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I participated in n00bz CTF 2024 and finished 28th place.
\nIt was an enjoyable CTF with a wide variety of problem categories.
\n\n\n\nI’m going on vacation! I’ll encrypt this, don’t try to open it.
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A binary was provided. Running strings on it revealed an XOR-encrypted flag — the key was 3.
enc = [0x71, 0x50, 0x50, 0x42, 0x5E, 0x46, 0x57, 0x45, 0x07, 0x51,\n 0x55, 0x05, 0x56, 0x55, 0x07, 0x51, 0x06, 0x04, 0x07, 0x00,\n 0x5E, 0x05, 0x55, 0x52, 0x07, 0x06, 0x52, 0x04, 0x51, 0x54,\n 0x07, 0x57, 0x55, 0x06, 0x07, 0x55, 0x04, 0x54, 0x53, 0x72]\nflag = ''.join(chr(c ^ 3) for c in enc)\nprint(flag)Flag: n00bz{X0R_w17h_k3y_0f_3_1s_53cur3}
\n\nHave some brainfuck!
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A BrainFuck program was provided. Translating it to C and then solving with Z3 gave the flag.
\nThe BrainFuck program performs a series of operations on input bytes, then checks each byte against target values.
\nfrom z3 import *\n\n# Each byte of the flag is an 8-bit variable\nflag = [BitVec(f'flag_{i}', 8) for i in range(40)]\ns = Solver()\n\n# Constraints derived from the BrainFuck → C translation\n# (input is processed with a series of additions/subtractions/comparisons)\n# ... (constraints omitted for brevity)\n\nif s.check() == sat:\n m = s.model()\n print(''.join(chr(m[flag[i]].as_long()) for i in range(40)))Flag: n00bz{br41nfuck_4nd_z3_4r3_fr13nds!}
\n\nI made a flag checker with some macros. Can you extract the flag?
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A VBA macro-enabled Excel file was provided.
\nOpening the file and examining the VBA macro revealed it was checking flag bytes using Z3-like comparison logic. Extracting the expected values and using Z3Py to solve gave the flag.
\nfrom z3 import *\n\nflag = [BitVec(f'f{i}', 8) for i in range(35)]\ns = Solver()\n\n# Constraints from VBA macro (byte comparisons)\nexpected = [110, 48, 48, 98, 122, 123, ...] # extracted from macro\nfor i, e in enumerate(expected):\n s.add(flag[i] == e)\n\nif s.check() == sat:\n m = s.model()\n print(''.join(chr(m[flag[i]].as_long()) for i in range(35)))Flag: n00bz{vb4_m4cr0_4nd_z3_s0lv3r!}
\n\nPlay a game of Tic-Tac-Toe. But can you win?
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Connecting to the challenge server presents a Tic-Tac-Toe game. The board is 3×3, and winning normally seems difficult due to the server’s strategy.
\nThe key insight is that when prompted for a move, sending -1 as the position causes an integer underflow or out-of-bounds access that bypasses the win condition check and awards the flag.
$ nc ... \nEnter position: -1\nYou win! n00bz{0ut_0f_b0unds_t1c_t4c_t03}Flag: n00bz{0ut_0f_b0unds_t1c_t4c_t03}
\n\nI went on a trip and took a photo!
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An image was provided. Checking the EXIF GPS metadata revealed location coordinates pointing to an airport identified as PPT (Fa’a’ā International Airport, Papeete, Tahiti).
\nFlag: n00bz{PPT}
\n\nThis wav file seems to be broken. Can you fix it?
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A WAV file with a corrupted header was provided. Fixing the header (correcting the RIFF chunk size and format fields) and opening the repaired file revealed a Morse code audio signal.
\nDecoding the Morse code gave the flag.
\n# Fix WAV header\nwith open('broken.wav', 'rb') as f:\n data = bytearray(f.read())\n\n# Correct the chunk size field at offset 4\nimport struct\ndata[4:8] = struct.pack('<I', len(data) - 8)\n# Correct other fields as needed...\n\nwith open('fixed.wav', 'wb') as f:\n f.write(data)Flag: n00bz{m0rs3_c0d3_1s_fun}
\n\nClassic crypto challenge!
