Euler Problem 37

http://projecteuler.net/index.php?section=problems&id=37

The number 3797 has an interesting property. Being prime itself, it is possible to continuously remove digits from left to right, and remain prime at each stage: 3797, 797, 97, and 7. Similarly we can work from right to left: 3797, 379, 37, and 3.

Find the sum of the only eleven primes that are both truncatable from left to right and right to left.

NOTE: 2, 3, 5, and 7 are not considered to be truncatable primes.

______

so the first thing I did wrong was to consider 1 to be a prime.. once I’d sorted that out I got this list.. which is 11 primes…  so those must be what they’re after!

23
37
53
73
313
317
373
797
3137
3797
739397

_________________

 

static void Main(string[] args)
       {
           long sum = 0;
           int found = 0;
           long i = 11;
           while (found<11)
           {
              int c=0;
              i++;
              int l = i.ToString().Length;

              for (int x = 0; x < l; x++)
              {

                  long t1 = Convert.ToInt64(i.ToString().Substring(x, l - x));
                  long t2 = Convert.ToInt64(i.ToString().Substring(0, x+1));

                  if (isprime(t1) && isprime(t2) && t1 != 1 && t2 != 1 )
                  {
                      c++;
                  }
                  else
                  {
                      c = 0;
                  }

              }
              
               if (c == i.ToString().Length)
                   {
                       sum += i;
                       found++;
                       Console.WriteLine(i);
                   }

           }

           Console.WriteLine("Answer>>>"+sum);
           Console.ReadLine();

       }

static bool isprime(long n)
       {
           if (n == 2 || n == 3 || n == 5) { return true; }

          if (n%2==0) {return false;}
          if (n%3 == 0) { return false;}
          if (n%5 == 0) { return false;}

          for (int t = 7; t < Math.Sqrt(n); t++)
          {
              if (n % t== 0)
              {
                  return false;
              }
          }

          return true;

       }