[AlexCTF] What is the encryption [crypto][150pts]

“What is the encryption” is a crypto challenge from the AlexCTF 2017 , it worth 150 points.

Description :

Fady assumed this time that you will be so n00b to tell what encryption he is using
he send the following note to his friend in plain sight :

p=0xa6055ec186de51800ddd6fcbf0192384ff42d707a55f57af4fcfb0d1dc7bd97055e8275cd4b78ec63c5d592f567c66393a061324aa2e6a8d8fc2a910cbee1ed9

q=0xfa0f9463ea0a93b929c099320d31c277e0b0dbc65b189ed76124f5a1218f5d91fd0102a4c8de11f28be5e4d0ae91ab319f4537e97ed74bc663e972a4a9119307

e=0x6d1fdab4ce3217b3fc32c9ed480a31d067fd57d93a9ab52b472dc393ab7852fbcb11abbebfd6aaae8032db1316dc22d3f7c3d631e24df13ef23d3b381a1c3e04abcc745d402ee3a031ac2718fae63b240837b4f657f29ca4702da9af22a3a019d68904a969ddb01bcf941df70af042f4fae5cbeb9c2151b324f387e525094c41

c=0x7fe1a4f743675d1987d25d38111fae0f78bbea6852cba5beda47db76d119a3efe24cb04b9449f53becd43b0b46e269826a983f832abb53b7a7e24a43ad15378344ed5c20f51e268186d24c76050c1e73647523bd5f91d9b6ad3e86bbf9126588b1dee21e6997372e36c3e74284734748891829665086e0dc523ed23c386bb520

He is underestimating our crypto skills!

It’s a pretty classical challenge of RSA, there’s is the script to find the decrypted flag

#!/usr/bin/python2
import gmpy2
import binascii
p =  0xa6055ec186de51800ddd6fcbf0192384ff42d707a55f57af4fcfb0d1dc7bd97055e8275cd4b78ec63c5d592f567c66393a061324aa2e6a8d8fc2a910cbee1ed9
q =  0xfa0f9463ea0a93b929c099320d31c277e0b0dbc65b189ed76124f5a1218f5d91fd0102a4c8de11f28be5e4d0ae91ab319f4537e97ed74bc663e972a4a9119307
e =  0x6d1fdab4ce3217b3fc32c9ed480a31d067fd57d93a9ab52b472dc393ab7852fbcb11abbebfd6aaae8032db1316dc22d3f7c3d631e24df13ef23d3b381a1c3e04abcc745d402ee3a031ac2718fae63b240837b4f657f29ca4702da9af22a3a019d68904a969ddb01bcf941df70af042f4fae5cbeb9c2151b324f387e525094c41
c =  0x7fe1a4f743675d1987d25d38111fae0f78bbea6852cba5beda47db76d119a3efe24cb04b9449f53becd43b0b46e269826a983f832abb53b7a7e24a43ad15378344ed5c20f51e268186d24c76050c1e73647523bd5f91d9b6ad3e86bbf9126588b1dee21e6997372e36c3e74284734748891829665086e0dc523ed23c386bb520
t = (p-1)*(q-1)
n = p*q

# returns d such that e * d == 1 modulo t, or 0 if no such y exists.
d = gmpy2.invert(e,t)

# Decryption
m = pow(c,d,n)
print "Solved ! m = %d" % m
print "Flag : %s " %  binascii.unhexlify(hex(m)[2:])
decrypt.py

The output of the script is

neolex@neolex-pc> crypto3:What_is_this_encryption[crypto150]$./decrypt.py 
Solved ! m = 2128227823044560158221770077085626332217552832266712453155349316324457362201641068235664016509
Flag : ALEXCTF{RS4_I5_E55ENT1AL_T0_D0_BY_H4ND} 
Resultat

So the flag is ALEXCTF{RS4_I5_E55ENT1AL_T0_D0_BY_H4ND}

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