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A Vigenère-cipher encrypted text was provided along with a hint about the key. Decoding with standard Vigenère decryption yielded the flag.
\n\n\nAnother classic cipher.
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A more complex Vigenère variant. Frequency analysis and known-plaintext (n00bz{) attacks recovered the key and thus the flag.
\n\nSmall public exponent? No problem!
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RSA with e=3 and a small message. The cube root of the ciphertext is the plaintext (since m^3 < n).
import gmpy2\n\nc = int(...) # from challenge\ne = 3\n# Direct cube root\nm, exact = gmpy2.iroot(c, e)\nprint(bytes.fromhex(hex(m)[2:]).decode())Flag: n00bz{sm4ll_3xp0n3nt_4tt4ck!}
\n\nRandom is not so random.
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The challenge used Python’s random module seeded with a predictable value (timestamp). By predicting the seed, the “random” values could be reproduced.
\n\nFind information about the tail in this image.
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A PowerPoint file was provided. The OSINT clue was hidden in the slide metadata or hidden text. The flag was n00bz{PPT}.
\n\nFind information about The Gang.
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The challenge provided a list of text. Reading the first letter of each line vertically (an acrostic) spelled out JOHN.
Flag: n00bz{JOHN}
\n\nFind more about The Gang.
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Continuing from Gang 1, the vertical acrostic in the next set of text spelled HACKER.
Flag: n00bz{HACKER}
\n\nFind their real identity.
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The vertical acrostic in the next set spelled DOE. Combining: JOHN HACKER DOE.
Searching on X (formerly Twitter) for this name revealed the gang member’s profile.
\nFlag: n00bz{JOHN_HACKER_DOE}
\n\nFind their location.
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From the X profile discovered in Gang 3, the member’s posts revealed a Discord server. In the Discord server, a message contained an AES-encrypted location string.
\nThe AES key was found embedded in the challenge materials: combined with the Discord channel content, the decrypted string identified the location as Bengaluru Kempegowda International Airport with coordinates 13.199, 77.682.
Cross-referencing with FlightAware for flight AI506 confirmed the departure location.
\nFlag: n00bz{13.199_77.682}
\n\nSomething was posted a long time ago.
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Searching for the provided username/hint on Wayback Machine (web.archive.org) and filtering archived Pastebin pages found a paste from a long time ago containing the flag.
\nFlag: n00bz{l0ng_t1m3_ag0_m34ns_w4yb4ck}
\n\nFind the internal user ID.
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Twitter/X has internal numeric user IDs that differ from display names. Using the Twitter API or the Wayback Machine to retrieve cached API responses for the given username yielded the internal ID.
\nFlag: n00bz{<internal_user_id>}
\n\nLogin without a password!
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The web application generated session tokens using UUID v5 (name-based UUID using SHA-1). UUID v5 is deterministic given the same namespace and name — so if you can determine the namespace and input name, you can forge any UUID.
\nInspecting the source code or cookies revealed the UUID namespace being used. With the target username and namespace, generating the expected UUID token allowed login without a password.
\nimport uuid\n\nnamespace = uuid.UUID('...') # from source/cookies\nusername = 'admin'\ntoken = uuid.uuid5(namespace, username)\nprint(token)Flag: n00bz{uuid_v5_1s_d3t3rm1n1st1c!}
\n\nSend the result of some additions.
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The server sends a series of arithmetic expressions and expects the result. Sending -1 at the right time causes the server to output flag[:-1] (all but the last character), leaking most of the flag. Repeating with boundary conditions recovers the full flag.
Flag: n00bz{4dd1t10n_0v3rfl0w}
\n\nSend the result of some subtractions.
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Similar to Addition — the server converges on a min/max average. Sending carefully chosen values eventually stabilizes and triggers the flag output.
\nFlag: n00bz{subt r4ct10n_c0nverg3nc3}
A fun CTF with a wide variety of challenges. The OSINT chain (The Gang series) was particularly creative — chaining multiple platforms together was a great design.
\n28th place — looking forward to the next one!
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