Crypto & Osint Sig21CTF IIUM
Crypto
Flag 1.
we got an image for this challenge with seemingly gebrish, but cmon the category is cryptography why not try substitution techniques? 7GEa'!v$L238032J=0D69G80A?20@C06388C2N Ok this looks like 47 due to its character composition. Now lets decode it. I used cyberchef for that. And heres the flag. 
Flag 2.
We got a file with the folling content
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n = 47871871319309860974932493994368503837616324093829993047813212088563420860561
c = 18061800228431074448341444333709757167868575268631469201579336930973186970176
e = 6553
So i wrote a simple python script to break this
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from Crypto.Util.number import *
from factordb.factordb import FactorDB
c = 18061800228431074448341444333709757167868575268631469201579336930973186970176
n = 47871871319309860974932493994368503837616324093829993047813212088563420860561
e = 65537
# init factordb connection
f = FactorDB((n))
f.get_factor_list()
f.connect()
# store res in result
result = f.get_factor_list()
p = result[0]
q = result[1]
phi = (p-1) * (q-1)
d = inverse(e, phi)
m = pow (c, d, n)
print(long_to_bytes(m))
And we got the flag. sig21CTF{crypt0_rs4_t00_3asy}
Flag 3.
We got 2 file. message.secret and key.pub
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message.secret content
HELLO!!!!!!.... I HEARD YOU WERE LEARNING SOME CRYPTOGRPHIC TRICKS..
CAN YOU GET THE FLAG..... I RECIEVED THIS MESSAGE BUT I CANT FIGURE IT OUT..
THEY ALSO LEFT THIS HINT `dGhpcyBtaWdodCB0YWtlIGEgd2hpbGUuLi4uIHNvIGRvbnQgZ2l2ZSB1cC4uIHdlIHVzZWQgdGhlIHVzdWFsIHZhbHVlIGZvciBl`
c = 36471761181664780564914260343964863418853945528543016847566551168186484704567
I decode the hint to this might take a while.... so dont give up.. we used the usual value for e
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key.pub content
-----BEGIN PUBLIC KEY-----
MDswDQYJKoZIhvcNAQEBBQADKgAwJwIgYoP1rrW69NmQ8LzNsTX6ongx3kS4IxCh
TlY1JynxjJUCAwEAAQ==
-----END PUBLIC KEY-----
From the hint we know its RSA, so we basically need to find p, q, phi, d, c, n but how can get n when its not give? the answer here is using rsaCTFtools to get n from .pub file, RsaCtfTool.py --dumpkey --key key.pub Now we got n and e
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n = 44559811764670192392592515903341200648507091942717623043713990581324912626837
e = 65537
c = 36471761181664780564914260343964863418853945528543016847566551168186484704567
Now the only thing left is getting p, q. Let use our previous script to get it
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from Crypto.Util.number import *
from factordb.factordb import FactorDB
n = 44559811764670192392592515903341200648507091942717623043713990581324912626837
e = 65537
c = 36471761181664780564914260343964863418853945528543016847566551168186484704567
f = FactorDB((n))
f.get_factor_list()
f.connect()
result = f.get_factor_list()
p = result[0]
q = result[1]
phi = (p-1) * (q-1)
d = inverse(e, phi)
m = pow (c, d, n)
print(long_to_bytes(m))
But hey.. why is the flag printing this 7\xec\x83\x98\x0c\x96\x7f%\x04\x07V".\x91\x83z\xea\xce\xf8\xb4\x805c\x86\xaf\x1c@\xfb0\xa5q\xf5 ? Well.. i made a mistake encrypting the wrong massage c. And i deeply apologize to the participants that spend hours trying to solve this challenge. Anyways u got the steps on how to solve this kinda challenges.
Osint
Link to the write up gdrive
Thanks for reading and i hope this helps :)