It’s a write-up about the challenge : Ringzer0CTF - Martian message 3

Challenge 26 - Martian message 3

1. Challenge

RU9CRC43aWdxNDsxaWtiNTFpYk9PMDs6NDFS

2. Solution

The cipher is only composed of : * alphanumeric characters and the number * the number of characters for the cipher is a multiple of 4

We put the code in a base64 format decoder on this website : https://www.base64decode.org/ Or we can do this command also :

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echo RU9CRC43aWdxNDsxaWtiNTFpYk9PMDs6NDFS | openssl enc -d -base64; echo

We found : EOBD.7igq4;1ikb51ibOO0;:41R

The decoded result has only printable characters and alphanumeric characters.

We knew ringer0team has many of the flags starting with FLAG. So, we thought that the first four characters of the code should be FLAG.

We tried with a XOR decoder on this website : http://xor.pw/ with the first four characters, and we obtained 3. Indeed : E ^ F = 3 Therefore, 3 is the key.

We wrote a simple python program flag.py :

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code = 'EOBD.7igq4;1ikb51ibOO0;:41R'

flag = ''

for i in code :
	x = ord(i) ^ 3
	flag = flag + chr(x)
	
print(flag)

We launched this program : python flag.py

The result is : FLAG-4jdr782jha62jaLL38972Q