Question

How to generate a code made of 4 hex digits?
the code meets the following conditions:
-first and third digits are the same
-second digit is even
-third digit is larger than 9
-fourth digit is random (has no conditions)

Answer 1

#include <stdio.h>
#include <stdlib.h>
#include <time.h>

int main()
{
  int code1, code2, code4;
  srand(time(NULL)); // set random seed
  code1 = random()%6+10;   // random between 10 and 15
  code2 = random()%8*2;    // even
  code4 = random()%16;     // random between 0-15
  printf("%X%X%X%X\n",code1, code2, code1, code4);
}

There are a few things to note here. The first is that I generate 3 ‘codes’, since the third is always the same as the first.

To generate a random number from 0-15, use random()%16 I am using type int to store these, it may not matter. When printing, I use %X and not the more common %d, as %X will print it in hex. I print four hex digits (reusing code1 of course).
The final part is to meet your conditions.
code1 is a random number in the range 0-5 and adds 10 to it (making it greater than 9)
code2 is random()%8*2 which generates a random number in the range 0-7 and then doubles it (making it even)
code4 is just a random number from 0-15

You did not explain what to do with the generated code, so I just printed it. You could use the sprintf function if you need to print it into a string for further work.