Release [Reverse Engineering] Challenge 1 - Crack me and get the right flag.

[1.315]

Wtf why are ppl leeching here? XD

ONLY DISCORD: zombie#7550 (ID: 785571908267671613)
ONLY TELEGRAM: @zombie_dev

I DONT DO MM. CONFIRM ALWAYS VIA PM ON CRACKED!

[1]

cuz we can lulululululu

[12]

Show ContentSpoiler:

[164]

I really miss you <3

[1.239]

any one? No one?

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[1.239]

@AntiLeech

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[481]

Trying to learn some reverse engineering, do not think I can do this but will give it a try.

[47]

Key is: AEEIIIIOO?OUUUI
@ForlaxPy????

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Well to begin with the first thing I did was to place a jmp to the following address

jmp > 0040197B >>> 004016AA

Already from there starting by obligation you have to skip all comparisons, if you do a simple jmp will not show anything on the flag.




another thing I noticed is that a payload is injected into the process  >> "rundll32.exe"




I made several jumps in the comparisons I saw.

jmp > 0040197B >>> 004016AA

jmp >004016AA >004016C9

jmp > 004016DD > 004016FD

jmp > 00401712 > 0040172E

jmp > 00401751 > 0040176E

jmp > 00401781 > 00401783

When I was advanced I made a mistake.



That can be solved with a simple nop "nop 00401791"

I also tried to make a jump to this address.

jmp > 0040197B >>> 00401839 

it would take me here, In that comparison I would put "0" So that it would let me access 



then I do the following.
I place these asm instructions at the following addresses
JZ > 00401896 
JZ > 0040183D
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Posible Key
"zakhstan"
"dmo3dmo3dmo3dmo3dmo3dmo3dmo3dmo3dmo3dmo3dmo3dmo3"
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The truth tries several things but does not succeed in solving it