[All WriteUps for this CTF here](https://github.com/ipv6-feet-under/WriteUps-S.H.E.L.L.CTF21): https://github.com/ipv6-feet-under/WriteUps-S.H.E.L.L.CTF21

# haxxor

We are given an encrypted string:
Encrypted string : 0x2-0x19-0x14-0x1d-0x1d-0x2a-0x9-0x61-0x3-0x62-0x15-0xe-0x60-0x5-0xe-0x19-0x4-0x19-0x2c

The title gives us the hint to use XOR. Lazy as I am i searched online for another solver and edited it to my pleasure:

##edited: https://ctftime.org/writeup/9230

cipher = '0219141d1d2a09610362150e60050e1904192c'.decode('hex')

for i in range(0x00,0xff):
result = ''
for j in cipher:
result += chr(i^ord(j))
if 'SHELL{' in result:
print 'flag :', result

That's enough to calc the flag:
└─$ python2.7 ./haxxor.py
flag : SHELL{X0R3D_1T_HUH}


Original writeup (https://github.com/ipv6-feet-under/WriteUps-S.H.E.L.L.CTF21/tree/main/Cryptography/haxxor).