JavaScript Challenge - Sum of Prime Factors

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Hi All,

Here is my solution to this challenge:

function sumOfPrimeFactors(n) {
  let factorsArray = [];
  for (let i = 1; i <= n; i++) {
    if (n % i == 0) {
      factorsArray.push(i);
    }
  }
  
  //now we have the factors array
  //we need to filter the prime numbers from the array
  
  let primeFactorsArray = [];
  factorsArray.forEach(factor => {
    let factorCounter = 0;
    for(let i = 1; i <= factor; i++) {
      if (factor % i == 0) {
        factorCounter++;
      }
    }
    if (factorCounter == 2) {
      primeFactorsArray.push(factor);
    }
  });

  //now we have an array of prime factors only
  //we just need to find the sum of them all

  let sum = 0;
  primeFactorsArray.forEach(factor => {
    sum += factor;
  });
  return sum; 
}

console.log(sumOfPrimeFactors(91));


// Leave this so we can test your code:
module.exports = sumOfPrimeFactors;
function sumOfPrimeFactors(n) {
  let sum = 0;
  primeLoop:for(let i = 2; i <= n; ++i){
    if(n % i != 0) continue;
    const s = Math.sqrt(i);
    if(Number.isInteger(s)) continue;
    for(let j = 2; j < s; ++j)
      if(i % j == 0)
        continue primeLoop;
    sum += i;
  }
  return sum;
}

module.exports = sumOfPrimeFactors;
const _ = require('lodash');

function sumOfPrimeFactors(n) {

  function isPrime(num) {
    for (let i = 2; i * i <= num; i++) {
        if (!(num % i)) return false; 
    }
    return num > 1;
  }

  return _.range(2, n + 1)
    .filter(x => isPrime(x) && !(n%x))
    .reduce((total, element) => total + element, 0)
}

console.log(sumOfPrimeFactors(91));
// Leave this so we can test your code:
module.exports = sumOfPrimeFactors;