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get the ctf flag from this CYBERCOM{}

Créé le : 25 août 2026

Répondu en utilisant GPT-5.6 Thinking par Chat01

Question

de9e3930-a7e5-4e60-8a44-2a5fca4aa3a0.txt

get the ctf flag from this CYBERCOM{}

Python

python
from pathlib import Path
import base64, re, binascii, codecs, json, string

path = Path("/mnt/data/de9e3930-a7e5-4e60-8a44-2a5fca4aa3a0.txt")
data = path.read_text().strip()
len(data), data[:100]


(5036,
 'MzAgMzEgMzAgMzEgMzAgMzAgMzEgMzEgMjAgMzAgMzEgMzAgMzAgMzAgMzEgMzAgMzEgMjAgMzAgMzEgMzAgMzEgMzEgMzAgMzEg')

Python

python
d1=base64.b64decode(data).decode()
len(d1), d1[:300]


(3776,
 '30 31 30 31 30 30 31 31 20 30 31 30 30 30 31 30 31 20 30 31 30 31 31 30 31 30 20 30 31 30 31 30 30 31 31 20 30 31 30 31 30 30 31 30 20 30 30 31 31 30 30 30 30 20 30 31 31 30 30 31 30 30 20 30 31 30 30 31 31 31 30 20 30 31 30 31 30 31 31 31 20 30 31 31 30 31 31 30 30 20 30 31 30 31 30 30 31 30 20 30 ')

Python

python
d2=bytes.fromhex(d1).decode()
len(d2), repr(d2[:300])


(1259,
 "'01010011 01000101 01011010 01010011 01010010 00110000 01100100 01001110 01010111 01101100 01010010 01001110 01010011 01010100 01010010 01000101 01010101 01010110 01110000 01001101 01010001 00110000 01100100 01010111 01010101 00110001 01010010 01001000 01010100 01010101 01111000 01001000 01010010 001'")

Python

python
d3=''.join(chr(int(b,2)) for b in d2.split())
len(d3), d3[:300], d3[-50:]


(140,
 'SEZSR0dNWlRNSTREUVpMQ0dWU1RHTUxHR0VZVElNSlhNVlNER05CU0dFWURPT0pZTUZRV0dZSlJHTVpUU05SVkhBMlRFWUpVTVJTVElPTEZNRlFUTVpKV0dZWkRTT0JVTVlaR0lZST0=',
 'RFWUpVTVJTVElPTEZNRlFUTVpKV0dZWkRTT0JVTVlaR0lZST0=')

Python

python
d4=base64.b64decode(d3).decode()
len(d4), d4


(104,
 'HFRGGMZTMI4DQZLCGVSTGMLGGEYTIMJXMVSDGNBSGEYDOOJYMFQWGYJRGMZTSNRVHA2TEYJUMRSTIOLFMFQTMZJWGYZDSOBUMYZGIYI=')

Python

python
d5=base64.b32decode(d4).decode()
len(d5), repr(d5)


(64, "'9bc33b88eb5e31f11417ed34210798aaca133965852a4de49eaa6e662984f2da'")

Python

python
raw = path.read_text()
raw.count('\n'), raw[-200:], raw.splitlines()[-1] if len(raw.splitlines())>1 else None


(1,
 'zAgMzEgMzEgMzAgMzAgMjAgMzAgMzEgMzAgMzEgMzEgMzAgMzEgMzAgMjAgMzAgMzEgMzAgMzEgMzAgMzAgMzEgMzEgMjAgMzAgMzEgMzAgMzEgMzAgMzEgMzAgMzAgMjAgMzAgMzAgMzEgMzEgMzAgMzAgMzAgMzAgMjAgMzAgMzAgMzEgMzEgMzEgMzEgMzAgMzE=\n',
 None)

Réflexion

Decoded layered Base64, binary, Base32, hex, and hash data

Worked for 21s

Réponse

The decoded CTF flag is:

CYBERCOM{9bc33b88eb5e31f11417ed34210798aaca133965852a4de49eaa6e662984f2da}

The file unwraps as Base64 → hex → binary → Base64 → Base32 → final hex string. (de9e3930-a7e5-4e60-8a44-2a5fca4aa3a0.txt)

Question

the challenge desription is ..... which decodes to 5

Réflexion

Investigated and decoded the five-layer clue

Worked for 9s

Réponse

Exactly — ..... is Morse code for 5, and it matches the file perfectly.

There are five decoding operations:

  1. Base64
  2. Hex / ASCII bytes
  3. Binary
  4. Base64
  5. Base32

After the fifth decode, the result is:

9bc33b88eb5e31f11417ed34210798aaca133965852a4de49eaa6e662984f2da

So the intended flag is:

CYBERCOM{9bc33b88eb5e31f11417ed34210798aaca133965852a4de49eaa6e662984f2da}

The CYBERCOM{} wrapper is just the required flag format, not a sixth decoding layer. (de9e3930-a7e5-4e60-8a44-2a5fca4aa3a0.txt)

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