Solution

The number of divisors of a natural number \(n\) is given by tau(n) or \(\tau(n)\) or sometimes \(\delta(n)\) as mentioned here already. Every natural number can be expressed as the product of their \(k\) prime factors like this:

\[n = \prod\limits_{i=1}^k p_i^{e_i}\]

Then the number of divisors is given by

\[\tau(n) = \prod\limits_{i=1}^k (e_i+1)\]

So we need a prime factorization and simply need a product with their exponents. By modifying the factorize function from Problem 3, we can calculate the product easily:

function tau(num) {

    var n = num;
    var i = 2;
    var p = 1;

    if (num === 1) return 1;

    while (i * i <= n) {
        var c = 1;
        while (n % i === 0) {
            n /= i;
            c++;
        }
        i++;
        p*= c;
    }

    if (n === num || n > 1)
        p*= 1 + 1;

    return p;
}

With this function, we can formulate the final solution:

function solution(x) {

    var n = 1;
    var d = 1;

    while (tau(d) <= x) {
        n++;
        d+= n;
    }
    return d;
}
solution(500);

One improvement could be to prepare an array of primes to loop over, to not linearly scan for primes.