AAAAAAAAEEEEEEEESSSSSSSS - CTF Writeup

Sun, 15 Feb 2026 06:00:00 +0000

AAAAAAAAEEEEEEEESSSSSSSS - Alpacahack challenge writeup

Category: Crypto
Difficulty: Medium
Author: hiikunz

Challenge Description

We’re given a Python script that encrypts a flag using AES-ECB mode with a twist - each character of the flag is repeated 8 times before encryption. We can also query an encryption oracle with arbitrary plaintexts.

for c in flag:
 ffffffffllllllllaaaaaaaagggggggg += bytes([c] * 8)
ciphertext = cipher.encrypt(ffffffffllllllllaaaaaaaagggggggg)

Initial Analysis

Let’s break down what the code does:

Flag format: 32 bytes total, format Alpaca{...}
Character repetition: Each character is repeated 8 times
Total plaintext: 32 characters × 8 repetitions = 256 bytes
Encryption: AES-ECB mode (Electronic Codebook)
Blocks: 256 bytes ÷ 16 bytes per block = 16 blocks

Understanding the Encoding

If the flag is Alpaca{test_flag_here______}, the plaintext becomes:

Ecb on 4mN3s14 | CTF Player & Student

AAAAAAAAEEEEEEEESSSSSSSS - CTF Writeup

AAAAAAAAEEEEEEEESSSSSSSS - Alpacahack challenge writeup

Challenge Description

Initial Analysis

Understanding the Encoding