Version
By |
Version |
Comment |
noraj |
1.0 |
Creation |
CTF
Description
Bulljog isn't much harder than this one.
encrypt.py encryptme.txt.out
Solution
encrypt.py
is a simple xoring:
Let's xortool
show us some probability:
xortool
tell us there is 21% chances of a 9 bytes length key. So let's try it:
One key seems nearly good:
So let's try it:
The output is 100% printable text, we have the good key, now I need to find the flag: