Write up: Crypto Hack General
This week I did a quiz from Infomation Security course. Through this I’ve learned a lot of knowdledge like how to use pwntools, pwnlib, more about XOR, modulo and RSA. This is my approach to solve 19 exercises in this quiz.
It in consists of:
Encoding
ASCII
Using the below integer array, convert the numbers to their corresponding ASCII characters to obtain a flag.
[99, 114, 121, 112, 116, 111, 123, 65, 83, 67, 73, 73, 95, 112, 114, 49, 110, 116, 52, 98, 108, 51, 125]
Use function chr() to convert integer to char
c = [99, 114, 121, 112, 116, 111, 123, 65, 83, 67, 73, 73, 95, 112, 114, 49, 110, 116, 52, 98, 108, 51, 125]
s = ""
for i in c:
s += chr(i)
print(s)
Flag:
crypto{You_will_be_working_with_hex_strings_a_lot}
Hex
Included below is a flag encoded as a hex string. Decode this back into bytes to get the flag.
63727970746f7b596f755f77696c6c5f62655f776f726b696e675f776974685f6865785f737472696e67735f615f6c6f747d
c = "63727970746f7b596f755f77696c6c5f62655f776f726b696e675f776974685f6865785f737472696e67735f615f6c6f747d"
s = bytes.fromhex(c)
print(s)
Flag:
crypto{You_will_be_working_with_hex_strings_a_lot}
Base64
Take the below hex string, decode it into bytes and then encode it into Base64.
72bca9b68fc16ac7beeb8f849dca1d8a783e8acf9679bf9269f7bf
c = "72bca9b68fc16ac7beeb8f849dca1d8a783e8acf9679bf9269f7bf"
s1 = bytes.fromhex(c)
print(s1)
# b'r\xbc\xa9\xb6\x8f\xc1j\xc7\xbe\xeb\x8f\x84\x9d\xca\x1d\x8ax>\x8a\xcf\x96y\xbf\x92i\xf7\xbf'
import base64
s2 = base64.b64encode(s1)
print(s2)
Flag:
crypto/Base+64+Encoding+is+Web+Safe/
Bytes and Big Integers
To illustrate:
message: HELLO
ascii bytes: [72, 69, 76, 76, 79]
hex bytes: [0x48, 0x45, 0x4c, 0x4c, 0x4f]
base-16: 0x48454c4c4f
base-10: 310400273487
Convert the following integer back into a message:
11515195063862318899931685488813747395775516287289682636499965282714637259206269
from Crypto.Util.number import *
c = "11515195063862318899931685488813747395775516287289682636499965282714637259206269"
long_to_bytes(c)
Flag:
crypto{3nc0d1n6_4ll_7h3_w4y_d0wn}
Encoding Challenge
Now you’ve got the hang of the various encodings you’ll be encountering, let’s have a look at automating it. Can you pass all 100 levels to get the flag?
The 13377.py file attached below is the source code for what’s running on the server. The pwntools_example.py file provides the start of a solution using the incredibly convenient pwntools library. which we recommend. If you’d prefer to use Python’s in-built telnetlib, telnetlib_example.py is also provided.
For more information about connecting to interactive challenges, see the FAQ. Feel free to skip ahead to the cryptography if you aren’t in the mood for a coding challenge!
Connect at nc socket.cryptohack.org 13377 Read file 13377.py I knew that server open on port 13377. When it’s connected, it would take a random word in /usr/share/dict/words then use base64, hex, rot13, bigint or utf-8 to encrypt it then send to client.
I write a simple code to recieved the string then decrypted it and send it back for 100 times.
from pwn import * # pip install pwntools
import json
from Crypto.Util.number import bytes_to_long, long_to_bytes
import base64
import codecs
import random
from binascii import unhexlify
r = remote('socket.cryptohack.org', 13377, level='debug')
def json_recv():
line = r.recvline()
return json.loads(line.decode())
def json_send(hsh):
request = json.dumps(hsh).encode()
r.sendline(request)
def list_to_string(s):
output = ""
return(output.join(s))
for i in range(0, 101):
received = json_recv()
if "flag" in received:
print("\n[*] FLAG: {}".format(received["flag"]))
break
print("\n[-] Cycle: {}".format(i))
print("[-] Received type: {}".format(received["type"]))
print("[-] Received encoded value: {}".format(received["encoded"]))
word = received["encoded"]
encoding = received["type"]
if encoding == "base64":
decoded = base64.b64decode(word).decode('utf8').replace("'", '"')
elif encoding == "hex":
decoded = (unhexlify(word)).decode('utf8').replace("'", '"')
elif encoding == "rot13":
decoded = codecs.decode(word, 'rot_13')
elif encoding == "bigint":
decoded = unhexlify(word.replace("0x", "")).decode(
'utf8').replace("'", '"')
elif encoding == "utf-8":
decoded = list_to_string([chr(b) for b in word])
print("[-] Decoded: {}".format(decoded))
print("[-] Decoded Type: {}".format(type(decoded)))
to_send = {
"decoded": decoded
}
json_send(to_send)
Flag:
crypto{3nc0d3_d3c0d3_3nc0d3}
XOR
XOR Starter
Given the string “label”, XOR each character with the integer 13. Convert these integers back to a string and submit the flag as crypto{new_string}.
