Author:
pourmeadrinkwhileimfloating

Language:
C/C++

Upload:
12:04 AM 03/17/2022

Download

Platform
Unix/linux etc.

Difficulty:
1.5

Quality:
3.4

Arch:
x86-64

Description

easily cracked by peeking at in Ghidra and analyzing the code : since count=0x32 is the password (total sum of digits in pass must == 0x32 (50 decimal) running the program with ./license_checker_3 999995 solves the puzzle :) void main(int argc,char **argv) { int integer; size_t strlen; long offset; char c; int count; int i; undefined8 off_plus_28; off_plus_28 = *(undefined8 *)(offset + 0x28); if (argc != 2) { printf("Usage : %s \n",*argv); // WARNING: Subroutine does not return exit(0); } count = 0; i = 0; while( true ) { strlen = ::strlen(argv[1]); if ((int)strlen

• Comments
• Solutions

legolas on 1:24 PM 05/01/2022: Disassemble and edit as the following 0x0000121d nop 0x0000121e nop

abspwr on 11:30 PM 05/02/2022: I wrote a solution with explanation and keygen here: https://abspwr.github.io/re/unix/licesnse_checker_3/license_checker_3.html Best regards

strawberry_cake on 1:32 PM 04/10/2023: 999995

chris.dinozzi on 12:39 PM 04/14/2023: cool room!

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