2023 BYUCTF - Leet 2

Overview

• Tag: Easy

Description

Just make 1337 (again)

nc byuctf.xyz 40001

Attached

import re

FLAG = open('flag.txt').read()

inp = input('> ')

if re.search(r'[123456789]', inp) or re.search(r'\(', inp) or eval(inp) != 1337:
    print('Nope')
else:
    print(FLAG)

Analyzation

We cannot use 123456789, and the open parenthese. Oh wait, the 0 digit has not been blocked!!!

Solution

Hexadecimal will work:

0xfed - 0xaaa - 0xa

The flag is:

byuctf{aaaaaaand_more_simple_bypasses_:)}