How to find prime numbers between 0 - 100?

Answer 1

Here's the very simple way to calculate primes between a given range(1 to limit).

Simple Solution:

    public static void getAllPrimeNumbers(int limit) {

        System.out.println("Printing prime number from 1 to " + limit);

        for(int number=2; number<=limit; number++){
            //***print all prime numbers upto limit***
            if(isPrime(number)){
                System.out.println(number);
            }
        }
    }


    public static boolean isPrime(int num) {
        if (num == 0 || num == 1) {
            return false;
        }
        if (num == 2) { 
            return true;
        }

        for (int i = 2; i <= num / 2; i++) {
            if (num % i == 0) {
                return false;
            }
        }
        return true;
    }

Answer 2

I modified Rinto's answer just for those who don't want to use the prompt method and just want to see the program print prime numbers . its working

for (n = 0; n < 100; n++) {
    var x = 1;
    if (n == 0 || n == 1) x = 0;
    for (i = 2; i < n; i++) {
        if (n % i == 0) {
            x = 0;
            break;
        }
    }
    if (x == 1) {
        // if prime print the numbers 
        document.write(n);
    } else {
        // if not prime print the number do nothing 
    }
}

Answer 3

Whatever the language, one of the best and most accessible ways of finding primes within a range is using a sieve.

Not going to give you code, but this is a good starting point.

For a small range, such as yours, the most efficient would be pre-computing the numbers.

Answer 4

check the number is prime or not with JS function

function isPrime(num)
        {
            var flag = true;
            for(var i=2; i<=Math.ceil(num/2); i++)
            {
                if((num%i)==0)
                {
                    flag = false;
                    break;
                }
            }
            return flag;    
        }

Answer 5

Here's an example of a sieve implementation in JavaScript:

function getPrimes(max) {
    var sieve = [], i, j, primes = [];
    for (i = 2; i <= max; ++i) {
        if (!sieve[i]) {
            // i has not been marked -- it is prime
            primes.push(i);
            for (j = i << 1; j <= max; j += i) {
                sieve[j] = true;
            }
        }
    }
    return primes;
}

Then getPrimes(100) will return an array of all primes between 2 and 100 (inclusive). Of course, due to memory constraints, you can't use this with large arguments.

A Java implementation would look very similar.

Answer 6

Here's how I solved it. Rewrote it from Java to JavaScript, so excuse me if there's a syntax error.

function isPrime (n)
{
    if (n < 2) return false;

    /**
     * An integer is prime if it is not divisible by any prime less than or equal to its square root
     **/

    var q = Math.floor(Math.sqrt(n));

    for (var i = 2; i <= q; i++)
    {
        if (n % i == 0)
        {
            return false;
        }
    }

    return true;
}

A number, n, is a prime if it isn't divisible by any other number other than by 1 and itself. Also, it's sufficient to check the numbers [2, sqrt(n)].