zh3r0-CTF (Pre-CTF)
is this the robo loanguage?
Crypto
Points:495
Description:
I have found some thing can you tell me what’s inside it.
Solution:
We are give a file which contains this:
--.- .-- -..- ... -.-- ... -... -- .- .-- ..... .-.. .. --. .-- -. .- --. ..-. .--. .. .... -. .--- -... ...-- -... .-.. -.-. -. .-. ...-
3,4,8,9,12,15,16,17,20,24,25,28,29,30,32
Decoding the morse code gives us a base64-like string!
QWXSYSBMAW5LIGWNAGFPIHNJB3BLCNRV
But decoding this, gives us junk! Now if we take the numbers given and play with the the character in those index we might get something!
We actually had to turn the letters at those indices to lowercase
a='QWXSYSBMAW5LIGWNAGFPIHNJB3BLCNRV'
b=[3,4,8,9,12,15,16,17,20,24,25,28,29,30,32]
c=[a[i].lower() if i+1 in b else a[i] for i in range(len(a))]
print(''.join(c))
We now have the correct base64… decoding it gives this Alla fine l'hai scoperto, which seems like another language but lets try translating it!
yoo! its translates to In the end you found out
Hence the flag is zh3r0{In_the_end_you_found_out}
AES
Crypto
Points: 567
Description:
You people kept asking about AES. So here it is
0d152c0ffefd78da21f04c769546b29b
enc(Key)=”dnmjrqsqfsfnfqgt”
Hint: no 13 may help you
Solution:
I’m mad!!! was sleeping after having enough of brainfuck in general, when they released the hint for this one!
Anyway, it’s AES decryption using ECB mode since we don’t have any IV here. As the challenge depicts the key in encoded! … well, 13! ROT13!
so ROT13(key)==qazwedfdsfsasdtg … and you don’t need my help but still!
from Crypto.Cipher import AES
c='0d152c0ffefd78da21f04c769546b29b'
Key=b"qazwedfdsfsasdtg"
cipher=AES.new(Key,AES.MODE_ECB)
a=cipher.decrypt(bytes.fromhex(c))
print(a)
You will get zh3r0{AES_1s_C00000l} as flag!
Ultra Strong RSA
Crypto
Points: 920
Description:
A simple RSA maybe be easy to crack, what if we use multilayered encryption, where we encrypt the flag in multiple methods. To crack an advanced RSA Script, study the script well before even trying to crack it.
Not always tools help.
HAPPY HACKING :)
Solution:
We are given a python script and a textfile…which has RSA encrypted text in it!
from base64 import b64encode as b64e
from Crypto.Util.number import *
from Crypto.PublicKey import RSA
from flag import flag, hint
pt=bytes_to_long(flag)
ht=bytes_to_long(hint)
p=getPrime(2048)
q=getPrime(2048)
r=getPrime(2048)
s=getPrime(2048)
x=getPrime(1024)
y=getPrime(1024)
e=65537
if x>y:
key=x-y
else:
key=y-x
key=b64e(str(key))
n=x*y
n1=p*q
n2=r*s
n3=n1*n2
if p>q:
a=p-q
else:
a=q-p
b=r+q
c=p+r
enh=pow(ht,e,n)
ct=pow(pt,e,n3)
opt="N : "+str(n)+"\ne : "+str(e)+"\nenc(Hint) : "+str(enh)+"\n\nN1 : "+str(n1)+"\nN2 : "+str(n2)+"\nN3 : "+str(n3)+"\nCipher Text : "+str(ct)+"\n\n\nkey : "+str(key)+"\n"
pvt="a = "+str(a)+"\nb = "+str(b)+"\nc = "+str(c)+"\n"
test="p="+str(p)+"\nq="+str(q)+"\nr="+str(r)+"\ns="+str(s)+"\nx="+str(x)+"\ny="+str(y)+"\n"
#print(opt,pvt)
ob=open("out.txt","w")
ob.write(opt)
ob.close()
ob=open("hint.txt","w")
ob.write(pvt)
ob.close()
ob=open("test.txt","w")
ob.write(test)
ob.close()
At first the hint which is important is encrypted with (e,n). Info about n is in key which is x-yory-x! Simple qudractic here… x**2+key*x-n=0 will give us x,y!
