# Strip

cryptoCTF 2020
Cryptography
ac01010
Aug. 15th, 2020
RSA

Sometimes it is the people no one imagines anything of who do the things that no one can imagine.

# Strip

## 285 points, 11 solves, First Blood

First, I noticed that a(n) was just a recursive function that generated the product of Mersenne numbers - {1, 3, 21, 315, 9765, 615195, 78129765...}. Knowing this, we can find a multiple of phi by multiplying the same Mersenne numbers together. Each factor of n can be represented in the form 2^n - 1 - it's corresponding factor in phi could be represented as (2^n - 1 - 1), which is just double the previous Mersenne number.

Knowing this, we can repeatedly attempt to decrypt c using an increasingly large phi. Searching for "CCTF{" in the decrypted cs, we find the flag.

``````enc = 616666....
temp = 2
phi = 1
for i in range(1000):
phi*=(2^temp)-1
a = hex(pow(enc,inverse_mod(65539,phi*2),phi))
if "434354467b" in a:
print(bytes.fromhex(a[2:]))
break
``````

Flag: CCTF{R3arR4n9inG_7He_9Iv3n_eQu4t10n_T0_7h3_mUcH_MOrE_traCt4bLe_0n3}