# TJCTF – RSABC
* **Category:** crpyot* **Points:** 50
## Challenge
> I was just listening to some [relaxing ASMR](https://youtu.be/J2g3lvNkAfI) when a notification popped up with this.>> ???>> Attachments:> > n=57772961349879658023983283615621490728299498090674385733830087914838280699121> >> > e=65537> >> > c=36913885366666102438288732953977798352561146298725524881805840497762448828130
## Solution
hmmm I guess its a regular RSA and we don't need any hints because the numbers are small enough to factorize
so I used this handy dandy (website)[https://www.alpertron.com.ar/ECM.HTM] to factorize N : (it took me a couple minutes though)
```p = 202049603951664548551555274464815496697q = 285934543893985722871321330457714807993```
and from there it's easy to decrypt the ciphertext, I used a python [script](https://github.com/0d12245589/CTF-writeups/raw/master/2020/TJCTF/crypto/rsabc/solve.py) to do that
```tjctf{BOLm1QMWi3c}```