# Packet capture challenges## Monstrum ex Machina1. In this challenge we are given a pcap file and are asked to locate a search query where the attacker searched for the victim's name.2. Since packet captures are long, it is generally a good idea to use the wireshark filters to narrow down your search space. In this example, we know that the search engines make a HTTP GET request. So we have to just look at the responses received for the HTTP GET requests.3. When going through the pcap file, the victim's name is present in packet number 7019, as shown in the screenshot below. ![image](https://user-images.githubusercontent.com/78410304/137625134-ba341e2a-227c-4b15-83c2-5989c4a7343f.png)
4. Flag : flag{charles geschickter}
# Programming challenges## The count1. In this challenge we are given the IP address and port of a remote server. When we connect to the port using netcat, the server prints a word and we have to calculate the sum of its digits (where a=0, b=1, etc.) and send it to the server within 5 seconds.2. This challenge can be solved using pwntools. My script is as follows:```pythonfrom pwn import *conn = remote('code.deadface.io',50000)
conn.recvuntil(b'Your word is: ')word = conn.recvline()word = word.decode('utf-8').strip()cnt=0for letter in word:    cnt+=(ord(letter)%97)conn.sendline(bytes(str(cnt), 'utf-8'))data=conn.recv()print(data)```3. flag: flag{d1c037808d23acd0dc0e3b897f344571ddce4b294e742b434888b3d9f69d9944}
## Trick or treat1. In this challenge, we are given a game where the player is supposed to dodge the multiple enemies headed its way. We have been given the source code of the game.2. When the player crashes into the enemy, a prompt saying "death" appears and the game exits. So the solution to this challenge is to stop the prompt appearing so that even if player crashes into the enemy, the exit condition is never triggered.3. In line 152, we can comment the do\_coll() function that is responsible for the collisions. Now run the game and the flag is printed on the terminal```pythondef get_intersect(x1, x2, w1, w2, y1, y2, h1):    if y1 < (y2 + h1):        if x1 > x2 and x1 < (x2 + w2) or (x1 + w1) > x2 and (x1 + w1) < (x2 + w2):            #do_coll()            pass```4. Flag : flag{CaNT\_ch34t\_d34th}
# Reversing challenges## Cereal1. In this challenge we are given an ELF file that asks for a passphrase and checks if it is the same as the flag or not. 2. In the following screenshot, we can see that the do while loop does some computation and stores it in a buffer.![image](https://user-images.githubusercontent.com/78410304/137625893-bba1b2ce-8e09-4d47-ac85-3c91bb445af0.png)3. The solution is to load this file into GDB, set a breakpoint anywhere after the flag is stored in the buffer, and read the flag.```gdb(gdb) x/32s $rsp0x7fffffffdc20: "flag{c0unt-ch0cula-cereal-FTW}"```4. Flag : flag{c0unt-ch0cula-cereal-FTW}
# SQL challenges## Demonne1. In this challenge, we are given a dump of a SQL database. We are asked to find the number of records in the customers table. 2. The solution is to load the file in Mysql database and count the rows in it.```mysqlmysql> create database demonne;Query OK, 1 row affected (0.05 sec)
mysql> use demonne;Database changedmysql> source demonne.sql;mysql> select count(*) from customers;+----------+| count(*) |+----------+|    10000 |+----------+1 row in set (0.01 sec)```3. Flag : flag{10000}
## Demonne 2 1. In this challenge, we have to use the same sql dump file from above and find the foreign key used in the loans table.2. Open the demonne.sql file and goto the query where loans table is created and look for foreign keys.3. Flag : flag{fk_loans_cust_id}