# EasyCTF - RSA 2
In this challenge, we are asked to decipher an RSA encoded message.We are given N, e and C.
N looks quite small, so there is a chance it has already been factored online.Using https://factordb.com, we find that `N = 338779901332693957541 * 381858611614052422763`.
There goes the rest of the challenge:
```python#!/usr/bin/python
import gmpy2
p = 338779901332693957541q = 381858611614052422763e = 0x10001c = 41786000149048295568588665335025364614298
phi = (p-1)*(q-1)d = gmpy2.invert(e, phi)m = pow(c, d, p*q)
print hex(m)[2:].replace('L','').decode('hex')```
The result is `flag{l0w_n_0868}`.