Tags: easy 


# crackme-py

Category: Reverse Engineering


## Solving

Open up the source code:
# Hiding this really important number in an obscure piece of code is brilliant!
# AND it's encrypted!
# We want our biggest client to know his information is safe with us.
bezos_cc_secret = "A:4@r%uL`M-^M0c0AbcM-MFE067d3eh2bN"

# Reference alphabet
alphabet = "!\"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ"+ \
Wow, okay. Jess Bezos! Now look a little lower:
def decode_secret(secret):
"""ROT47 decode

NOTE: encode and decode are the same operation in the ROT cipher family.
So a ROT47, we can just use [cyberchef](https://gchq.github.io/CyberChef/) and decode the secret:
Here we go.

Original writeup (https://github.com/xnomas/PicoCTF-2021-Writeups/tree/main/crackme_py).