Reverse Engineering - patchme.py - writeup

description

Can you get the flag?

Run this Python program in the same directory as this encrypted flag.

writeup

Let’s take a look at the encrypted flag…

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cat flag.txt.enc

CR1@    UYX+
6UB
P\E

That does not help.

Let’s look at the python code…

Mh that is peculiar:

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if( user_pw == "ak98" + \
                   "-=90" + \
                   "adfjhgj321" + \
                   "sleuth9000")

Let’s fire up bpython and input this code:

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"ak98" + \
                   "-=90" + \
                   "adfjhgj321" + \
                   "sleuth9000"

This is the result:

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'ak98-=90adfjhgj321sleuth9000'

Once I try to supply the aforementioned string as the password I get this output:

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Please enter correct password for flag: ak98-=90adfjhgj321sleuth9000
Welcome back... your flag, user:
picoCTF{p47ch1ng_l1f3_h4ck_e40c120e}

so the flag is:

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picoCTF{p47ch1ng_l1f3_h4ck_e40c120e}

This is probably not the way that was intended by the author to solve this challenge since the name of it is ‘patchme.py’. But I got the flag nevertheless.