Normal way, when xor we would padding or xor one-by-one but the function XOR by pwnlib included flat (padding) so we just use it.
import pwnlib
s = pwnlib.util.fiddling.xor("label", 13).decode()
Flag:
crypto{aloha}
XOR Properties
… undo the encryption in the final line to obtain the flag.
KEY1 = a6c8b6733c9b22de7bc0253266a3867df55acde8635e19c73313
KEY2 ^ KEY1 = 37dcb292030faa90d07eec17e3b1c6d8daf94c35d4c9191a5e1e
KEY2 ^ KEY3 = c1545756687e7573db23aa1c3452a098b71a7fbf0fddddde5fc1
FLAG ^ KEY1 ^ KEY3 ^ KEY2 = 04ee9855208a2cd59091d04767ae47963170d1660df7f56f5faf
The FLAG was xor for K1, K2, K3 so we would xor again with K1, K2 and K3 to reveal the FLAG again.
KEY1 = bytes.fromhex("a6c8b6733c9b22de7bc0253266a3867df55acde8635e19c73313")
KEY21 = bytes.fromhex("37dcb292030faa90d07eec17e3b1c6d8daf94c35d4c9191a5e1e")
KEY23 = bytes.fromhex("c1545756687e7573db23aa1c3452a098b71a7fbf0fddddde5fc1")
FK123 = bytes.fromhex("04ee9855208a2cd59091d04767ae47963170d1660df7f56f5faf")
F = pwnlib.util.fiddling.xor(FK123, KEY23, KEY1)
print(F)
Flag:
crypto{x0r_i5_ass0c1at1v3}
Favourite byte
I’ve hidden some data using XOR with a single byte, but that byte is a secret. Don’t forget to decode from hex first.
73626960647f6b206821204f21254f7d694f7624662065622127234f726927756d
After decode the text from hex to bytes, I bruteforce it with a integer (limit in value of 1 byte).
c = "73626960647f6b206821204f21254f7d694f7624662065622127234f726927756d"
s1 = bytes.fromhex(c)
for i in range(pow(2,9)):
s = pwnlib.util.fiddling.xor(s1, i)
s = s.decode()
if s.startswith("crypto{"):
print(s)
print(i)
# crypto{0x10_15_my_f4v0ur173_by7e}
# 16
Flag:
crypto{0x10_15_my_f4v0ur173_by7e}
You either know, XOR you don’t
I’ve encrypted the flag with my secret key, you’ll never be able to guess it.
0e0b213f26041e480b26217f27342e175d0e070a3c5b103e2526217f27342e175d0e077e263451150104
The flag must start with “crypto{“, I XOR the encrypted message with this word to have the key. The key is “myXORke”, but when XOR it with cipher I didn’t have } at the last position. Then I append the y character to key. Tada !!!
c = "0e0b213f26041e480b26217f27342e175d0e070a3c5b103e2526217f27342e175d0e077e263451150104"
s1 = bytes.fromhex(c)
key = pwnlib.util.fiddling.xor(s1[:7], b'crypto{').decode()
print(key)
# myXORke
key += 'y'
s = pwnlib.util.fiddling.xor(key.encode(), s1).decode()
print(s)
Flag:
crypto{1f_y0u_Kn0w_En0uGH_y0u_Kn0w_1t_4ll}
Lemur XOR
I’ve hidden two cool images by XOR with the same secret key so you can’t see them!
- Lemur image
- Flag image
The png file is a structure file, we only need to xor the 2d-array pixel in data part. I use cv2 bitwise xor to do.
import cv2
img1 = cv2.imread('flag.png')
img2 = cv2.imread('lemur.png')
solve = cv2.bitwise_xor(img1, img2)
cv2.imwrite('solve.png', solve)
Solve image
Flag:
crypto{X0Rly_n0t!}
Mathematics
Greatest Common Divisor
Now calculate gcd(a,b) for a = 66528, b = 52920 and enter it below.
def gcd(a, b):
if a==0:
return b
else:
return gcd(b%a, a)
a = 66528
b = 52920
gcd(a,b)
# 1512
Result: 1512
Extended GCD
Using the two primes p = 26513, q = 32321, find the integers u,v such that:
p * u + q * v = gcd(p,q)
Enter whichever of u and v is the lower number as the flag.
