crypto
Hexahedron
Solved by: Taz34
 We are give values of n,e and c.

Decode them into simple numeric form using python.
 it looks like RSA, we used RsaCtfTool to decipher it.
: https://github.com/Ganapati/RsaCtfTool
python3 RsaCtfTool.py n 112339816301925396926211289689793745814213925314273886071305785874178028552510482239036537066616690493241410015435402110525284201411608164205573122430898583517515498250410244592963132324072861567753086739636553410154316180827724708002409356129254383468446158145079982391991062389788544378839486986385137994309 e 3 uncipher 2217344750798178599616518881851238192046537371134831984828894413752520937378161486880269974456574131502921272953104454680926482208357166098075344508240480152890914678813031666242202555794691235412837030045499161787224264164243336308650477343133919653356349913604131486721125
here we have the flag.
N1TP
Solved By : choco
We are given an encrypted flag with randomized key. The hint is that this uses a one time pad encryption
The general gist of one pad encryption: You have a plain text: P = “HELLO” (8 5 12 12 15) We are given a key: K = “DREFX” (4 18 5 6 24)
The encryption comes in when we add both P and K to get encryption C = “L W Q R M” (12 23 17 18 13(39%26))
So P + K = (C%N) where N is the letter range To get the decipher text we simple subtract C with K P = “HELLO” (8 5 12 12 15(11%26)
So (P%N) = C  K
The vulnerability comes in if the same key K is used to encrypt another plain text provided we know the range of text used along with the length
Suppose P1 = “AAAAA” (1 1 1 1 1) P1 + K = C1 C1 = “ETFGY” (5 19 6 7 25)
We could easily get the key if it is within the modulus range C1  P1 = K K = “DREFX” (4 18 5 6 24)
We can now use this to get P with the given C
But another way to find P is using simple logic
C1  P1 = C  P So P = C  C1 + P1 All we need to do is find C  C1 to 0 and we can guess P
Using this logic we will find the flag The code for difference C  C1 is given below and difference is added to P1 and manually repeated until we get C  C1 as 0
import binascii
A = "dc0de91facbe90ee6f652167906ee0d17123cf9e746a63db4b4e7d93040f59331ead9be0b2fe"
Ahex = binascii.unhexlify(A)
B = "dc0de91facbe90ee6f652167906ee0d17123cf9e746a63db4b4e7d93040f59331ead9be0b2fe"
Al = list(bytearray(Ahex))
Bl = list(bytearray(binascii.unhexlify(B)))
C = "flag{00000000000000000000000000000000}"
Chex = list(bytearray(C))
print(Chex)
d = []
for i in range(len(Al)):
d.append(Al[i]  Bl[i])
print(d)
k = []
c = []
for i in range(len(Al)):
k.append(Chex[i] + d[i])
if((48 <= k[i] <= 57) or (97 <= k[i] <= 102) or k[i] == 108 or k[i]==103 or k[i]==123 or k[i]==125):
c.append(chr(k[i]))
else:
c.append("#")
print(k)
print(c)
flag: flag{9276cdb76a3dd6b1f523209cd9c0a11b}
TRIFORCE
Solved By : choco
This is a CBC AES encryption (Chained Block Cipher)
Suppose the plain text P is 64 bits long It is then split into 4 parts (16 bit each)
There is an initializer iv that is 16 bits long and key k (16 bits long) The cipher text for first part of P is
C0 = E(P0 ^ iv,k) and following chain goes like this Ci = E(Pi ^ Ci1,k) i > 0
The decryption does like this
Where P0 = D(C0,k) ^ iv and following decryption goes Pi = D(Ci,k) ^ Ci1
If the key k and vi are same, a vulnerability arises During decryption, P0 = D(C0,k) ^ k P1 = D(C1,k) ^ C0 P2 = D(C2,k) ^ C1
if we replace C2 with C0 P2 = D(C0,k) ^ C1
xor the result with any of the two inputs will give the other input D(C0,k) = P2 ^ C1 P0 = D(C0,k) ^ k k = D(C0,k) ^ P0 therefore, k = P2 ^ C1 ^ P0 so since we know P0,P2 and C1, we can easily find k
Hence all we need to do is to replace the 3rd block of ciphertext with the first block of cipher text, Xor the result of 3rd plain text block with 2nd cipher text block and xor that result with 1st plaintext block to get the key
Do this three times with the different keys above to get the flag
input: 616161616161616161616161616161616161616161616161616161616161616161616161616161616161616161616161
Code for replacing ciphertext:
def listToString(s):
str1 = ""
for ele in s:
str1 += ele
return str1
def chunk(in_string,num_chunks):
chunk_size = len(in_string)//num_chunks
if len(in_string) % num_chunks: chunk_size += 1
iterator = iter(in_string)
for _ in range(num_chunks):
accumulator = list()
for _ in range(chunk_size):
try: accumulator.append(next(iterator))
except StopIteration: break
yield ''.join(accumulator)
string = "e5cc7f05f279ca152d3fbd32d899b55fec8de5d9ae3d62bac51195ddedc0863462021d3c51a4a56555ca95ff4303b764a02a2f44327cb2007d46de117900720a"
l = list(chunk(string,4))
l[2] = l[0]
print(l)
out = listToString(l)
print(out)
code for getting hex for part of flag:
def chunk(in_string,num_chunks):
chunk_size = len(in_string)//num_chunks
if len(in_string) % num_chunks: chunk_size += 1
iterator = iter(in_string)
for _ in range(num_chunks):
accumulator = list()
for _ in range(chunk_size):
try: accumulator.append(next(iterator))
except StopIteration: break
yield ''.join(accumulator)
string = "6227616161616161616161616161616161616161616161616161616161616161e8c6e5dfb46432e2c2499084e899d462e6af4534aed0627d75f825c096970f36"
l = list(chunk(string,4))
c = "ec8de5d9ae3d62bac51195ddedc08634"
print(hex(int(l[2],16)))
d = int(l[2],16) ^ int(c,16)
print(hex(d))
e = d ^ int(l[0],16)
print(hex(e)) ![](https://i.imgur.com/TgyoAnU.jpg)
do this and replace each cipher for each key
each keys in hex = 666c61677b3831396639643864383337
32316163346334343262313635396633
36646632647d20202020202020202020
flag: **flag{819f9d8d83721ac4c442b1659f36df2d}**