663757a26b8bd8ddfe33c420

About

This was a very fun crackme, link is here. This is for someone who is going from beginner to intermediate level. you basically are expected to find the password to this crackme.

Diving in

Using IDA to disassemble this, the code block shown in the figure is the main block i focused on since it literally had everything in it.

input string length should be 25 to enter the first if statement. This gets directly reflected in the v16 == 25 statement which is weird because IDA doesn’t show any dependency between v16 and input str but fuck that we ball. coming directly to the SHIDWORD statement, now that is what will reveal the password for obvious reasons. so i bypassed it using windbg and printed out v5[SHIDWORD(v10)].

64, 43,44 etc is aur password but in ascii. using basic python string manipulation, password comes out to be “dCD11yGqoTJWxcwIZotS8AeOn”