from Crypto.Util.number import inverse,long_to_bytes
from sympy import solve,Symbol
hint_cp=<REDACTED>
N1=<REDACTED>
N2=<REDACTED>
N3=<REDACTED>
N=<REDACTED>
KEY=<REDACTED>
C=<REDACTED>
e=65537
y=Symbol('y')
Y=solve(y**2+KEY*y-N,'y')[1]
X=N//Y
tot_h=(X-1)*(Y-1)
d_h=inverse(e,tot_h)
hint=long_to_bytes(pow(hint_cp,int(d_h),N))
#print(hint)
#hint:https://pastebin.com/Ss2RBhVN
We get a link, and the data for the next part which is similar to the first one! Quadratics here too!
a = <REDACTED>
b = <REDACTED>
c = <REDACTED>
#a=q-p,b=r+q,c=r+p
q=Symbol('q')
P=solve(q**2+a*q-N1,'q')[1]
Q=a+P
R=c-P
S=N2//R
tot=(P-1)*(Q-1)*(R-1)*(S-1)
d=inverse(e,tot)
m=pow(C,int(d),N3)
print(long_to_bytes(m))
Finally you get zh3r0{Y0u_h4v3_b34ten_RSA} !!!
BaseXOR
Crypto
Points: 481
Description:
The author seemed to find a strange string can u help him in decoding them
FQ0ACgAUVgcCSTAdAwpvGA0AFm8EVkonAVo6BkhCWBg=
Solution:
Decode the base64, you get this!
b"\x15\r\x00\n\x00\x14V\x07\x02I0\x1d\x03\no\x18\r\x00\x16o\x04VJ'\x01Z:\x06HBX\x18"
Now lets XOR the first few bytes with the flag-format zh3r0
We get oe3x0… this might possible be the key we need! Let’s XOR it with the whole thing!
Yoo! We get zh3r0{34zy_x0r_wh3n_k3y_15_50r7}!! Our flag!
Are you the Master? 2
Master
Points: 983
Description:
Enter key as the flag. (part1) folder in drive may help you
Solution:
We get another RSA encrypted file with the source code! It’s similar to the Ultra RSA chall but yet different! See here, we have a maths problem which can mislead us to may places!
So, the important info given is:
N1=p*q N2=r*s A=p+q+r+s B=(p-q)-(r-s)
Now after sneding few hours trying bi-quadractics, getting iotas and skimming through my old high-school notes, I got it!
See, this way we get a quadratic in sq which fun fact… also gives pr as its other root! So, when we get pr and sq, we can perform GCD on it with N1 and N2 and obtain p,q,r and s!!
from sympy import Symbol,solve
from Crypto.Util.number import *
N1=<REDACTED>
N2=<REDACTED>
E =65537
ct1 =<REDACTED>
ct2 =<REDACTED>
#p+q+r+s
A=<REDACTED>
#(p-q)-(r-s)
B=<REDACTED>
#t1t2=A+B+2-C
x_coeff=(A**2-B**2)//4-(N1+N2)
x=Symbol('x')
sol=solve(x**2-x_coeff*x+N1*N2,'x')
X,Y=sol[0],sol[1]
p=GCD(X,A)
q=GCD(Y,A)
tot1=(p-1)*(q-1)
d1=inverse(E,tot1)
m=pow(ct1,int(d1),A)
print(long_to_bytes(m))
zh3r0{Th1s_1s_4_K3y_gfh} the flag or a key?
Are you the Master? 3
Crypto
Points: 1199
Description:
You must me the master to get till here.
now give me the flag.
drive part 2 folder.
Solution:
In the drive folder we get a binary name iv and a img.txt. Now iv can only mean the initialization vector of an AES encrypted data!
The flag from Master_3 says its a key… so maybe thats all we need! Running the IV tho… prints the message It’s a string, so maybe we have to do strings on it to get the iv!
The H character her seems unecessary so let’s remove it! We get Here is you IV: ryq9rkoU+XhtIuoFAvujDQ==
Its seems we are ready to decrypt!!!
from Crypto.Cipher import AES
from base64 import b64decode
iv=b64decode('ryq9rkoU+XhtIuoFAvujDQ==')
key='zh3r0{Th1s_1s_4_K3y_gfh}'
dat=b64decode(open('img.txt','rb').read())
cipher=AES.new(key,AES.MODE_CBC,iv)
plaintext=b64decode(cipher.decrypt(dat))
#print(plaintext)
f=open('flag.png','wb').write(plaintext)
In the plaintext, we see corrupted PNG headers and also other corruptions. Lets fix those with ghex!
Now we see the flag on opneing the file!
zh3r0{You_are_master_Of_All_Arts} Cheers!