Author:
RainerZimmerman

Language:
C/C++

Upload:
8:38 PM 03/04/2023

Download

Platform
Unix/linux etc.

Difficulty:
2.5

Quality:
4.2

Arch:
x86-64

Description

Pretty simple key checking algorithm, reverse it and generate a key for it - There exists many valid keys.

• Comments
• Solutions

RainerZimmerman on 12:37 AM 03/08/2023: Just realized there is a very easy solution I forgot to remove! If you find it, please ignore it. :)

strawberry_cake on 7:16 PM 04/07/2023: 000-000000000

RedVi on 10:13 AM 06/22/2023: Thx For Fun Challenge 000-000000000

RedVi on 1:12 AM 06/24/2023: nice challenge flag: flag{_y0u_f0und_key_} python script: first = [0x51,0x53,0x4E,0x11,0x50,0x3D,0x51,0x11,0x41,0x50,0x11,0x52,0x3D,0x49,0x11,0x57,0x3D,0x0F,0x11,0x11,0x15] second = [0x37,0x3F,0x2F,0x76,0x2B,0x62,0x28,0x21,0x34,0x0F,0x77,0x62,0x48,0x27,0x75,0x8,0x56,0x6A,0x68,0x4E,0x68] for i in range(0,len(first)): print(chr(first[i] ^ second[i]),end="")

driftxmaster on 7:49 PM 07/31/2023: very nice challenge, it was fun

v0da on 9:04 AM 08/01/2023: there is also another small bug. 000-000999981 is valid where 9+9+9+8+1 = 45 , also 000-999990000 etc

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