How to find prime numbers between 0 - 100?

Answer 1

This is my solution

//find all prime numbers
function showMePrimeNumbers(start, end){
    var primes = [];
    for(var number = start; number < end; number++){
        var primeNumberDividers = []; //there should only be 2: 1 & number
        for(var divider = 1; divider <= number; divider++){
            if(number % divider === 0){
                primeNumberDividers.push(divider);
            }      
        }
        if(primeNumberDividers.length === 2){
            primes.push(number);
        }
    }
    return primes;
}

console.log(showMePrimeNumbers(1, 100));           

Answer 2

Luchian's answer gives you a link to the standard technique for finding primes.

A less efficient, but simpler approach is to turn your existing code into a nested loop. Observe that you are dividing by 2,3,4,5,6 and so on ... and turn that into a loop.

Given that this is homework, and given that the aim of the homework is to help you learn basic programming, a solution that is simple, correct but somewhat inefficient should be fine.

Answer 3

• Why try deleting by 4 (and 6,8,10,12) if we've already tried deleting by 2 ?
  Why try deleting by 9 if we've already tried deleting by 3 ?
  Why try deleting by 11 if 11 * 11 = 121 which is greater than 100 ?
  Why try deleting any odd number by 2 at all?
  Why try deleting any even number above 2 by anything at all?

Eliminate the dead tests and you'll get yourself a good code, testing for primes below 100.

And your code is very far from being the worst code ever. Many many others would try dividing 100 by 99. But the absolute champion would generate all products of 2..96 with 2..96 to test whether 97 is among them. That one really is astonishingly inefficient.

Sieve of Eratosthenes of course is much better, and you can have one -- under 100 -- with no arrays of booleans (and no divisions too!):

console.log(2)
var m3 = 9, m5 = 25, m7 = 49, i = 3
for( ; i < 100; i += 2 )
{
    if( i != m3 && i != m5 && i != m7) console.log(i)
    else
    {
        if( i == m3 ) m3 += 6
        if( i == m5 ) m5 += 10
        if( i == m7 ) m7 += 14
    }
} "DONE"

---

This is the sieve of Eratosthenes, were we skip over the composites - and that's what this code is doing. The timing of generation of composites and of skipping over them (by checking for equality) is mixed into one timeline. The usual sieve first generates composites and marks them in an array, then sweeps the array. Here the two stages are mashed into one, to avoid having to use any array at all (this only works because we know the top limit's square root - 10 - in advance and use only primes below it, viz. 3,5,7 - with 2's multiples, i.e. evens, implicitly skipped over in advance).

In other words this is an incremental sieve of Eratosthenes and m3, m5, m7 form an implicit priority queue of the multiples of primes 3, 5, and 7.

Answer 4

Sieve of Eratosthenes. its bit look but its simple and it works!

function count_prime(arg) {

    arg = typeof arg !== 'undefined' ? arg : 20; //default value
    var list = [2]
    var list2 = [0,1]
    var real_prime = []

    counter = 2
    while (counter < arg ) {
        if (counter % 2 !== 0) {
            list.push(counter)
        }
        counter++
    }

    for (i = 0; i < list.length - 1; i++) {
        var a = list[i]
        for (j = 0; j < list.length - 1; j++) {
            if (list[j] % a === 0 && list[j] !== a) {
                list[j] = false;       // assign false to non-prime numbers
            }
        }
        if (list[i] !== false) { 
            real_prime.push(list[i]);  // save all prime numbers in new array
        }
    }
 }
window.onload=count_prime(100);

Answer 5

And this famous code from a famous JS Ninja

var isPrime = n => Array(Math.ceil(Math.sqrt(n)+1)).fill().map((e,i)=>i).slice(2).every(m => n%m);

console.log(Array(100).fill().map((e,i)=>i+1).slice(1).filter(isPrime));

Answer 6

Use following function to find out prime numbers :

function primeNumbers() {
    var p
    var n = document.primeForm.primeText.value
    var d
    var x
    var prime
    var displayAll = 2 + " "
    for (p = 3; p <= n; p = p + 2) {
        x = Math.sqrt(p)
        prime = 1
        for (d = 3; prime && (d <= x); d = d + 2)
        if ((p % d) == 0) prime = 0
        else prime = 1
        if (prime == 1) {
            displayAll = displayAll + p + " "
        }
    }
    document.primeForm.primeArea.value = displayAll
}