Find the first Triangular number which has 50 factors?

Question

-----Modification of code requested --------

Question : Count the Fast Triangular Series Number which is having 50 Factors ?

Elaborated : Let's say there is a series

   1 : 1
   3 : 1+2
   6 : 1+2+3 
   10 : 1+2+3+4
   15 : 1+2+3+4+5
   21 : 1+2+3+4+5+6
   28 : 1+2+3+4+5+6+7

here 1,3,6,10,15,21,28 are the numbers coming under triangular series.

lets see the factors of the number

    Number factors         Count
    1     : 1               1              
    3     : 1,3             2
    6     : 1,2,3,6         4
    10    : 1,2,5,10        4
    15    : 1,3,5,15        4
    21    : 1,3,7,21        4
    28    : 1,2,4,7,14,28   6

here 6 is the first triangular number which is having 4 factors. even if 10,15,21 also having 4 factors but they are not the 1st one. Like that lets take a number as 2 which is having 2 factors as 1 and 2 same for number 3 also having 2 factors as 1 and 3

but as per question 3 will be the answer not 2 because 2 is not coming under Triangular series number list even if it is faster than 3.

Answer 1

Triangle number #2591 = 3357936 is the first one that has exactly 50 factors: 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 36, 48, 54, 72, 81, 108, 144, 162, 216, 324, 432, 648, 1296, 2591, 5182, 7773, 10364, 15546, 20728, 23319, 31092, 41456, 46638, 62184, 69957, 93276, 124368, 139914, 186552, 209871, 279828, 373104, 419742, 559656, 839484, 1119312, 1678968, 3357936

Triangle number #12375 = 76576500 is the first one that has at least 500 factors (actually 576 factors): 1, 2, 3, 4, 5, 6, 7, 9, 10, 11, ..., 19144125, 25525500, 38288250, 76576500

Triangle number #1569375 = 1231469730000 is the first one that has exactly 500 factors

The solution code itself is very easy, providing you can get divisors:

   public static long Solution(int factorsCount) {
      for (long i = 1; ; ++i) {
        long n = i * (i + 1) / 2;

        IList<long> factors = GetDivisors(n);

        // This code tests if a triangle number has exactly factorsCount factors
        // if you want to find out a triangle number which has at least factorsCount factors
        // change "==" comparison to ">=" one:
        // if (factors.Count >= factorsCount)  
        if (factors.Count == factorsCount) 
          return n;
      }
    }

  ...

  long solution = Solution(50);

If you haven't got a routine to get number's factors, you can use this one:

// Get prime divisors 
private static IList<long> CoreGetPrimeDivisors(long value, IList<int> primes) {
  List<long> results = new List<long>();

  int v = 0;
  long threshould = (long) (Math.Sqrt(value) + 1);

  for (int i = 0; i < primes.Count; ++i) {
    v = primes[i];

    if (v > threshould)
      break;

    if ((value % v) != 0)
      continue;

    while ((value % v) == 0) {
      value = value / v;

      results.Add(v);
    }

    threshould = (long) (Math.Sqrt(value) + 1);
  }

  if (value > 1)
    results.Add(value);

  return results;
}

/// <summary>
/// Get prime divisors 
/// </summary>
public static IList<long> GetPrimeDivisors(long value, IList<int> primes) {
  if (!Object.ReferenceEquals(null, primes))
    return CoreGetPrimeDivisors(value, primes);

  List<long> results = new List<long>();

  while ((value % 2) == 0) {
    results.Add(2);

    value = value / 2;
  }

  while ((value % 3) == 0) {
    results.Add(3);

    value = value / 3;
  }

  while ((value % 5) == 0) {
    results.Add(5);

    value = value / 5;
  }

  while ((value % 7) == 0) {
    results.Add(7);

    value = value / 7;
  }

  int v = 0;
  long n = (long) (Math.Sqrt(value) / 6.0 + 1);
  long threshould = (long) (Math.Sqrt(value) + 1);

  for (int i = 2; i <= n; ++i) {
    v = 6 * i - 1;

    if ((value % v) == 0) {
      while ((value % v) == 0) {
        results.Add(v);

        value = value / v;
      }

      threshould = (long) (Math.Sqrt(value) + 1);
    }

    v = 6 * i + 1;

    if ((value % v) == 0) {
      while ((value % v) == 0) {
        results.Add(v);

        value = value / v;
      }

      threshould = (long) (Math.Sqrt(value) + 1);
    }

    if (v > threshould)
      break;
  }

  if (value > 1) {
    if (results.Count <= 0)
      results.Add(value);
    else if (value != results[results.Count - 1])
      results.Add(value);
  }

  return results;
}

/// <summary>
/// Get all divisors
/// </summary>
public static IList<long> GetDivisors(long value, IList<int> primes) {
  HashSet<long> hs = new HashSet<long>();

  IList<long> divisors = GetPrimeDivisors(value, primes);

  ulong n = (ulong) 1;
  n = n << divisors.Count;

  for (ulong i = 1; i < n; ++i) {
    ulong v = i;
    long p = 1;

    for (int j = 0; j < divisors.Count; ++j) {
      if ((v % 2) != 0)
        p *= divisors[j];

      v = v / 2;
    }

    hs.Add(p);
  }

  List<long> result = new List<long>();

  result.Add(1);

  var en = hs.GetEnumerator();

  while (en.MoveNext())
    result.Add(en.Current);

  result.Sort();

  return result;
}

/// <summary>
/// Get all divisors
/// </summary>
public static IList<long> GetDivisors(long value) {
  return GetDivisors(value, null);
}