# supercomputer

by [haskal](https://awoo.systems)

crypto / 457 pts / 85 solves

>I ran this code with a supercomputer from the future to encrypt the flag, just get your own and
>decryption should be simple!

provided files: [supercomputer.py](supercomputer.py) [output.txt](output.txt)

## solution


first, do some source review. there is not a lot of it
```python
def v(p, k):
    ans = 0
    while k % p == 0:
        k /= p
        ans += 1
    return ans
```
this seems to be computing the number of times `p` is a factor of `k`

```python
p, q, r = getPrime(2048), getPrime(2048), getPrime(2048)
print(p, q, r)
n = pow(p, q) * r
```
this is taking a very very large power and multiplication (don't actually run this it'll never
finish lol -- it was run on a supercomputer from the future but it can't be run on our computers)

```python
a1 = random.randint(0, n)
a2 = n - a1
assert a1 % p != 0 and a2 % p != 0
t = pow(a1, n) + pow(a2, n)
print(binascii.hexlify(xor(flag, long_to_bytes(v(p, t)))))
```

and here we generate a random number, subtract it from n, take the power and then use the
factor-counting function to xor the flag

now i don't know how to math at all so here i decided to run this with some much smaller primes and
see what happens (dynamic analysis ftw). try replacing `p, q, r` with combinations of 3, 5, 7, and
11.  you'll notice immediately that the result is always `2q`. so let's try it

```python
[ins] In [1]: import ast

[ins] In [2]: with open("output.txt") as f:
         ...:     [p,q,r] = [int(x) for x in f.readline().strip().split(" ")]
         ...:     enc = ast.literal_eval(f.readline())

[ins] In [3]: from pwn import xor

[ins] In [4]: from binascii import unhexlify

[ins] In [5]: enc = unhexlify(enc)

[ins] In [6]: v = 2 * q

[ins] In [7]: xor(enc, v.to_bytes(v.bit_length()//8+1, "big"))
Out[7]: b'corctf{1_b3t_y0u_d1dnt_4ctu411y_d0_th3_m4th_d1d_y0u?}\ncorctf{1_b3t_y0u_d1dnt_4ctu411y_d0_th3_m4th_d1d_y0u?}\ncorctf{1_b3t_y0u_d1dnt_4ctu411y_d0_th3_m4th_d1d_y0u?}\ncorctf{1_b3t_y0u_d1dnt_4ctu411y_d0_th3_m4th_d1d_y0u?}\ncorctf{1_b3t_y0u_d1dnt_4ctu411y_d0_th3_m4'
```

yea