2023 BYUCTF - Leet 1

• Tag: Easy

Description

Just make 1337

nc byuctf.xyz 40000

Attached

import re

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

inp = input('> ')

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

Analyzation

We cannot just input 1337 because re.search(r'\d', inp) blocks numbers. But just number!

Solution

Use ascii code instead

ord(f'z') + ord(f'z') + ord(f'z') + ord(f'z') + ord(f'z') + ord(f'z') + ord(f'z') + ord(f'z') + ord(f'z') + ord(f'z') + ord(f'u') 

And we got the flag:

byuctf{simple_bypasses!}