# Massive RSA
## Description
I was scared that my RSA would be broken, so I made sure that the numbers were massive.
[massive.txt](massive.txt)
## Solution
The n is actually a prime.
So phi(n) = n-1. Perform the modded inverse of e and decrypt the message.
```d=modinv(e,n-1)import codecsprint(codecs.decode(hex(pow(c,d,n))[2:],'hex'))```
The flag is```hsctf{forg0t_t0_mult1ply_prim3s}```