Crazy palindrome

Let the number be $x_1x_2x_3...x_{n}x_{n}...x_3x_2x_1$

This number is divisible by 11 because the sum of alternating digits is the same.

Also, the number is prime which implies that the number must be $\boxed{11}$

The divisibility rule of 11 is that the sum of alternating digits, starting with the first, minus the sum of the other digits, gives a result that’s divisible by 11.

Let’s consider a palindrome $abba$ . It’s divisible by 11 because $a-b+b-a=0$ . 0 is divisible by 11.

The same applies to all other crazy palindromes, EXCEPT for 11 itself, which is prime.

Therefore, the answer is $\boxed{11}$ .

Let's consider a crazy palindrome of 4 digits $abba$

$abba=1001a+110b=11(91a+20b)$

So,it isn't prime as it is divisible by 11

6 digits : $abccba$

$abccba=100001a+1001b+11c=11(991a+91b+c)$

It is also not prime

Same holds true for other even numbers greater than 2 because of divisibility rule of 11 .

2 digits : $aa$

$aa=11a$ , when $a=1$ we get $11$ which is prime

So $\boxed{m=11}$

Surprisingly, it is the only crazy prime in any base