# Easy RSA> Points: 407
## Description> Just a easy and small E-RSA for you :)>>[File](https://mega.nz/file/600TkQbK#0o6mqJjLxReiBoP3HAtsYj8ulp9K99246EdzmeVNiS4)
## Solution
A very simple RSA form :) The modulo **N** isn't given. Why?
Because we don't need it!
Assuming the **N** to be a big 2048-bit number (general format) and my plaintext (flag) to be relatively small it's clear that `(pt ^ e) < N`
This is the vulnerabilty as `a mod b = a when a < b` so ct = (pt ^ e) mod N becomes equivalent to ct = (pt ^ e).
Taking e-th root of ciphertext will retrieve the plaintext (flag).
```py#!/bin/env python3
from Crypto.Util.number import long_to_bytesimport gmpy2
ct = 70415348471515884675510268802189400768477829374583037309996882626710413688161405504039679028278362475978212535629814001515318823882546599246773409243791879010863589636128956717823438704956995941e = 3
# Calculating e-th root of ciphertextpt = gmpy2.iroot(ct,e)[0]print("Flag is : " + str(long_to_bytes(pt).decode()))```
## Flag> darkCTF{5m4111111_3_4tw_xD}