x86 Assembly: Crack Challenge 1, Dynamic Analyses 2

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

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s