Finding prime number in a range in javascript

Question

I'm trying to solve this problem where you loop through the numbers between 1 to 60 to find the total prime numbers. Here is what I've come up with:

var totalPrimeNumber = 0;
for(let i=1; i<=60; i++) {
    for(let j= 2; j< i; j++) {
        if(i%j ==0) {
            console.log(i, "is not a prime number");
        }
        console.log(i, "is a prime number");
        totalPrimeNumber+=1;
    }
}

But the output is wrong:

3 is a prime number
4 is not a prime number
4 is a prime number
4 is a prime number
5 is a prime number
5 is a prime number
5 is a prime number
6 is not a prime number
6 is a prime number
6 is not a prime number
6 is a prime number
6 is a prime number
6 is a prime number
7 is a prime number
7 is a prime number
7 is a prime number

Do you know how I can fix this problem?

You need to `break;` out of the loop once you've established that a number is not prime. Also for checking primes, `let j=2; j<=Math.sqrt(i); j++` is sufficient. — Niet the Dark Absol, Jun 30 '21 at 09:53
A better approach is to write a function `isPrime(number)` that returns a boolean and to call it in the loop for each value of `i`. Split your problems into smaller problems and handle them one at a time. Something like this: https://jsfiddle.net/nha613vg/ — , Jun 30 '21 at 10:07
Does this answer your question? [Number prime test in JavaScript](https://stackoverflow.com/questions/40200089/number-prime-test-in-javascript) — ceving, Jun 30 '21 at 10:17

score 2 · Answer 1 · edited Jun 30 '21 at 10:27

You should break the inner loop once you know that the number is not a prime!

You can also improve your code by only checking numbers up to the square-root as divider. E.g. if you check 100 to be a prime, then you only need to check numbers up to 10 since higher numbers will have a lower multiplicator (e.g. you don't need to check 50 since you already checked 2 and 2*50 = 100)

var totalPrimeNumber = 0;
for (let i = 1; i <= 60; i++) {
  let prime = i > 1;
  for (let j = 2; j <= Math.sqrt(i); j++) {
    if (i % j == 0) {
      prime = false;
      break;
    }
  }
  if (prime) {
    console.log(i, "is a prime number");
    totalPrimeNumber += 1;
  } else {
    console.log(i, "is not a prime number");
  }
}

Note that the SO console misses everything below 11 but they are there! — Jamiec, Jun 30 '21 at 10:29

score 1 · Answer 2 · answered Jun 30 '21 at 10:33

You could also try Array.filter to select only those numbers in an Array that are primes, for example:

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

const count = 60;
const numbers = Array.from({ length: count}, (v,k) => k + 1);
const primes = numbers.filter(n => isPrime(n))
console.log("Primes:", primes)

score 0 · Answer 3 · answered Sep 19 '22 at 18:33

If your interest is mainly in the result, and not implementation, then the optimum solution would be via prime-lib library:

import {generatePrimes, stopOnValue} from 'prime-lib';

const i = generatePrimes(); // primes iterator

const arr = [...stopOnValue(i, 60)]; // produce primes up to 60

console.log(arr); //=> [2, 3, 5, 7, 11, 13, ..., 59]

score -1 · Answer 4 · answered Sep 19 '22 at 23:48

-1

function primeFinder(n) {
  // Write your code here
  if(n === 2){
    return true;
  }
  if(n < 2 || isNaN(n) || (n%2===0)){
    return false;
  }
  for(let i = 3; i < n; i++){
    if(n%i === 0){
      return false;
    }
  }
  return true;
}

for(let i = 0; i <= 11; i++){
  if(primeFinder(i) === true){
    console.log(i);
  }
}

console.log(`${primeFinder(11)}`)

// Leave this so we can test your code:
module.exports = primeFinder;

answered Sep 19 '22 at 23:48

Paschalyn Ukwuani

7
2

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Finding prime number in a range in javascript

4 Answers4