## Prime number check in Java

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I'm new to programming and I'm learning Java at present.Could anyone help me on how to check whether a number is prime using for-loop?thanks in advance!

Here is a Simple Programme to Check Whether a Number is Prime or Not ;

import java.util.*; class Example{ public static void main(String args[]){ Scanner input=new Scanner(System.in); int count=0; System.out.print("Input a Number : "); int num=input.nextInt(); for(int i=2;i<num;i++){ if(num%i==0){count++;} } if(num>1&count==0){ System.out.print("\nIt Is a Prime Number."); } else{ System.out.print("\nIt Is a Not Prime Number."); } }

}

**Java program to check prime number,** check if a number is prime. import java.util.*;. import java.lang.*;. class GFG {. //check for number prime or not. static boolean isPrime( int n) {. //check if n is a Prime number in Java: Prime number is a number that is greater than 1 and divided by 1 or itself only. In other words, prime numbers can't be divided by other numbers than itself or 1. For example 2, 3, 5, 7, 11, 13, 17. are the prime numbers.

From this site.

We learned numbers are prime if the only divisors they have are 1 and itself. Trivially, we can check every integer from 1 to itself (exclusive) and test whether it divides evenly.

For example, one might be tempted to run this algorithm:

//checks whether an int is prime or not. boolean isPrime(int n) { for (int i = 2; i < n; i++) { if (n % i == 0) return false; } return true; }

This doesn't seem bad at first, but we can make it faster - much faster. Consider that if 2 divides some integer n, then (n/2) divides n as well. This tells us we don't have to try out all integers from 2 to n. Now we can modify our algorithm:

//checks whether an int is prime or not. boolean isPrime(int n) { for (int i = 2; 2 * i < n; i++) { if (n % i == 0) return false; } return true; }

With some more efficient coding, we notice that you really only have to go up to the square root of n, because if you list out all of the factors of a number, the square root will always be in the middle (if it happens to not be an integer, we're still ok, we just might over-approximate, but our code will still work).

Finally, we know 2 is the "oddest" prime - it happens to be the only even prime number. Because of this, we need only check 2 separately, then traverse odd numbers up to the square root of n. In the end, our code will resemble this:

//checks whether an int is prime or not. boolean isPrime(int n) { //check if n is a multiple of 2 if (n % 2 == 0) return false; //if not, then just check the odds for (int i = 3; i * i <= n; i += 2) { if (n % i == 0) return false; } return true; }

As you can see, we've gone from checking every integer (up to n to find out that a number is prime) to just checking half of the integers up to the square root (the odd ones, really). This is a huge improvement, especially considering when numbers are large.

**Java Program to Check Whether a Number is Prime or Not,** The main method contains a loop to check prime numbers one by one. The main method calls the method CheckPrime to determine whether a A prime number (or a prime) is a natural number greater than 1 that has no positive divisors other than 1 and itself. How to check if a number is a prime number in Java? If we cannot divide a number (greater than 1) by any positive number other than 1 or itself, then the number is a prime number.

I will suggest you to read @Neng Liu's answer from up to bottom and try to understand all the algorithms. However, You can check this code for second algorithm from @Neng Liu's answer.

import java.util.Scanner; public class PrimeNumber{ public static void main(String [] args){ int number,i,rem; boolean flag = true; Scanner input = new Scanner(System.in); System.out.print("Enter an integer to check if its PRIME number : "); number = input.nextInt(); if(number >= 2){ for(i = 2; 2 * i < number; i++){ rem = number % i; if(rem == 0){ flag = false; break; } } if(flag){ System.out.println(number+" is a Prime number"); }else{ System.out.println(number+" is not Prime number"); } }else{ System.out.println("1 & Negative numbers can't be prime ! "); } } }

**Prime Numbers,** Program Logic: We need to divide an input number, say 17 from values 2 to 17 and check the remainder. If remainder is 0 number is not prime Check If a Number Is Prime in Java 1. Introduction. First, let's go over same basic theory. 2. A Custom Implementation. With this approach, we can check if a number between 2 and 3. Using BigInteger. BigInteger class is generally used for storing large sized integers, i.e., 4. Using Apache

**Prime Number From 1 to 100 Program in Java,** However, the basic logic remains the same i.e. you need to check whether the entered number (or already defined in the program) has some A prime number is a number that is only divisible by 1 or itself. For example, 11 is only divisible by 1 or itself. Other Prime numbers 2, 3, 5, 7, 11, 13, 17. Note: 0 and 1 are not prime numbers. 2 is the only even prime number.

**Java Program to Check Prime Number,** The isPrime(int n) method is used to check whether the parameter passed to it is a prime number or not. If the parameter passed is prime, then it 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.

**Prime Number Program in Java,** Write a Java Program to Check Prime Number using For Loop, While Loop, and Functions. Prime Numbers is any natural number that is not divisible by any The main issue here is that you overwrite the value of isPrime in every iteration, so if the last divisor you check does not divide num, you interpret it as a prime. A better approach would be to assume a number is a prime until proven otherwise (i.e., until you find a divisor for it).

##### Comments

- We can help you getting started, but giving away answer won't help you learn.
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