In RSA, a small e value can be problematic, but what about N? Can you decrypt this?

How RSA works

1. $c^d\equiv m(\mathrm{mod}n)$
2. $m^e\equiv c(\mathrm{mod}n)$
3. $d\equiv e^{-1}(\mathrm{mod}\varphi (n))$

here

$m$ is message(plaintext), and

$c$ is ciphertext

If

$n$ is small, we can crack it
(tool 1: https://www.dcode.fr/euler-totient tool 2: http://factordb.com/)
and know what

$p$ and

$q$ are.
After acquiring

$p$ and

$q$, we can calculate

$\varphi (n)=(p-1)(q-1)$, then calculate d=pow(e,-1,phin) using Python.

proof of RSA

(source: https://drive.google.com/file/d/1EURAqSaG2Mn0w0j70rI0Ph100CzXANty/view)

Writeup

1. Use FactorDB to get the value of
   $p$ and
   $q$
   
2. Use Python to decode it

from Crypto.Util.number import *
p=1593021310640923782355996681284584012117
q=521911930824021492581321351826927897005221
c=240986837130071017759137533082982207147971245672412893755780400885108149004760496
e=65537

n=p*q
phin=p*q-p-q+1#(p*q-1)-(q-1)-(p-1)
d=pow(e,-1,phin)
m=pow(c,d,n)
print(m)
print(long_to_bytes(m))

Output:

┌──(kali㉿kali)-[~]
└─$ /bin/python /home/kali/code/rsa.py
13016382529449106065927291425342535437996222135352905256639592405461024281868413
b'picoCTF{sma11_N_n0_g0od_23540368}'

Easier way

note that ✔ P,Q computed with N (FactorDB database) is mentioned