Aperi’CTF 2019 - Simple Basique
Challenge details
| Event | Challenge | Category | Points | Solves |
|---|---|---|---|---|
| Aperi’CTF 2019 | Simple Basique | Cryptography | 50 | 60 |
Un apprenti cryptologue souhaite communiquer avec vous:
%51%56%42%53%53%33%74%43%4e%48%4d%78%63%58%55%7a%58%31%4d%78%62%58%42%73%5a%53%46%39%43%67%3d%3d
Methodology
Looking at the charset, we got an hexadecimals ciphertext, lets decode it on a python shell:
Python 2.7.13 (default, Sep 26 2018, 18:42:22)
[GCC 6.3.0 20170516] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> import urllib
>>> urllib.unquote_plus("%51%56%42%53%53%33%74%43%4e%48%4d%78%63%58%55%7a%58%31%4d%78%62%58%42%73%5a%53%46%39%43%67%3d%3d")
'QVBSS3tCNHMxcXUzX1MxbXBsZSF9Cg=='
>>> import base64
>>> base64.b64decode(urllib.unquote_plus("%51%56%42%53%53%33%74%43%4e%48%4d%78%63%58%55%7a%58%31%4d%78%62%58%42%73%5a%53%46%39%43%67%3d%3d"))
'APRK{B4s1qu3_S1mple!}\n'
Flag
APRK{B4s1qu3_S1mple!}