Asmuth Shares

Description

Asmuth wants to share something with you!

• chal.py

• out.txt

TL;DR

• Use Chinese Remainder Theorem on all shares.

• Use result from CRT and perform mod m0 to get long value of FLAG.

• Stonks

Solution

Analysis

out.txt

Public (m0): 0x61f42f24878101c5186ec677cd44a2b14071b9a69260a026fb385482123f2bd647f74e8c569d3cad1c2d48ad6da58ceddd021c5d

Share 0 (s0, m1): (0x60ec3bb546f5905293052cc6154d632942ef447f3983665f34651637c34c384970abf40095368ffd426f5139ec3d1e07a868a830, 0x6307ff90da6fa375ccf1803b158219da8b9e44fb6859c2f18ce7095584136dfd805a5ac4ecea326c10863ed6e2e39ac82abc9c49)
Share 1 (s1, m2): (0x58675225153f997cd4f4d5bcc0ec9febf60ca7e7aa1759ae8a7037ad0dc4ce45ec79e49c3a1f8bb011adadc57e7d8a39bb845cad, 0x6cdc758a8d408a4ef39acd33f9677c268d685c9ddb3276fb970ad6fe2ebe549daace27fd7b05ca495ba3fe7fef9f9bbd23baf44d)
Share 2 (s2, m3): (0x3d045f39ea1b2f9a5486b16fef3b9a99fd68de9d2d9235c14b6254d3dedb40f7bc8f2087de7a111671426d83b6c9f1709ab90d89, 0x727113b40e9c974febb4e7cb2952083f96a97d1b17093963ee7c474fe780f5bf31241b540470c949593b363ca2c84d73584eb2f3)
Share 3 (s3, m4): (0x3e27c86088d7c7725814c9efe17909c408a1ddca3f0397c81d142b03d6329ef30941b558f22dc8bc20e752e4a6088a7067e76565, 0x7739b291c0ba89eec860dc8a6b018acb5d05d879504285a079b2785edac2cd543c6a86589071835d74f63524952910944b0977e7)

chal.py

We are given the values of quite a few variables in the out.txt file.

Reading the code for chal.py, we can see that it encrypts the flag with the use of nearly all values given in out.txt. Should be fairly simple to decrypt the flag.

Analyzing chal.py

The flag is first converted into a long integer. Then a value N is assigned <number of bits of the flag> + 2**7. It then randomly generates 5 different prime numbers with N bits and stores them into a sorted list m.

FLAG = bytes_to_long(open("flag.txt", "rb").read())
N = FLAG.bit_length() + 2**7

m = sorted([ getPrime(N) for _ in range(5) ])

Then is just loops through until m[4] is a prime number such that m[0]*m[3]*m[4] is more than m[1]*m[2]*m[3], and m[4] is more than m[3]. This part isn't really important.

while ((m[1]*m[2]*m[3] < m[0]*m[3]*m[4]) and m[4]<m[3]):
	m[4] = getPrime(N)

It then generates a number and assigns it to the variable alpha. We can't really find the value of alpha as it is a random number and M is a pretty large number.

M = m[1]*m[2]
alpha = getRandomRange(2, M)
k = 4

4 values are then generated and stored in s. This is the important part of the encryption. Last few lines are just to write out the values to out.txt.

for i in range(k):
    s = (FLAG + alpha*m[0]) % m[i+1]
    share.append(s)

From this part of the code, we have all the values of every variable except for alpha.

Take note that the portion s = (FLAG + alpha*m[0]) % m[i+1] repeats 4 times, with different s results, using m[1] to m[4] as the modulus at each iteration, and (FLAG + alpha*m[0]) being the same at each iteration (the number being modulo is the same each time).

So we can form 4 equations given the values we got from out.txt:

• s0 = (FLAG + alpha*m[0]) % m1

• s1 = (FLAG + alpha*m[0]) % m2

• s2 = (FLAG + alpha*m[0]) % m3

• s3 = (FLAG + alpha*m[0]) % m4

If you are unfamiliar with these equations, we can basically solve for the value of (FLAG + alpha*m[0]) using the Chinese Remainder Theorem (CRT).

Since I haven't officially study Number Theory and I mostly read about these algorithms online, I may not be the best at explaining it. You can search on Youtube or Wikipedia on how it works.

We can easily implement CRT in Python by doing so:

from sympy.ntheory.modular import crt
x = crt(mod,res)[0] 	# mod and res has to be lists

We can substitute mod for a list containing m1,m2,m3,m4.

We can substitute res for a list containing s0,s1,s2,s3.

With this, we retrieved the value of (FLAG + alpha*m[0]).

To get value of FLAG, notice how m0 is an extremely large number and how it is multiplied to alpha. Thus, we can get rid of the alpha*m[0] part just by performing mod m0 on the value!

(FLAG + alpha*m0) % m0 = FLAG % m0 + (alpha*m0) % m0
(alpha*m0) % m0 = 0
FLAG % m0 = FLAG       # Since it is most likely that FLAG < m0 as m0 is an extremely large number
# Therefore
(FLAG + alpha*m0) % m0 = FLAG

Solve.py

#!/usr/bin/env python3

from sympy.ntheory.modular import crt
from Crypto.Util.number import long_to_bytes

s0 = 0x60ec3bb546f5905293052cc6154d632942ef447f3983665f34651637c34c384970abf40095368ffd426f5139ec3d1e07a868a830
m1 = 0x6307ff90da6fa375ccf1803b158219da8b9e44fb6859c2f18ce7095584136dfd805a5ac4ecea326c10863ed6e2e39ac82abc9c49

s1 = 0x58675225153f997cd4f4d5bcc0ec9febf60ca7e7aa1759ae8a7037ad0dc4ce45ec79e49c3a1f8bb011adadc57e7d8a39bb845cad
m2 = 0x6cdc758a8d408a4ef39acd33f9677c268d685c9ddb3276fb970ad6fe2ebe549daace27fd7b05ca495ba3fe7fef9f9bbd23baf44d

s2 = 0x3d045f39ea1b2f9a5486b16fef3b9a99fd68de9d2d9235c14b6254d3dedb40f7bc8f2087de7a111671426d83b6c9f1709ab90d89
m3 = 0x727113b40e9c974febb4e7cb2952083f96a97d1b17093963ee7c474fe780f5bf31241b540470c949593b363ca2c84d73584eb2f3

s3 = 0x3e27c86088d7c7725814c9efe17909c408a1ddca3f0397c81d142b03d6329ef30941b558f22dc8bc20e752e4a6088a7067e76565
m4 = 0x7739b291c0ba89eec860dc8a6b018acb5d05d879504285a079b2785edac2cd543c6a86589071835d74f63524952910944b0977e7

m0 = 0x61f42f24878101c5186ec677cd44a2b14071b9a69260a026fb385482123f2bd647f74e8c569d3cad1c2d48ad6da58ceddd021c5d

mod = [m1,m2,m3,m4]
res = [s0,s1,s2,s3]

x = crt(mod,res)[0]

print(long_to_bytes(x%m0))

# flag: STANDCON22{CRT_F0r_7h3_W1n_38a9c56b}