Lacks's Ultimate Boss Level

Author:
Lacks

Language:
C/C++

Upload:
2026-07-10 20:41

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Platform:
Windows

Difficulty:
3.0

Quality:
3.0

Arch:
x86-64

Downloads:
257

Size:
29.50 KB

Writeups:
4

Comments:
4

Description

The 2nd Last ultimate crack me im uploading

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Comments (4)
Writeups (4)

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nke on 2026-07-11 06:47: quick solve, always fun though. thanks for posting! ^.^

karity on 2026-07-11 11:57: [Click to reveal]

kotsuge on 2026-07-23 19:29: [Click to reveal]

ludzof on 2026-07-29 00:14: What an absolutely incredible finish to this series! The author threw everything at us for this final crackme. Here is a breakdown of all the protections and how the algorithm works: 1. The Distractions & Anti-Debugging Maze If you look for strings, you'll only find standard C++ exception messages (like "bad allocation") left behind as a distraction. The program uses std::cin and std::cout and dynamically decrypts its strings on the fly. Before you even get to the password check, you have to survive a gauntlet of anti-debugging traps: IsDebuggerPresent & CheckRemoteDebuggerPresent GetThreadContext: Checks your CPU's debug registers to see if you placed any Hardware Breakpoints. Memory Check: Checks if the first byte of the password function is 0xCC, meaning it detects if you placed a software breakpoint (F2) on it! GetTickCount64 Timing Check: Measures how long the password check takes. If it takes longer than 500ms, it assumes you are single-stepping in a debugger and fails you! 2. Control Flow Flattening The actual license check is hidden inside a function that uses a technique called Control Flow Flattening. The code is chopped up into blocks and scrambled using a giant state machine (jmp rcx), making the assembly look like a complete mess. 3. The Algorithm Once you piece the flattened blocks back together, you'll find the author used advanced Intel SIMD/Vector instructions (paddd xmm) to add up the ASCII characters of your Username 16 bytes at a time! The math boils down to this: Part 1 must exactly equal Username_Sum * 31 (hex 1F) Part 2 must exactly equal (Part 1 XOR 0x1337) + 123 (hex 7B) 4. The Final Troll If you write a Keygen for the above math, the program unlocks, but the success message prints as scrambled garbage (┤îö├Üîû├...)! Why? Because the decryption key for the success message is mathematically tied to your Part 2! Any valid pair of numbers bypasses the check, but only a Username whose ASCII characters sum to exactly 246, 502, or 758 generates the correct key to properly decrypt the final flag. Valid License: Username: Zackm (Sum = 502) Part 1: 15562 Part 2: 12408 Decrypted Flag: Wow you did it! This was a 10/10 crackme series. Thanks for the amazing challenge!

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Solution by BigLongs on 2026-07-11 05:12:
Easy

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Solution by nke on 2026-07-11 06:45:
comprehensive writeup (radare2)

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Solution by 0x3xp on 2026-07-11 20:30:
Lacks's Ultimate Boss Level - Final: Crackme Writeup

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Solution by MoriartyPuth on 2026-07-14 18:32:

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crackmes.one

crackmes.one

Lacks's Ultimate Boss Level - Final