import java.io.*;

/*
Copyright Mark Campbell, distributed under any GPL license.
*/

public class Problem4
{

 public static void main (String[] args)
 {

  BufferedReader in = new BufferedReader(new InputStreamReader(System.in));

  System.out.print("Enter the number of digits that you wish to calculate to: ");

  long digits = 0L;

  try
  {
   digits = Long.parseLong(in.readLine());
  }
  catch (Exception ex)
  {
   System.out.println("\nNot a valid number. Going with 3 digits.");

   digits = 3L;
  }

  digits = (long)Math.pow(10.0, digits);

  long largest = 0L;
  long count = 0L;
  long largestI = 0L;
  long largestJ = 0L;

  long jCount = 0L;

  //For timing
  long time = System.currentTimeMillis();

  for (long i = digits / 10L; i < digits; i++)
  {
   for (long j = (digits / 10L) + jCount; j < digits; j++)
   {
    if (isPalindrome(Long.toString(i * j)))
     if (((long)(i)) * ((long)(j)) > largest)
     {
      count++;
      largestI = i;
      largestJ = j;
      largest = i * j;
     }
   }

   jCount++;

  }

  time = System.currentTimeMillis() - time;

  System.out.println("\n\nThe largest number was:\t\t\t" + largest);
  System.out.println("Number of 'largest' numbers found:\t" + count);
  System.out.println("The two miraculous numbers found:\t" + largestI + " * " + largestJ);

  System.out.println("\nTime it took for the computation:\t" + time + "ms");

  System.out.print("Press any key to continue... ");

  try
  {
   in.readLine();
  }
  catch (IOException ex)
  {

  }


 }



 public static boolean isPalindrome(String val)
 {
  char[] num = val.toCharArray();
  int length = num.length;

  //Check if it's even
  if (length % 2 == 0)
  {


   for (int i = 0; i <= length / 2; i++)
   {
    if (num[length - i - 1] != num[0 + i])
    {
     return false;
    }
   }

  }
  else //for odd
  {

   for (int i = 0; i < length / 2; i++)
   {
    if (num[length - i - 1] != num[0 + i])
    {
     return false;
    }
   }
  }

  return true;
 }

}