def gcdExtended(a, b):
if a == 0 :
return b,0,1
gcd,x1,y1 = gcdExtended(b%a, a)
x = y1 - (b//a) * x1
y = x1
return gcd,x,y
p = 26513
q = 32321
gcd, u, v = gcdExtended(p, q) # (1, 10245, -8404)
print(min(u, v)) # -8404
Result: -8404
Modular Arithmetic 1
Calculate the following integers:
11 ≡ x mod 6
8146798528947 ≡ y mod 17
The solution is the smaller of the two integers.
r1 = 11 % 6
r2 = 8146798528947 % 17
print(min(r1, r2))
# 4
Result: 4
Modular Arithmetic 2
Calculate 27324678765465536 mod 65537
Fermat little Theorem: If p is a prime and a is any integer not divisible by p, then a^(p − 1) − 1 is divisible by p.
273246787654^65536 - 1 === 0 mod 65537 so 273246787654^65536 === 1 mod 65537
273246787654 % 65537 == 31167
# True
# Not divisible
Result: 1
Modular Inverting
Example: 7 * 8 = 56 ≡ 1 mod 11 What is the inverse element: 3 * d ≡ 1 mod 13?
gcdExtended(3, 13)
# (1, -4, 1)
Result: -4
DATA FORMATS
Privacy-Enhanced Mail?
Extract the private key d as a decimal integer from this PEM-formatted RSA key. privacy_enhanced_mail.pem
PEM is a format file to encapsulate the encrypted key. To get the key we read the pem file on asn1 format. RSA_DER
Run command below to extract info from pem file. It’s include:
!openssl asn1parse -i -in privacy_enhanced_mail.pem
# Result
0:d=0 hl=4 l=1187 cons: SEQUENCE
4:d=1 hl=2 l= 1 prim: INTEGER :00
7:d=1 hl=4 l= 257 prim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
268:d=1 hl=2 l= 3 prim: INTEGER :010001
273:d=1 hl=4 l= 256 prim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
533:d=1 hl=3 l= 129 prim: INTEGER :EDFB4715EBA93BC4C2CBE712C8081027CC86A8D28D2C78C9720E6DE6F68031E0E34FFA5EEF0FD1D085AE49C0A800388BF7EE98A94A77E1181E603924B3B3BB9DCE97B80062F2830C8F11983DFADD55F1F9CE5362992E14C25F776EF7DACEEB719E1CF9F2F62F4BA6D003DE4D427EEB5A4D9815644FCE1255931BDF2BA37FC7A7
665:d=1 hl=3 l= 129 prim: INTEGER :DE9DB5C35D2562F1CD3622342818C7BEBA0333207EDFDBC3F2648E6D1410B8914974A5AE32AFA8E4EAE40B42ADA5867E1B0E332FD0D0A2C8A9DE1ADBEDBD81F9BAB4C8FEC8CE3E660155E2CD04C6925B93FD88AFBE05DCC552A836E353A931209B23A13E7EB0F8FA919C44AC485CE37D6ADA8530AB56899C6669D44C5874AEFD
797:d=1 hl=3 l= 128 prim: INTEGER :444CDEBCFAD2AA35B15685EE0CFCCB6E30B3E115F4B073C614F6F131DD43338D808FBEA2AA67D6E6CAC717A1B455C3E4DFF6595814E84CF0F81ED3A7A5EF8A8422FBC6324E339DCAE7F0BBC9E60ACA14D58612C67482163126E70731195A53965B33A3C4C84510A8428129B6F0C3AE564F78BB82FBA87FF8916CE96303DCB377
928:d=1 hl=3 l= 129 prim: INTEGER :D3F4F73E16EEE4E173510A89FC6F73A79E3633B4C9F85CA7999FB2981AD5BCD5E049A70250123E4E0F73A7610A32A2F668CE4160528283AB694926EBA5D59CEE689D7F0E4FA5477619E96B73670BA60879C49923335B2393E11A76804584BF58DB3DB665E97C98E30246F67FCEBA5A836C7CB8F9D8F92136FFAFDDC9FF22C205
1060:d=1 hl=3 l= 128 prim: INTEGER :76BC5D830BCC7EB721E87AF55645FFB8CEDDDDE56782E5304613D0117BB329DF7BDABAC7BB3489AF5B7FAFD00A498EC4F0BCEBCAA138C8124B8F0BF9330A9903504A6F5BF68CB620B94B034283B17E4EFC5A328B3D6C730AFB9E1EAD67EB5540246F16F88810691A5DD12204DE1E4DB7237DCE6677FBBD780E4DDB53F381DFC6
I use Crypto librabry to convert hex in line 273 to long.
