Take each number mod 37 and map it to the following character set: 0-25 is the alphabet (uppercase), 26-35 are the decimal digits, and 36 is an underscore.
simple code:
m = [387 ,248 ,131 ,272 ,373, 221,161 ,110 ,91 ,359 ,390 ,50, 225 ,184 ,223 ,137 ,225 ,327, 42, 179, 220 ,365]
import string
alpha = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
test = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25]
flag = ''
for i in m:
if int(i%37) in test:
for j in alpha:
if alpha.index(j) == int(i%37) :
flag += j
elif int(i%37)!= 36:
flag += str(int(i%37)%26)
else:
flag += '_'
print(flag)
#R0UND_N_R0UND_B0D5F596
basic-mod2
Following the decription:
Take each number mod 41 and find the modular inverse for the result. Then map to the following character set: 1-26 are the alphabet, 27-36 are the decimal digits, and 37 is an underscore.
solve:
m = [145 ,126, 356, 272, 98 ,378 ,395 ,352, 392 ,215 ,446, 168 ,180 ,359 ,51, 190, 404, 209, 185, 115 ,363, 431 ,103 ]
import string
alpha = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ '
test = [0,1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25]
flag = ''
enc = [28, 13, 21, 30, 17, 32, 30, 11, 24, 37, 7, 31, 17, 3, 37, 30, 34, 31, 1, 4, 34, 1, 1] #-1
# [28, 14, 22, 30, 18, 32, 30, 12, 25, 37, 8, 31, 18, 4, 37, 30, 34, 31, 2, 5, 34, 2, 2]
for i in enc:
if i in test:
for j in alpha:
if alpha.index(j) == i :
flag += j
elif i != 37:
flag += str(i %27 )
else:
flag += '_'
print(flag)
# 1NV3R53LY_H4RD_374BE7BB
heTfl g as iicpCTo{7F4NRP051N5_16_35P3X51N3_V8450214}1
Notice that:
"but every block of 3 got scrambled around!"
if we shift char of "iicpCTo" first [0] to [3] we can get "piciCTo" than "picoCTi" continue do that until the last char is '" i ".
some code for loving:
m ='iicpCTo{7F4NRP051N5_16_35P3X51N3_V8450214}1'
m = list(m)
def solve(s):
s = list(s)
for i in range(0, len(s)-1,3):
s[i], s[i+3] = s[i+3], s[i]
return ''.join(s)
print(solve(m))
#picoCTF{7R4N5P051N6_15_3XP3N51V3_58410214}i
Vigenere
decrypt Vigenere with key: "CYLAB". That's quite easy.
picoCTF{D0NT_US3_V1G3N3R3_C1PH3R_0df54reb}
diffie-hellman
Actually this chall want us to find key by diffie-hellman then decrypt Caesar with that key
However, we can brute force they key without using diffie-hellman so that this chall have been deleted in picoCTF
At this point, my ancestor told me to do anything so I won't write anything from now on,
thank you for reading!
flag: picoCTF{Yu_toi_nho_em!}
Web
Includes
Inspect HTML
Local Authority
Search source
Power Cookie
Roboto Sans
SQLiLite
wscCTF 2022
Crypto
ANYTHING
This could be encrypted with ANYTHING! wfa{oporteec_gvb_ogd}
Vernam Cipher (One Time Pad Vigenere) =>flag: WSC{VIGENERE_NOT_BAD}
RSA With The Dogs
source: gen.sage
from random import getrandbits
from Crypto.Util.number import bytes_to_long
p = random_prime(2^(1024//2),False,2^(1023//2))
q = random_prime(2^(1024//2),False,2^(1023//2))
n = p*q
phi = (p-1) * (q-1)
done = False
while not done:
d = getrandbits(1024//4)
if (gcd(d,phi) == 1 and 36*pow(d,4) < n):
done = True
Flag = open('flag.txt').read().encode()
m=bytes_to_long(Flag)
e = Integer(d).inverse_mod(phi)
c=pow(m,e,n)
print("n =",n)
print("e =",e)
print("c =",c)
n = 80958280137410344469270793621735550547403923964041971008952114628165974409360380289792220885326992426579868790128162893145613324338067958789899179419581085862309223717281585829617191377490590947730109453817502130283318153315193437990052156404947863059961976057429879645314342452813233368655425822274689461707
e = 3575901247532182907389411227211529824636724376722157756567776602226084740339294992167070515627141715229879280406393029563498781044157896403506408797685517148091205601955885898295742740813509895317351882951244059944509598074900130252149053360447229439583686319853300112906033979011695531155686173063061146739
c = 80629080505342932586166479028264765764709326746119909040860609021743893395577080637958779561184335633322859567681317501709922573784403504695809067898870536224427948000498261469984511352960143456934810825186736399371084350678586129000118485271831798923746976704036847707653422361120164687989605124465224952493
assert(int(pow(c,d,n)) == m)
Notice: 36*pow(d,4) < n => P,Q computed with N,E (Wiener's attack)
flag: wsc{w13n3r5_wer3_bre4d_t0_hunt_b4dger5!}
EAV-Secure Diffie–Hellman?
source: key_exchange.py
from Crypto.Util.number import bytes_to_long
# I love making homespun cryptographic schemes!
def diffie_hellman():
f = open("flag.txt", "r")
flag = f.read()
a = bytes_to_long(flag.encode('utf-8'))
p = 320907854534300658334827579113595683489
g = 3
A = pow(g,a,p) #236498462734017891143727364481546318401
if __name__ == "__main__":
diffie_hellman()
# EAV-Secure? What's that?
Workflow:
A = pow(g,a,p) of course that's discrete log, i used sage math to calculate easily and get this result:
Nice, let's decrypt and gonna flag
Hmm this one's no meaning. May i am wrong in somewhere ?
No, i ensure my result !
At this time i review the code and notice that:
f = open("flag.txt", "r")
flag = f.read()
a = bytes_to_long(flag.encode('utf-8'))
Implement the idea!
from Crypto.Util.number import *
flag = 67514057458967447420279566091192598301
p = 320907854534300658334827579113595683489
g = 3
A = 236498462734017891143727364481546318401
for i in range(10000000):
flag_here = long_to_bytes(flag+(i*(p-1)))
if b'wsc{' in flag_here:
print(flag_here,'ehehhehhehhehe')
break
print(i)
After bruteforcing 8300951 times, you will get the flag :))))))
Web
Warmup: Burp
Just check history of burpsuite
We can see the redirect, send the request with cookie to get flag
SSRF 101
Notice the port, that's quite interting when private 1's port is 1001 and private2's is 10011 so that we can bypass with /ssrf?path=1/flag/
