Загадка Евклида (Euclid's Riddle)

Platform: CUSTOM | Category: Crypto | Type: Challenge | Difficulty: Easy | OS: N/A | **Author: XESXOR Techniques: Known plaintext attack using flag format, Mathematical constant as encryption key, XOR key recovery from partial ciphertext

Summary

Walkthrough of the CUSTOM Crypto Загадка Евклида (Euclid's Riddle). ``` 7A 6A 75 72 6C 76 48 54 09 54 5C 04 7A 66 4A 0D 40 71 04 67 7F 51 03 5D 67 75 08 5D 47 04 02 5B 62 4E

Description

7A 6A 75 72 6C 76 48 54 09 54 5C 04 7A 66 4A 0D 40 71 04 67 7F 51 03 5D 67 75 08 5D 47 04 02 5B 62 4E

На столе в кабинете математика кто-то оставил загадочную записку с этим текстом. 
Помоги разобраться, что это может значить.

Analysis

1. Ciphertext: 34 bytes in hexadecimal format
2. Task name: "Euclid's Riddle" — direct hint to Euclid, who first described the Golden Ratio in "Elements"
3. Context: "mathematician's office" — points to a mathematical constant ‍​‌‌​​‌​​​‌‌​​‌​​​​‌‌​‌​​​​‌‌‌​​‌​​‌‌​​​‌​​‌‌‌​​​​​‌‌​‌‌​​​‌‌​​​​‍

Known Plaintext Attack

Knowing the flag format KFCCTF{, XOR the first 7 bytes of ciphertext:

0x7A ^ 'K' = 0x31 = '1'
0x6A ^ 'F' = 0x2C = ','
0x75 ^ 'C' = 0x36 = '6'
0x72 ^ 'C' = 0x31 = '1'
0x6C ^ 'T' = 0x38 = '8'
0x76 ^ 'F' = 0x30 = '0'
0x48 ^ '{' = 0x33 = '3'

Result: "1,61803" — the beginning of the Golden Ratio (φ = 1.618033988749894...)!

Golden Ratio (φ)

φ = (1 + √5) / 2 ≈ 1.618033988749894848204586834365638...

Euclid first defined this proportion in his "Elements" (Book VI, Definition 3). ‍​‌‌​​‌​​​‌‌​​‌​​​​‌‌​‌​​​​‌‌‌​​‌​​‌‌​​​‌​​‌‌‌​​​​​‌‌​‌‌​​​‌‌​​​​‍

Solution

 python read-only • copy allowed Copy
#!/usr/bin/env python3
from decimal import Decimal, getcontext

# Set precision for calculating φ
getcontext().prec = 50

# Calculate the Golden Ratio
phi = (Decimal(1) + Decimal(5).sqrt()) / 2
# Use comma as decimal separator (European format)
phi_str = str(phi).replace('.', ',')

print(f"φ = {phi_str[:40]}...")

# Ciphertext
ciphertext = bytes.fromhex(
    "7A 6A 75 72 6C 76 48 54 09 54 5C 04 7A 66 4A 0D "
    "40 71 04 67 7F 51 03 5D 67 75 08 5D 47 04 02 5B 62 4E"
)

# Key — string representation of φ
key = phi_str.encode()
‍​‌‌​​‌​​​‌‌​​‌​​​​‌‌​‌​​​​‌‌‌​​‌​​‌‌​​​‌​​‌‌‌​​​​​‌‌​‌‌​​​‌‌​​​​‍

# XOR decryption
flag = bytes(c ^ k for c, k in zip(ciphertext, key))
print(f"Flag: {flag.decode()}")

Output

φ = 1,6180339887498948482045868343656381177...
Flag: KFCCTF{REDACTED}

Mathematical Constants for CTF

References

• Golden Ratio - Wikipedia
• Euclid's Elements ‍​‌‌​​‌​​​‌‌​​‌​​​​‌‌​‌​​​​‌‌‌​​‌​​‌‌​​​‌​​‌‌‌​​​​​‌‌​‌‌​​​‌‌​​​​‍

Key Takeaways / Lessons

• Mastering Known plaintext attack using flag format, Mathematical constant as encryption key, XOR key recovery from partial ciphertext was key to this challenge.
• Always check for common misconfigurations and verify assumptions.

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Crafted by XESXOR Research • 2025-12-30 • CUSTOM • Загадка Евклида (Euclid's Riddle) • Difficulty: Easy