DozorRage on 2025-05-02 10:29:
[Click to reveal]0000-0000-BEEF
DozorRage on 2025-05-02 10:31:
print(f"{(a:=0x1111):04X}-{(b:=0x2222):04X}-{(48879 - (a ^ b)):04X}")
ultrazvukoff on 2025-05-03 19:34:
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 2025-05-05 08:27:
[Click to reveal]1234-5678-7AA3
or 0000-0000-BEEF
0xCC on 2025-05-06 11:20:
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.
5d79 on 2025-07-31 22:09:
This crackme was really fun but took me a moment to figure out the license check but that made it even better. Enjoyed solving it and wrote a simple keygen