Primes

All p prime numbers with more than 2 digits ends in 1, 3, 7 or 9 and as such, our six digits number can only have one or more of those digits. So, if you start with 999999 which obviously is not a prime, you after some testing, will eventually end up with 999331 which is our answer.

These primes can be found here: https://oeis.org/A068652 (Cyclic permutation primes).

Here's my solution with C. Though the code is a bit long but it was fast to execute as I only checked the cases in which all the digits were odd

#include <stdio.h>
#include <stdbool.h>
bool isprime(long long int x);
int main(void)
{
  long long int a, b;
  int ar[6];
  for (a = 100000; a < 1000000; a++)
  {
    b = a;
    for (int i = 0; i < 6; i++)
    {
      ar[i] = b % 10;
      b /= 10;
    }

    for (int z = 0; z < 1; z++) 
    {
      if ((ar[0] % 2) == 0 || (ar[1] % 2) == 0 || (ar[2] % 2) == 0 || (ar[3] % 2) == 0 || (ar[4] % 2) == 0 || (ar[5] % 2) == 0)
    {
      break;
    }

    long long int kar[6];
    kar[0] = a;
    kar[1] = 100000 * ar[0] + 10000 * ar[5] + 1000 * ar[4] + 100 * ar[3] + 10 * ar[2] + ar[1];
    kar[2] = 100000 * ar[1] + 10000 * ar[0] + 1000 * ar[5] + 100 * ar[4] + 10 * ar[3] + ar[2];
    kar[3] = 100000 * ar[2] + 10000 * ar[1] + 1000 * ar[0] + 100 * ar[5] + 10 * ar[4] + ar[3];
    kar[4] = 100000 * ar[3] + 10000 * ar[2] + 1000 * ar[1] + 100 * ar[0] + 10 * ar[5] + ar[4];
    kar[5] = 100000 * ar[4] + 10000 * ar[3] + 1000 * ar[2] + 100 * ar[1] + 10 * ar[0] + ar[5];
    if (isprime(kar[0]) && isprime(kar[1]) && isprime(kar[2]) && isprime(kar[3]) && isprime(kar[4]) && isprime(kar[5]))
    {
      printf("%lld\n", a);
    }
    }
  }
}
bool isprime(long long int x)
{
  long long int i;
      for (i = 2; i < x; i++)
      {
        if ((x % i) == 0)
        {
          break;
        }
      }
      if (i == x)
      {
        return true;
      }
      else
      {
        return false;
      }
}

I'm beginner in Python , my code take a quite long time , but the result is correct .

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from collections import deque

def isPrime(number):
    if (number<1):
        return False
    else:
        for i in range(2,int(math.sqrt(number))+1):
            if (number % i)==0:
                return False
    return True

problem_list=[]
for i in range(100000,1000000):
    if  (isPrime(i)):
        problem_list.append(i)

def checkPrime(number):
    count=0
    a=deque(str(number))
    for i in range(0,len(a)):
        tmp=int("".join(a))

        if (isPrime(tmp)):
            count=count+1
        else:
            return 0
        a.rotate(1)
    return    count

result_list=[]
for i in problem_list:
    if (checkPrime(i)!=0):
        result_list.append(i)

print(max(result_list))

I'm just a beginner at Java Programming, so the code is very very messy I believe, but here it is:

package javaapplication6;

public class JavaApplication6 {

private static boolean isPrime(Integer num) {

    if (num < 2) return false;
    if (num == 2) return true;
    if (num % 2 == 0) return false;
    for (int i = 3; i * i <= num; i += 2)
        if (num % i == 0) return false;
    return true;

}

public static void main(String[] args) {

  Integer temp;
  int countPrime=0;
  int flag=1;
  Integer j;
  Integer count;
  for(count = 100001;count<1000000;count+=2)
  {

      if(isPrime(count)!=true)
         continue;

      String str = count.toString();
      for(j=0;j<6;j++)
         {
            if(!(str.charAt(j)=='1'||str.charAt(j)=='3'||str.charAt(j)=='7'||str.charAt(j)=='9'))
               flag = 0;              
         }

     String newString = str;
      for(j=0;j<6;j++)
    {
        if(flag==0)
          break;

      newString = newString.substring(1,6)+newString.charAt(0);
      temp=Integer.parseInt(newString);

      if(isPrime(temp)==true)
          countPrime++;   
      else
          break;

    } 
   if(countPrime==6){
         System.out.println(count);
     }

countPrime= 0;      
 flag=1;  }

} }