256 words

1 minute

World Wide CTF 2024 Writeup

2025-10-12

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Misc

Name	Category	Solved?
Simpler RSA	Beginners	✅
Bongcloud	Miscellaneous	✅
World Wide Flags	Miscellaneous	✅

Beginners#

Simpler RSA [50 Pts]#

Description#

Author: yun

RSA is so complicated! I made it simpler.

Solution#

We are given a python file

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from secret import flag
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from Crypto.Util.number import bytes_to_long, getPrime
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flag = bytes_to_long(flag)
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p = getPrime(2048)
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q = getPrime(2048)
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c = pow(flag, p, q)  # i believe this is the fancy rsa encryption?
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print(f'{p=}')
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print(f'{q=}')
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print(f'{c=}')

This code calculates using the equation below:

c = \text{flag}^p \bmod q

We can retrieve the flag by inversing it to:

\text{flag} = c^{\text{inverse}\_p} \bmod q

Solver:

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from Crypto.Util.number import inverse, long_to_bytes
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p = 20322136122026329892580404875086132520732558134579258531781672192065024437324055172065343417524169304918928056147680414370351055409439818026607876517460045945556933456319117456860928521423787112252544266864178773974904640732880445449138842965327995838722222110164109025916914430044528254715080648900354468118393295346137198518513075775514617222780524163798065365970392865107270392212968677531885628998155305428785133820145555740608026626724539584106018453003156159305252013173659975815845286802275956807162426425721298560633326719023970391963404981189820163950120529861779878077006530640930032570206978446007206971761
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q = 19097560527100693557502945814016176943507375936656621847599300620729196257594977906326233653252987169303598004653720974045696589437233399711658994040877123702369987961301047714594623670674571987772814959679153558360152976652255742578324469478560556855210734037861198243000935281050776548747455717266013266531885744852759548255091579407464355390341944708706006878618904548103612995804547530724085856234186750409404880456083750984829553552127853848824218180459231650990529456828407224866655873224370892839628814748212142246752082561042142636866939231370987974125358875253454199574864895153300338298982667319003886687691
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c = 4281681357519343869235268029657832985104802601857889851833662824770073601279722389949102805423012693423900316266993146428480448851806951090530135683459342224839031144425810971344588481297094697047852347659595441639804230546879345999083627138617034295731725402645279785129174304818023129638779656619113578465655082808462489379872294929944719545647280271454196700396004152529288987570497804498041888697213294509916951489315431831556860863264254674452235360890586742441263188663158067860877772336480637257856658858967478284817730555629113613134338975168062044831796369552664256963808360408525644200922627703094455580032
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p_inv = inverse(p, q - 1)
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flag = pow(c, p_inv, q)
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decoded_flag = long_to_bytes(flag)
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print(decoded_flag)

Flag

wwf{ju57_u53_l1br4r135}

Miscellaneous#

Bongcloud [153 Pts]#

Description#

Author: yun

Are you a true chess master?

nc chess.chal.wwctf.com 1337

Solution#

When I connect to the netcat connection, it says that we need to win 3 games of chess to get the flag. Since it has no timelimit to the connection, let’s just spam the best move for each condition of the board with the help of Chess Bot

chess

Flag

wwf{y0u_4r3_4_7ru3_ch355_m4573r}

World Wide Flags [150 Pts]#

Description#

“Oh, you are a WWCTF player? Name every flag.” Identify flags with increasing difficulty!

http://flags.chal.wwctf.com:1337/

Solution#

From the Website, we need to guess 100 flags that appeared. But something interesting when I solve a couple of them, the answer is not entirely correct, for example the flag France will be incorrect if we submit France as the answer.