> Surrounded by an untamed forest and the serene waters of the Primus river, your sole objective is surviving for 24 hours. Yet, survival is far from guaranteed as the area is full of Rattlesnakes, Spiders and Alligators and the weather fluctuates unpredictably, shifting from scorching heat to torrential downpours with each passing hour. Threat is compounded by the existence of a virtual circle which shrinks every minute that passes. Anything caught beyond its bounds, is consumed by flames, leaving only ashes in its wake. As the time sleeps away, you need to prioritise your actions secure your surviving tools. Every decision becomes a matter of life and death. Will you focus on securing a shelter to sleep, protect yourself against the dangers of the wilderness, or seek out means of navigating the Primus’ waters?

Let's take a look at what we have:

// output.txt

```plaintext
n = 144595784022187052238125262458232959109987136704231245881870735843030914418780422519197073054193003090872912033596512666042758783502695953159051463566278382720140120749528617388336646147072604310690631290350467553484062369903150007357049541933018919332888376075574412714397536728967816658337874664379646535347
e = 65537
c = 15114190905253542247495696649766224943647565245575793033722173362381895081574269185793855569028304967185492350704248662115269163914175084627211079781200695659317523835901228170250632843476020488370822347715086086989906717932813405479321939826364601353394090531331666739056025477042690259429336665430591623215
```

The source file contains this:

```python
import math

from Crypto.Util.number import getPrime, bytes_to_long

from secret import FLAG

m = bytes_to_long(FLAG)

n = math.prod([getPrime(1024) for _ in range(2**0)])

e = 0x10001

c = pow(m, e, n)

with open('output.txt', 'w') as f:

    f.write(f'{n = }\n')
    f.write(f'{e = }\n')
    f.write(f'{c = }\n')
```

This is an implementation of the RSA algorithm, where we get the public key `n` and `e`, and the ciphertext `c`. The security issue lies in the fact that the public key `n` is generated by multiplying one prime number (as $2^0$ = 1), which makes it very easy to factorize the public key `n` into its prime factors, and then decrypt the ciphertext `c` using the private key. Read up on [RSA](https://ctf101.org/cryptography/what-is-rsa/) if you have zero clue what any of this means. We'll use Euler's totient function to decrypt the ciphertext.

### Solution

```python
from Crypto.Util.number import long_to_bytes
from sympy import factorint

n = 144595784022187052238125262458232959109987136704231245881870735843030914418780422519197073054193003090872912033596512666042758783502695953159051463566278382720140120749528617388336646147072604310690631290350467553484062369903150007357049541933018919332888376075574412714397536728967816658337874664379646535347

e = 65537

c = 15114190905253542247495696649766224943647565245575793033722173362381895081574269185793855569028304967185492350704248662115269163914175084627211079781200695659317523835901228170250632843476020488370822347715086086989906717932813405479321939826364601353394090531331666739056025477042690259429336665430591623215

# Factor n into p and q
factors = factorint(n)

p = next(iter(factors.keys()))
q = next(iter(factors.keys()))

# calculating the private exponent d
phi = (p - 1) * (q - 1)
d = pow(e, -1, phi)

# decrypting ciphertext c using private exponent

m = pow(c, d, n)

# decoding the flag to utf-8
flag = long_to_bytes(m)
print(flag.decode())
```

And the flag is:

```text
HTB{0h_d4mn_4ny7h1ng_r41s3d_t0_0_1s_1!!!}
```