My code

… was written in C++ and can be compiled with G++, Clang++, Visual C++. You can download it, too. Or just jump to my GitHub repository.

       
#include <iostream>
       
#include <vector>
       
 
       
int main()
       
{
       
  // find the smallest number with at least 1000 divisors
       
  // (due to Hackerrank's input range)
       
  const unsigned int MaxDivisors = 1000;
       
 
       
  // store [divisors] => [smallest number]
       
  std::vector<unsigned int> smallest;
       
  smallest.push_back(0); // 0 => no divisors
       
 
       
  // for index=1 we have triangle=1
       
  // for index=2 we have triangle=3
       
  // for index=3 we have triangle=6
       
  // ...
       
  // for index=7 we have triangle=28
       
  // ...
       
  unsigned int index    = 0;
       
  unsigned int triangle = 0; // same as index*(index+1)/2
       
  while (smallest.size() < MaxDivisors)
       
  {
       
    // next triangle number
       
    index++;
       
    triangle += index;
       
 
       
    // performance tweak (5x faster):
       
    // I observed that the "best" numbers with more than 300 divisors end with a zero
       
    // that's something I cannot prove right now, I just "saw" that debugging my code
       
    if (smallest.size() > 300 && triangle % 10 != 0)
       
      continue;
       
 
       
    // find all divisors i where i*j=triangle
       
    // it's much faster to assume i < j, which means i*i < triangle
       
    // whenever we find i then there is a j, too
       
    unsigned int divisors = 0;
       
    unsigned int i        = 1;
       
    while (i*i < triangle)
       
    {
       
      // divisible ? yes, we found i and j, that's two divisors
       
      if (triangle % i == 0)
       
        divisors += 2;
       
      i++;
       
    }
       
    // if i=j then i^2=triangle and we have another divisor
       
    if (i*i == triangle)
       
      divisors++;
       
 
       
    // fill gaps:
       
    // e.g. 10 is the smallest number with 4 divisors
       
    //      28 is the smallest number with 6 divisors
       
    // there is no number between 10 and 28 with 5 divisors
       
    // therefore 28 is the smallest number with AT LEAST 5 divisors, too
       
    while (smallest.size() <= divisors)
       
      smallest.push_back(triangle);
       
  }
       
 
       
  unsigned int tests;
       
  std::cin >> tests;
       
  while (tests--)
       
  {
       
    unsigned int minDivisors;
       
    std::cin >> minDivisors;
       
 
       
    // problem setting asks for "over" x divisors => "plus one"
       
    std::cout << smallest[minDivisors + 1] << std::endl;
       
  }
       
 
       
  return 0;
       
}

This solution contains 9 empty lines, 24 comments and 2 preprocessor commands.