How to find prime numbers between 0 - 100?

Answer 1

Here is a way to test if number is prime number.

function isPrime(numb){
  if (numb % 2 == 0) return false;
  for (var i=3; i<= Math.sqrt(numb); i = i + 2) {
    if (numb % i == 0) {
     return false;
    }
  }
  return true;
}

Answer 2

<code>
<script language="javascript">
   var n=prompt("Enter User Value")
     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)
             {
                alert(n +" "+" is prime");
             }
             else
             {
                alert(n +" "+" is not prime");
             }


          </script>

Answer 3

Here is the live demo of this script: http://jsfiddle.net/K2QJp/

First, make a function that will test if a single number is prime or not. If you want to extend the Number object you may, but I decided to just keep the code as simple as possible.

function isPrime(num) {
    if(num < 2) return false;
    for (var i = 2; i < num; i++) {
        if(num%i==0)
            return false;
    }
    return true;
}

This script goes through every number between 2 and 1 less than the number and tests if there is any number in which there is no remainder if you divide the number by the increment. If there is any without a remainder, it is not prime. If the number is less than 2, it is not prime. Otherwise, it is prime.

Then make a for loop to loop through the numbers 0 to 100 and test each number with that function. If it is prime, output the number to the log.

for(var i = 0; i < 100; i++){
    if(isPrime(i)) console.log(i);
}

Answer 4

A list built using the new features of ES6, especially with generator. Go to https://codepen.io/arius/pen/wqmzGp made in Catalan language for classes with my students. I hope you find it useful.

function* Primer(max) { 
  const infinite = !max && max !== 0;
  const re = /^.?$|^(..+?)\1+$/; 
  let current = 1;
 
  while (infinite || max-- ) {
      if(!re.test('1'.repeat(current)) == true) yield current;
      current++
  };
};


let [...list] = Primer(100); 
console.log(list);

Answer 5

A number is a prime if it is not divisible by other primes lower than the number in question.

So this builds up a primes array. Tests each new odd candidate n for division against existing found primes lower than n. As an optimization it does not consider even numbers and prepends 2 as a final step.

var primes = [];
for(var n=3;n<=100;n+=2) {
  if(primes.every(function(prime){return n%prime!=0})) {
    primes.push(n);
  }
}
primes.unshift(2);

Answer 6

            var n=100;
            var counter = 0;
            var primeNumbers = "Prime Numbers: ";
            for(var i=2; i<=n; ++i)
            {
                counter=0;
                for(var j=2; j<=n; ++j)
                {
                    if(i>=j && i%j == 0)
                    {
                        ++counter;
                    }

                }
                if(counter == 1)
                    {
                        primeNumbers = primeNumbers + i + "  ";
                    }

            }
            console.log(primeNumbers);