root💀bl4ck4non-sec:~#

Hack. Eat. Sleep. Repeat!!!

View on GitHub

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Challenges Solved

Beginner

Forensics

Beginner

Secure Shell

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Navigating to the webpage

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I got this, now from the looks of things here I knew immediately that it has something to do with command injection

Running the ls command, I got this

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There’s an index.php file

But then this command cd ../../../ && ls -la gets flagged

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This didn’t bother me actually because I saw something similar when I played picoctf 2023, you can read the writeup here

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This should work actually

Now, instead of using the cd ../../../ && ls -la command, I did this instead echo "$(cd ../../../ && ls -la)"

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All that’s left now is to read the flag, this command should do it echo "$(cd ../../../ && cat flag)"

It is evident from the above screenshot that the flag file most likely contains the flag but then it has read-only access, while the readflag is an executable that likely allows me to read the flag file.

All I did to get the flag was to run readflag executable using the command echo "$(cd ../../../ && ./readflag)"

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Got the flag

FLAG:-wwf{th3_os_c0mm4nd_1nj3ct10n!}

Simpler RSA

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Checking the content of the txt file, I saw this

┌──(bl4ck4non💀bl4ck4non-sec)-[~/…/CTF/wwctf/beginners/simpler_rsa]
└─$ cat out.txt                   
p=20322136122026329892580404875086132520732558134579258531781672192065024437324055172065343417524169304918928056147680414370351055409439818026607876517460045945556933456319117456860928521423787112252544266864178773974904640732880445449138842965327995838722222110164109025916914430044528254715080648900354468118393295346137198518513075775514617222780524163798065365970392865107270392212968677531885628998155305428785133820145555740608026626724539584106018453003156159305252013173659975815845286802275956807162426425721298560633326719023970391963404981189820163950120529861779878077006530640930032570206978446007206971761
q=19097560527100693557502945814016176943507375936656621847599300620729196257594977906326233653252987169303598004653720974045696589437233399711658994040877123702369987961301047714594623670674571987772814959679153558360152976652255742578324469478560556855210734037861198243000935281050776548747455717266013266531885744852759548255091579407464355390341944708706006878618904548103612995804547530724085856234186750409404880456083750984829553552127853848824218180459231650990529456828407224866655873224370892839628814748212142246752082561042142636866939231370987974125358875253454199574864895153300338298982667319003886687691
c=4281681357519343869235268029657832985104802601857889851833662824770073601279722389949102805423012693423900316266993146428480448851806951090530135683459342224839031144425810971344588481297094697047852347659595441639804230546879345999083627138617034295731725402645279785129174304818023129638779656619113578465655082808462489379872294929944719545647280271454196700396004152529288987570497804498041888697213294509916951489315431831556860863264254674452235360890586742441263188663158067860877772336480637257856658858967478284817730555629113613134338975168062044831796369552664256963808360408525644200922627703094455580032

The file contains 3 large integers p, q and c.

  1. p and are large prime numbers used to calaculate the modulus n where n = p * q
  2. c represents the ciphertext

So the task basically is to decrypt the ciphertext.

Upon solving this the way I know how, I got some gibberish stuff

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Apparently, I wasn’t the only one😂

I actually noticed my mistake after solving this chall, when I solved initally I calculated the value for n, also I assumed e to be 65537, all these wasn’t needed actually.

Now, determining e is unnecessary, as d can be directly computed using the private key. Specifically, d is obtained as the modular multiplicative inverse of p modulo q−1 (mod q - 1).

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To solve this, I used this python script

from Crypto.Util.number import inverse, long_to_bytes

p=20322136122026329892580404875086132520732558134579258531781672192065024437324055172065343417524169304918928056147680414370351055409439818026607876517460045945556933456319117456860928521423787112252544266864178773974904640732880445449138842965327995838722222110164109025916914430044528254715080648900354468118393295346137198518513075775514617222780524163798065365970392865107270392212968677531885628998155305428785133820145555740608026626724539584106018453003156159305252013173659975815845286802275956807162426425721298560633326719023970391963404981189820163950120529861779878077006530640930032570206978446007206971761
q=19097560527100693557502945814016176943507375936656621847599300620729196257594977906326233653252987169303598004653720974045696589437233399711658994040877123702369987961301047714594623670674571987772814959679153558360152976652255742578324469478560556855210734037861198243000935281050776548747455717266013266531885744852759548255091579407464355390341944708706006878618904548103612995804547530724085856234186750409404880456083750984829553552127853848824218180459231650990529456828407224866655873224370892839628814748212142246752082561042142636866939231370987974125358875253454199574864895153300338298982667319003886687691
c=4281681357519343869235268029657832985104802601857889851833662824770073601279722389949102805423012693423900316266993146428480448851806951090530135683459342224839031144425810971344588481297094697047852347659595441639804230546879345999083627138617034295731725402645279785129174304818023129638779656619113578465655082808462489379872294929944719545647280271454196700396004152529288987570497804498041888697213294509916951489315431831556860863264254674452235360890586742441263188663158067860877772336480637257856658858967478284817730555629113613134338975168062044831796369552664256963808360408525644200922627703094455580032

d = inverse(p, q-1)
pt  = long_to_bytes(pow(c, d, q))
print(pt)

The script decrypts a ciphertext c using modular arithmetic. It calculates d, the modular inverse of p modulo q−1, then computes c d mod q to derive the plaintext. The plaintext is converted to a readable byte string using long_to_bytes.

Running the script,

┌──(bl4ck4non💀bl4ck4non-sec)-[~/…/CTF/wwctf/beginners/simpler_rsa]
└─$ python solve.py
b'wwf{ju57_u53_l1br4r135}'

I got the flag

FLAG:-wwf{ju57_u53_l1br4r135}

Forensics

┌──(bl4ck4non💀bl4ck4non-sec)-[~/…/CTF/wwctf/forensics/black_meets_wukong]
└─$ file Evidence.ad1
Evidence.ad1: data

The writeup for this will be added later

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Later means either when I get wine fixed or when I get a windows box to run the ftkimager on.

Bruhh, I had to wait till midnight to download the 1.8gb file, only to find out that I can only solve it with a windows box just because I can’t get wine to work💀

Till Next Time :xD

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