.

“Miller-Rabin” test iterates a fixed number of times to determine the primality of number and this iteration count is determined by a simple check which involves the bit length of the number and the certainty value passed to the API:

public boolean isPrime(int number)

4. Using Apache Commons Math

Apache Commons Math API provides a method named org.apache.commons.math3.primes.Primes, which we will use for checking the primality of a number.

First, we need to import the Apache Commons Math library by adding the following dependency in our pom.xml:

 org.apache.commons commons-math3 3.6.1 

The latest version of the commons-math3 can be found here.

We could do the check just by calling the method:

5. Conclusion

In this quick write-up, we have seen three ways of checking for the primality of the number.

The code for this can be found in the package com.baeldung.algorithms.primechecker over on Github.

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How to determine a prime number efficiently?

Learn about an efficient algorithm to determine a given number is prime or not. Also learn to implement prime number algorithm in Java 8 program.

A prime number is a natural number greater than 1 that cannot be formed by multiplying two smaller natural numbers other than 1.

In other words, a prime number (P) is a number greater than 1 whose only factors are 1 and the number (P) itself.

For example, 3 is a prime number because it can be divided only by 1 and 3 (itself). However, 4 is NOT prime number because it can be divided by 2 i.e. it can be written as 2 x 2 .

There are infinitely many prime numbers. Another way of saying this is that the sequence ‘2, 3, 5, 7, 11, 13, . ‘ of prime numbers never ends.

2. Algorithm to calculate prime number

Please note that there is no known efficient formula (mathematically proven) to determine a number is prime or not.

Generally, we can determine a number is prime or not in below steps:

1. 2 is only prime number which is also even number. So, if given number N is 2 the it is PRIME number.
2. If given number N is even number then it is NOT PRIME number.
3. Find out square root on N. Traverse all odd numbers up to the sqrt(N) and try to devide the N with current odd number. If remainder is 0 for any odd number then number is NOT PRIME.
4. Else – number is PRIME.

boolean isPrime(int number) < if(number n % 2 != 0) .noneMatch(n -> (number % n == 0)); >

3. Java program to determine a prime number

Let’s implement above prime number algorithm in Java 8. We have used IntStream which helps in generating a sequence of integers supporting sequential and parallel aggregate operations.

package com.howtodoinjava.example; import java.util.stream.IntStream; public class Main < public static void main(String[] args) < System.out.println("2 is prime number :: " + isPrime(2)); System.out.println("3 is prime number :: " + isPrime(3)); System.out.println("4 is prime number :: " + isPrime(4)); System.out.println("5 is prime number :: " + isPrime(5)); System.out.println("6 is prime number :: " + isPrime(6)); System.out.println("7 is prime number :: " + isPrime(7)); System.out.println("8 is prime number :: " + isPrime(8)); System.out.println("9 is prime number :: " + isPrime(9)); System.out.println("10 is prime number :: " + isPrime(10)); System.out.println("11 is prime number :: " + isPrime(11)); >static boolean isPrime(int number) < if(number n % 2 != 0) .noneMatch(n -> (number % n == 0)); > >

2 is prime number :: true 3 is prime number :: true 4 is prime number :: false 5 is prime number :: true 6 is prime number :: false 7 is prime number :: true 8 is prime number :: false 9 is prime number :: false 10 is prime number :: false 11 is prime number :: true

Drop me your questions related to how to determine a given number is prime in Java.

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