We denote the number of divisors of a number x, as d(x). It can be proven that d(x) = (e₁ + 1)·(e₂ + 1)·...·(e_k + 1) , where x = p₁^e₁·p₂^e₂·...·p_k^e_k is the prime factorization of x.

We notice that it's optimal for p₁ to be as small as it can be, so it should be equal to 2. We notice the same for p₂, so it should be equal to 3. It is optimal to take the smallest possible prime every time. So we should only take the first k primes. But consider the case where e_i = 1 i. We see that if we take more than p_k = 41 (approximately), we are way over our limit of 10¹⁸ for our answer. So our list of primes should be up to 41.

Now we try to construct the desired number by some kind of trial-and-error. As long as we can multiply our current number with the current prime raised to some power e_i, we test for this situation, going to the next prime, knowing the current number's divisors and its value. Notice that every prime should not fit more than times in some number X, and we only have about 15 primes, so operations aren't too much.

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