Tags: easy
Rating:
# crackme-py
Category: Reverse Engineering
AUTHOR: SYREAL
## Solving
Open up the source code:
```py
# 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"+ \
"[\\]^_`abcdefghijklmnopqrstuvwxyz{|}~"
```
Wow, okay. Jess Bezos! Now look a little lower:
```py
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:
```
picoCTF{1|\/|_4_p34|\|ut_ef5b69a3}
```
Here we go.
