在间隔[1000,9999]中找到质数的更好解决方案，其中第一和第二的总和

Question

I want to Find prime numbers in interval [1000,9999] where the sum of first and second digits is equal to sum of third and fourth digits of the number.

For example 3517 is prime number and 3 + 5 = 1 + 7

My solution is the following:

for (int i = 1001; i <= 9999; i += 2)
{
     if (NumberOfDivisorsOf(i) == 2)
     {
        string x = i.ToString();
        int n1 = Convert.ToInt32(x[0]); 
        int n2 = Convert.ToInt32(x[1]); 
        int n3 = Convert.ToInt32(x[2]); 
        int n4 = Convert.ToInt32(x[3]); 

         if ((n1 + n2) == (n3 + n4))
              Console.WriteLine(i);
      }
}

NumberOfDivisorsOf method looks like this:

static int NumberOfDivisorsOf(int a)
{
    int count = 2;
    int i = 2;

    for (; i < a/i ; i++)
    {
       if (a / i * i == a)
           count += 2;
    }

    if (i * i == a)
        count++;

    return count;
}

I think NumberOfDivisorsOf method is fine and improvement needs to the solution itself.

I don't want to use LINQ. I want to get them with simple steps..

---

EDIT:

Depending On OMG answer I improve code as shown below:

for (int i = 1001; i <= 9999; i += 2)
{
    if (Isprime(i))
    {
       int n1 = i / 1000;
       int n2 = (i % 1000) / 100;
       int n3 = (i % 100) / 10;
       int n4 = i % 10;

       if ((n1 + n2) == (n3 + n4))
       {
          Console.WriteLine(i);
       } 
    }
}

I changed NumberOfDivisorOf method to IsPrime method:

static bool Isprime(int n)
{
    if (n == 2)
       return true;
    if (n == 3)
       return true;
    if ((n % 2) == 0)
       return false;
    if (n % 3 == 0)
        return false;

    int i = 5;
    int w = 2;

    while (i * i <= n)
    {
        if (n % i == 0)
           return false;

           i += w;
           w = 6 - w;
    }

    return true;
}

---

EDIT:

Depending on Siye answer I changed my code as shown below (it made executions speed 3x faster)

for (int i = 1001; i <= 9999; i += 2)
{
    int n1 = i / 1000;
    int n2 = (i % 1000) / 100;
    int n3 = (i % 100) / 10;
    int n4 = i % 10;

    if ((n1 + n2) == (n3 + n4))
    {
       if (Isprime(i))
       {
            Console.WriteLine(i);
       }
    }
}

Answer 1

To find the prime number efficiently, there exist some good solutions(such as this ). To better your approach to find the sum, it would be better using % or using .ToString() just once in your code as the overhead of it can effect on the performance in high scale.

To use % you can find them by:

int n4 = i % 10;
int n3 = (i % 100) / 10;
int n2 = (i % 1000) / 100;
int n1 = (i % 10000) / 1000;

Also, instead of NumberOfDivisorsOf it would be better to use is_prime by using break in the loop when you found the number is prime.

Answer 2

for (int i = 1001; i <= 9999; i += 2)
{
    int n1 = i.ToString()[0];
    int n2 = i.ToString()[1];
    int n3 = i.ToString()[2];
    int n4 = i.ToString()[3];
    if ((n1 + n2) == (n3 + n4)) {
        if (NumberOfDivisorsOf(i) == 2)
            Console.WriteLine(i);
        i = (i / 10 + 1) * 10;  // A little improvement
        continue;
    }
    if ((n1 + n2) > (n3 + n4)) {
        i = (i / 10 + 1) * 10;
        continue;
    }   
}

This may make a little improvment.

By the way, Sieve of Eratosthenes may improve the performance of method NumberOfDivisorsOf.