Dear Brother,
In this post, I will continue to write about the debuging to analyze the application dynamically with IDA
We will continue to analyze the 4 and 5 condition in the appliation. Here is the static analyses from the application
Lets set break point to check the result
Lets input the application using the value below
We can see that below, the value of returned char at indext 0B = 11 is equal to 20h since it is the same for both value (jnz) then it will go down
The nex condition is checking about the element no 8 shall be ‘.’ so by inputing the correct value then it will goes to the next condition
So we can conclude that up to condition no 5 is the key would be like Y1234567.90 2345
Cyber Security Architect | Red/Blue Teaming | Exploit/Malware Analysis 