ImaginaryCTF-2024 Writeup

ImaginaryCTF-2024

FORENSICS

bom

Description

楣瑦筴栴瑟楳渷彣桩渳獥

ictf{th4t_isn7_chin3se}

packed

Description

Packet Tracer is a cross-platform visual simulation tool designed by Cisco Systems that allows users to create network topologies and imitate modern computer networks. The software allows users to simulate the configuration of Cisco routers and switches using a simulated command line interface. (
same attachment as routed)

在线工具提取

ictf{ab4697882634d4aeb6f21141ea2724d0}

cartesian-1

Description

THIS IS AN OSINT CHALLENGE.

Greetings. You have been tasked with investigating the whereabouts of a potential recruit. To perform this top-secret background check, we must gather as much information as possible.

Please investigate Terrence Descartes. We know that his social accounts are relatively new, so please do not attack anything out of scope. For the sake of this series of challenges, nothing pertaining to Terry has existed before July 17, 2024. Do not investigate anything posted online before then.

查到 github 账户

继续搜索 ins 账号

查看最新 动态

ictf{i_love_revealing_info_on_the_internet}

routed

Description

Can you dig a bit deeper? (same attachment as packed)

There are some fake flags; this is unintentional. The real flag is not in plaintext.

1
md5(flag) = 17ef83e05f7f5410443994ac434b5823

Attachments

https://cybersharing.net/s/266989fb9d6c56b1

使用 Cisco Packet Tracer 打开 routed.pkz 查看历史命令 明文的flag 都不对

解密 password 7 使用最后一次登录的密码

ictf{hidden_not_hashed}

crash

Description

I didn’t save my work…

Attachments

https://storage.googleapis.com/ictf-2024-files/dump.vmem

MemProcFS-Analyzer 打开内存镜像 在 timeline_web.csv 中发现流量了 base64 解密网页

ictf{aa0eb707a41b2ca6}

REVERSING

unoriginal

Description

i like elf reversing

Attachments

https://cybersharing.net/s/8330af8ef6c673a8

xor 5 得到flag

1
2
3
4
a = 'lfqc~opvqZdkjqm`wZcidbZfm`fn`wZd6130a0`0``761gdx'

for i in a:
    print(chr(ord(i) ^ 5), end='')

ictf{just_another_flag_checker_a3465d5e5ee234ba}

CRYPTO

base64

Description

yet another base64 decoding challenge

Attachments

https://cybersharing.net/s/8c2a3e4e78a0161f

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
# from Crypto.Util.number import bytes_to_long
#
# q = 64
#
# flag = open("flag.txt", "rb").read()
# flag_int = bytes_to_long(flag)
#
# secret_key = []
# while flag_int:
#     secret_key.append(flag_int % q)
#     flag_int //= q
#
# print(f"{secret_key = }")


from Crypto.Util.number import long_to_bytes

q = 64
secret_key = [10, 52, 23, 14, 52, 16, 3, 14, 37, 37, 3, 25, 50, 32, 19, 14, 48, 32, 35, 13, 54, 12, 35, 12, 31, 29, 7, 29, 38, 61, 37, 27, 47, 5, 51, 28, 50, 13, 35, 29, 46, 1, 51, 24, 31, 21, 54, 28, 52, 8, 54, 30, 38, 17, 55, 24, 41, 1]

flag_int = 0
for i, value in enumerate(secret_key):
    flag_int += value * (q ** i)

flag_bytes = long_to_bytes(flag_int)
print(flag_bytes.decode())

ictf{b4se_c0nv3rs1on_ftw_236680982d9e8449}