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.
static void Main(string[] args) { DateTime dt1 = DateTime.Now; int truncPrimeCount = 0; int truncSum = 0; for (int j = 1; j < 1000000; j++) { if (isPrime(j)) { string specialPrimeNumer = "" + j; bool primedelims = true; while (primedelims == true) { for (int i = 1; i < specialPrimeNumer.Length; i++) { int prime_chck = int.Parse(specialPrimeNumer.Remove(specialPrimeNumer.Length - i)); if (!isPrime(prime_chck)) primedelims = false; } for (int i = 1; i < specialPrimeNumer.Length; i++) { int prime_chck = int.Parse(specialPrimeNumer.Remove(0, i)); if (!isPrime(prime_chck)) primedelims = false; } if (primedelims == true) { truncPrimeCount++; Console.WriteLine("Trunc prime # : " + specialPrimeNumer); if (truncPrimeCount > 4) truncSum += int.Parse(specialPrimeNumer); if (truncPrimeCount == 15) j = 1000001; primedelims = false; } } } } Console.WriteLine(" Sum :" +truncSum); DateTime dt2 = DateTime.Now; Console.WriteLine("Time : " +(dt2 - dt1)); } public static bool isPrime(int n) { if (n == 1) return false; if (n == 2) return true; for (int i = 2; i < n; ++i) { if ((n % i) == 0) return false; } return true; } |