## [Original/Source Writeup](https://bigpick.github.io/TodayILearned/articles/2020-10/b01lersbootcamp#clear-the-mind)
We’re given:
```pythonn = 102346477809188164149666237875831487276093753138581452189150581288274762371458335130208782251999067431416740623801548745068435494069196452555130488551392351521104832433338347876647247145940791496418976816678614449219476252610877509106424219285651012126290668046420434492850711642394317803367090778362049205437
c = 4458558515804625757984145622008292910146092770232527464448604606202639682157127059968851563875246010604577447368616002300477986613082254856311395681221546841526780960776842385163089662821
e = 3```
So, this time, it is RSA, but with a poor public exponent (e = 3). We can crack this by simply taking the cube root, assuming that it is unpadded.
```python#!/usr/bin/env python3
# Inspired by / taken from:#  https://baotdvi.wordpress.com/2018/11/28/safe-rsa-picoctf-2018/
import gmpy2import binascii
N = 102346477809188164149666237875831487276093753138581452189150581288274762371458335130208782251999067431416740623801548745068435494069196452555130488551392351521104832433338347876647247145940791496418976816678614449219476252610877509106424219285651012126290668046420434492850711642394317803367090778362049205437ct = 4458558515804625757984145622008292910146092770232527464448604606202639682157127059968851563875246010604577447368616002300477986613082254856311395681221546841526780960776842385163089662821e = 3
with gmpy2.local_context(gmpy2.context(), precision=300) as ctx:    cube_root = gmpy2.cbrt(ct)    print(f"Cube root: {cube_root}")    hex0 = str(hex(int(cube_root)))    print(f"Hex: {hex0}")    ascii_ = binascii.unhexlify(str(hex0)[2:len(hex0)]).decode()    print(f"Plaintext: {ascii_}")```
Which running, gives us:
```python cube.pyCube root: 164587995846552213349276905669580061809447554828318448024777341.000000Hex: 0x666c61677b77335f6e6565645f376f5f67305f6433657033727dPlaintext: flag{w3_need_7o_g0_d3ep3r}```
Flag is `flag{w3_need_7o_g0_d3ep3r}`.