DozorRage on 10:31 AM 05/02/2025: print(f"{(a:=0x1111):04X}-{(b:=0x2222):04X}-{(48879 - (a ^ b)):04X}")
ultrazvukoff on 7:34 PM 05/03/2025: Great crackme challenge! The logic was clear, and the use of XOR and arithmetic checks made it engaging. Including sub_140001EE0 added a nice layer of complexity. A hint about case sensitivity would improve clarity. Thanks for the fun reverse engineering task!
khoinguyenquang16 on 8:27 AM 05/05/2025: 1234-5678-7AA3
or 0000-0000-BEEF
0xCC on 11:20 AM 05/06/2025: flagOfFirstSegment ^ flagOfSecondSegment + flagOfThirdSegment = 48879. If let any of the three be 0, get a valid licence number would be easier. BEEF-0000-0000 or 0000-BEEF-0000 or 0000-0000-BEEF will all do.