# Little e | Crypto
### Solution
From ctf we have got encrypted.txt file.
##### encrypted.txt```txtN: 17260541145579198891286838820507756585494408484294770862002805660577661138753926064444981930310528890773266098356882517290033235056654103412024620204547445159627671127518965960486480229617902782023368077819854820837387791591683428592246121552228695417314295153721144499366280389254552597040315734269703314601367296670296596449184388246232676724558126845148454433428619114310396032130452938580427564701080866025573039407415733982384144637567337164211240182263026767455207192185903776322079215198832973810587735067694801737731617941390472537291532113711292822595269219795255224521641891049508289784761579371125213474439e: 3ct: 1112413624683819960899152482895461211039349964898672381675850025556800617245120168928400758297834676330400246617472191750627367991315450127361583383350639760738254818244740474313061192563860605923503717```
In this challenge we have to do something called **low public exponent attack**. I started from searching python script for it, and I found one:```pyimport gmpy2
n = 17260541145579198891286838820507756585494408484294770862002805660577661138753926064444981930310528890773266098356882517290033235056654103412024620204547445159627671127518965960486480229617902782023368077819854820837387791591683428592246121552228695417314295153721144499366280389254552597040315734269703314601367296670296596449184388246232676724558126845148454433428619114310396032130452938580427564701080866025573039407415733982384144637567337164211240182263026767455207192185903776322079215198832973810587735067694801737731617941390472537291532113711292822595269219795255224521641891049508289784761579371125213474439e = 3ct = 1112413624683819960899152482895461211039349964898672381675850025556800617245120168928400758297834676330400246617472191750627367991315450127361583383350639760738254818244740474313061192563860605923503717
m, result = gmpy2.iroot(ct,e)
try:    flag = binascii.unhexlify(format(m, 'x')).decode()except Exception as e:    flag = m
hexed = hex(flag)[2:]print(hexed)string = bytearray.fromhex(hexed).decode()print(string)```
### Flag
bcactf{R54_N0T_50_S3CUR3_33}
#### Credits
- Writeup by [HuntClauss](https://ctftime.org/user/106464)- Solved by [HuntClauss](https://ctftime.org/user/106464)- WaletSec 2021
#### License
**CC BY 4.0** WaletSec + HuntClauss