Revised Multi-Factor Authentication

Challenge description:

Okay, the MFA should have been fixed this time…

This is the follow up to Multi-Factor Authentication, so let’s see how much better it got.

❯ ./main
Enter value: 1000

Hm, not even an Incorrect! this time. It just exits. Let’s see what Ghidra thinks, and following entry like last time gets us to main.

undefined8 main(void)

{
  long in_FS_OFFSET;
  undefined4 value;
  long local_10;
  
  local_10 = *(long *)(in_FS_OFFSET + 0x28);
  printf("Enter value: ");
  __isoc23_scanf(&DAT_00102012,&value);
  check_value(value);
  if (local_10 != *(long *)(in_FS_OFFSET + 0x28)) {
                    /* WARNING: Subroutine does not return */
    __stack_chk_fail();
  }
  return 0;
}

So it reads a value, and then hands it off to a function I’ve named check_value. Let’s look at that.

void check_value(uint value)

{
  uint countdown;
  int i;
  int j;
  int k;
  int l;
  
  if (-1 < (int)value) {
    if ((int)value < 0x3e9) {
      countdown = value;
      if ((int)value < 0x1f5) {
        if (200 < (int)value) {
          for (i = 0; i < 10; i = i + 1) {
            for (j = 0; j < 5; j = j + 1) {
              if (i == j) {
                if (value == 0x539) {
                  win(0x539);
                }
                else if ((value & 1) == 0) {
                  FUN_001012b0(value);
                }
                for (k = 0; k < 3; k = k + 1) {
                  for (l = 0; l < 2; l = l + 1) {
                    if (k == l) {
                      FUN_00101264(value);
                    }
                  }
                }
              }
            }
          }
        }
      }
      else {
        for (; 0 < (int)countdown; countdown = countdown - 1) {
          if (countdown == 9999) {
            FUN_00101246(9999);
          }
        }
      }
    }
    else {
      if ((((int)value % 7 == 0) && ((int)value % 0xbf == 0)) && (value == 0x539)) {
        win(0x539);
      }
      if ((int)value % 5 == 0) {
        FUN_00101293(value);
      }
      if ((int)value % 7 == 0) {
        FUN_001012ed(value);
      }
      else if ((int)value % 0xb == 0) {
        FUN_00101201(value);
      }
    }
  }
  return;
}

Well that’s a mess. There’s also about 30 tiny functions in this binary that get called from here, or don’t get called at all. Most of them look like this:

int FUN_00101389(int param_1)

{
  return param_1 + -0x1e;
}

They do a little bit of math, and then nothing happens with the result, so we can ignore them. Rather than go through this from the top, let’s work backwards. We’re looking for whatever decrypts the flag, and only one of these functions stands out from the rest, so I named it win.

undefined8 win(void)

{
  decrypt_flag();
  return 1;
}

Instead of doing math, it calls a function I named decrypt_flag, which is the big function that has all the encrypted data in it. So that’s our target. It gets called in two places in that mess above, so let’s check both.

The Fake Path

The first spot is deep in those nested loops:

if (value == 0x539) {
  win(0x539);
}

So the password needs to be 0x539, or 1337, because of course it is. But look at how we’d get there. To reach this code the value has to be less than 0x1f5, which is 501. There’s no way for 1337 to be less than 501, so this code can never run. All those loops are just there to look scary.

The Real Path

So let’s look at the other spot:

if ((((int)value % 7 == 0) && ((int)value % 0xbf == 0)) && (value == 0x539)) {
  win(0x539);
}

This one is in the else of the check for (int)value < 0x3e9, so our value needs to be at least 1001. 1337 is good there. Then it has three more checks:

• It has to be divisible by 7
• It has to be divisible by 0xbf, which is 191
• It has to be 0x539

The last one is all that really matters, and the other two are there to make you second guess yourself. They’re true anyways, since 7 * 191 = 1337.

❯ ./main
Enter value: 1337
STOUTCTF{bJgmJ8ID106VVesrL38BXR34PXOdBOB0}

FLAG: STOUTCTF{bJgmJ8ID106VVesrL38BXR34PXOdBOB0}