# Task
I set up a secret sharing system with my friends, but now my friends all hate me. Can you help me get the secret anyway?
*Attachment with the source code:*
```pythonimport random, sysfrom crypto.util.number import long_to_bytes
def bxor(ba1,ba2):	return bytes([_a ^ _b for _a, _b in zip(ba1, ba2)])
bits = 128shares = 30
poly = [random.getrandbits(bits) for _ in range(shares)]flag = open("/challenge/flag.txt","rb").read()
random.seed(poly[0])print(bxor(flag, long_to_bytes(random.getrandbits(len(flag)*8))).hex())
try:	x = int(input('Take a share... BUT ONLY ONE. '))except:	print('Do you know what an integer is?')	sys.exit(1)if abs(x) < 1:	print('No.')else:	print(sum(map(lambda i: poly[i] * pow(x, i), range(len(poly)))))```
# Task analysis
First thing that catches the eye is that this "secret sharing system" isn't actually a polite one - it responses with a heartfelt insult if we dare to send it something but integer. Maybe that's the real reason why all friends hate you now, Shamir? Anyway, after hours of googling what an integer is and overcoming a sudden impostor syndrom attack I was able to finally recover from such a terrible assault and start investigating source code. Below is a scratch of how it works:- 30 random numbers (`shares`) 128 bits long are generated when we connect to server and are put into array called `poly`- First `share` of `poly` is used as seed for a subsequent number generation- Each flag character is XOR ed with randomly generated bits and result is printed to us as a hex value- Then we are requested to enter aforementioned integer `x`- `x` is used to calculate sum `poly[i] * x ** i`, where `i = 0, 1 ... 29`. This sum is also printed
# Finding vulnerability
So far, we are given two values:1. Encrypted flag2. Some sum of `shares` that depends on our input `x`
To get the flag we need to decrypt it. To decrypt the flag we need to know the key. Once we get it, we can simply XOR it with the encrypted flag for decryption. Key is some randomly generated number. Fortunately for us, first `share` of `poly` is used as a seed for key generation. So if we know the `poly[1]` then we'll be able to replicate key generation.
We know that `poly[0]` along with other `shares` is used to calculated the sum of shares. This sum also depends on `x`, our input. Let's investigate deeper the relation between `x` and sum of `shares` and see what we can get out of it.
For demonstration purpose, let's consider:`poly = [42, 33, 54]`for `x = 3````42 * (3^0) = 4233 * (3^1) = 9954 * (3^2) = 486sum = 42 + 99 + 486 = 627```Hope that now it's clear how the sum is calculated. But `x = 3` is not that useful for us - there's no way we can get `poly[1]` which is `42` from `627`. Let's consider more interesting case where `x = 1000````42 * (1000^0) =       4233 * (1000^1) =    3300054 * (1000^2) = 54000000sum = 42 + 33000 + 54000000 = 54033042```Bingo! You see it? Last two digits of sum is actually the first `share` that we are looking for. So, by entering `x = 10^n` where `n > len(poly[0])` we are able to easily extract first `share` from the resulting sum.
# Assembling exploit
Let's get `poly[1]` first. I am a lazy one, so I just did it manually: ```f79ace6c50045d9617387178738bc492c8a36bce6f62065ffd1712060127afTake a share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```
now all that is left to do is to generate key using `poly[0]` and XOR it with a given encrypted flag:
```python#!/usr/bin/python3
import randomfrom Crypto.Util.number import long_to_bytes, bytes_to_long
def bxor(ba1, ba2):    return bytes([_a ^ _b for _a, _b in zip(ba1, ba2)])
encrypted_flag = 'f79ace6c50045d9617387178738bc492c8a36bce6f62065ffd1712060127af'poly_0 = 5071636503793964919135745354381215807
random.seed(poly_0)key = long_to_bytes(random.getrandbits(len(encrypted_flag)*4))print(key)flag = bxor(long_to_bytes(int(encrypted_flag, 16)), key)print(flag.decode())```
After running this script we will get the flag:rarctf{n3v3r_trust_4n_1nt3g3r}
## Interesting factFormula for sum calculation is very similar to the positional system formula (https://en.wikipedia.org/wiki/Numeral_system#Positional_systems_in_detail) but the latter one has some restrictions