from Crypto.Util import number
hex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
number.bytes_to_long(bytes.fromhex(hex))
# 15682700288056331364787171045819973654991149949197959929860861228180021707316851924456205543665565810892674190059831330231436970914474774562714945620519144389785158908994181951348846017432506464163564960993784254153395406799101314760033445065193429592512349952020982932218524462341002102063435489318813316464511621736943938440710470694912336237680219746204595128959161800595216366237538296447335375818871952520026993102148328897083547184286493241191505953601668858941129790966909236941127851370202421135897091086763569884760099112291072056970636380417349019579768748054760104838790424708988260443926906673795975104689
CERTainly not
Find the modulus of the certificate, giving your answer as a decimal.
2048b-rsa-example-cert.der At first, I convert der file to pem file then get the public key. As the above challenges, I parst the string in line 19 then turn it to long number.
!openssl x509 -inform der -in 2048b-rsa-example-cert.der -out cert.pem
!openssl x509 -pubkey -in cert.pem -out 2048b-rsa-example-cert.pem
!openssl asn1parse -i -in 2048b-rsa-example-cert.pem
# 0:d=0 hl=4 l= 290 cons: SEQUENCE
# 4:d=1 hl=2 l= 13 cons: SEQUENCE
# 6:d=2 hl=2 l= 9 prim: OBJECT :rsaEncryption
# 17:d=2 hl=2 l= 0 prim: NULL
# 19:d=1 hl=4 l= 271 prim: BIT STRING
!openssl asn1parse -i -in 2048b-rsa-example-cert.pem -strparse 19
# 0:d=0 hl=4 l= 266 cons: SEQUENCE
# 4:d=1 hl=4 l= 257 prim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
# 265:d=1 hl=2 l= 3 prim: INTEGER :010001
hex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
number.bytes_to_long(bytes.fromhex(hex))
# 22825373692019530804306212864609512775374171823993708516509897631547513634635856375624003737068034549047677999310941837454378829351398302382629658264078775456838626207507725494030600516872852306191255492926495965536379271875310457319107936020730050476235278671528265817571433919561175665096171189758406136453987966255236963782666066962654678464950075923060327358691356632908606498231755963567382339010985222623205586923466405809217426670333410014429905146941652293366212903733630083016398810887356019977409467374742266276267137547021576874204809506045914964491063393800499167416471949021995447722415959979785959569497
SSH Keys
Extract the modulus n as a decimal integer from Bruce’s SSH public key.
I use ssh-keygen to get the pem format. Then do the same CERTainly not challenge.
!ssh-keygen -f bruce_rsa.pub -e -m PKCS8 >> bruce_rsa.pem
!openssl asn1parse -i -in bruce_rsa.pem
# 0:d=0 hl=4 l= 418 cons: SEQUENCE
# 4:d=1 hl=2 l= 13 cons: SEQUENCE
# 6:d=2 hl=2 l= 9 prim: OBJECT :rsaEncryption
# 17:d=2 hl=2 l= 0 prim: NULL
# 19:d=1 hl=4 l= 399 prim: BIT STRING
!openssl asn1parse -i -in bruce_rsa.pem -strparse 19
# 0:d=0 hl=4 l= 394 cons: SEQUENCE
# 4:d=1 hl=4 l= 385 prim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
# 393:d=1 hl=2 l= 3 prim: INTEGER :010001
hex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
number.bytes_to_long(bytes.fromhex(hex))
# 3931406272922523448436194599820093016241472658151801552845094518579507815990600459669259603645261532927611152984942840889898756532060894857045175300145765800633499005451738872081381267004069865557395638550041114206143085403607234109293286336393552756893984605214352988705258638979454736514997314223669075900783806715398880310695945945147755132919037973889075191785977797861557228678159538882153544717797100401096435062359474129755625453831882490603560134477043235433202708948615234536984715872113343812760102812323180391544496030163653046931414723851374554873036582282389904838597668286543337426581680817796038711228401443244655162199302352017964997866677317161014083116730535875521286631858102768961098851209400973899393964931605067856005410998631842673030901078008408649613538143799959803685041566964514489809211962984534322348394428010908984318940411698961150731204316670646676976361958828528229837610795843145048243492909
Transparency
Find the subdomain of cryptohack.org which uses these parameters in its TLS certificate, and visit that subdomain to obtain the flag.
TLS search commonly used sha256 encryption, em will generate sha256 file after converting pem file to der file. Then I use https://search.censys.io/ to find the domain using this certificate.
Web: Subdomain
Flag: crypto{thx_redpwn_for_inspiration}