mathematics - Madam I m Adam..please don’t get mad..you will no longer be prime

Take away $ \overline{MAD} $ from $ \overline{MADMADAM} $ to make her prime. $ \overline{ADAM} $ is prime too if he does not get mad: $ \overline{MADADAM}. $

$ M, A, D $ are 3 distinct digits that are to be determined by you so that the statements in this puzzle are true.

Answer

So, we have two prime numbers (MADMADAM-MAD, ADAM).

Consequently, from a more mathematical standpoint,

$M*10,010,001 + A* 1,001,010 + D*100,100 - (M*100 + A*10 + D)$ is prime, as is $A*1,010 + D*100 + M$.

I ran this through a little program that I wrote, and I discovered that there are three distinct solution sets. They are:

1. 
   

   
   

   M: 1 A: 5 D: 8 --> $15,815,851 - 158 = 15,815,693‬$, which is prime. $5851$ is also prime.

   
   

   
   


2. 
   

   
   
   

   M: 3 A: 1 D: 6 --> $31,631,613 - 316 = 31,631,297‬$, which is prime. $1613$ is also prime.

   
   

   
   


3. 
   

   
   

   M: 3 A: 5 D: 6 --> $35,635,653 - 356 = 35,635,297$, which is prime. $5653$ is also prime.

   
   

   
   

If you're interested, here's the Java program that I wrote to find the answer. It probably isn't super efficient, but it works.

public class Test {

  public static void main(String[] args) {
      int big = 0;
      for(int i = 1; i<=9; i++) {
          for(int j = 1; j<=9; j++) {
              for(int k = 0; k<=9; k++) { //k (or d) is the only digit which could be 0.
                  if(isPrime(i*10010001 + j* 1001010 + k*100100 - (i*100 + j*10 + k))&&

                          isPrime(j*1010 + k*100 + i)) {
                      big = i*10010001 + j* 1001010 + k*100100 - (i*100 + j*10 + k);
                      System.out.println("M: " + i + " A: " + j + " D: " + k);
                      System.out.println(""+i+j+k+i+j+k+j+i+"-"+ i+j+k+"="+big+
                              ", which is prime. "+j+k+j+i+" is also prime.\n");
                  }
              }
          }
      }
  }

  public static boolean isPrime(int i) {
      boolean prime = true;
      for(int j = 2; j<=i/2; j++) {
          if (i%j==0) {
              prime = false;
              break;
          }
      }
      return prime;
  }

}