# Really Smart Acronym

16 Jul 2020### Challenge Description:

```
Man, oracles are weird.
nc challenges1.hexionteam.com 5000
```

**Solving :**

Really Smart Acronym, of course, is RSA. Looking at the code, it uses PyCrypto to generate a RSA key to encrypt the flag. You also get one encryption and 1024 decrypts, but you only get the last bit of the decrypts. At first we thought it could be Franklin-Reiter related-message attack, but there is not enough information for that.

This thread was one of the resources that we used to solve this problem. From there, the method is given, but we still need to find e and N for the attack to work. e is easy. Since it uses PyCrypto, `e = 65537`

. As for N, that’s what the one encrypt is for. If you realize you can pass negative numbers for the encrypt, then it becomes easy. The encrypted message is `m^e mod N`

, so if you pass `-1`

, it becomes $-1^{65537}$ mod N = N – 1`.

**Solve Script :**

```
from pwn import *
sh = remote('challenges1.hexionteam.com', 5000)
sh.recvuntil('Flag: ')
cipher = int(sh.recvline().decode().strip())
#print(cipher)
sh.recvuntil('m => ')
sh.sendline('-1')
n = int(sh.recvline().decode().strip()) + 1
mult = pow(2, 65537, n)
#print(mult)
def get_lsb(num):
sh.recvuntil('> ')
sh.sendline(str(num))
return int(sh.recvline().decode().strip())
high = n
low = 0
for i in range(1024):
cipher *= mult
cipher %= n
lsb = get_lsb(cipher)
if lsb == 0:
high = (high + low) // 2
else:
low = (high + low) // 2
print(high)
```

Flag:

hexCTF{n1c3_r5a_tr1ck5_m4t3}