# Challenge Name: baby
![date](https://img.shields.io/badge/date-10.07.2021-brightgreen.svg)  ![solved in time of CTF](https://img.shields.io/badge/solved-in%20time%20of%20CTF-brightgreen.svg)   ![crypto category](https://img.shields.io/badge/category-Crypto-blueviolet.svg)   ![value](https://img.shields.io/badge/value-102-blue.svg)  
## Description
I want to do an RSA!
Author : EvilMuffinHa
[output.txt](output.txt)
## Detailed solution
We have an RSA encryption```n: 228430203128652625114739053365339856393e: 65537c: 126721104148692049427127809839057445790```   An RSA modulus is the product of two secret prime numbers p and q : N = pq. for factoring we gonna use factordb
![image](https://user-images.githubusercontent.com/72421091/125488201-965b12bd-a637-4295-9d30-a7443e2066a5.png)
We need to calcul :   `phi = ( q — 1 ) * ( p — 1 )` Euler’s Totient for the number N.  `d = inverse( e, phi )` the private key with the MODULAR INVERSE of e and phi.  
To decrypt the message :
`m ≡ c^d (mod n)`
I'm gonna use the python library **Pycryptodome** [baby.py](baby.py)
```pythonfrom Crypto.Util.number import long_to_bytes, inverse
n = 228430203128652625114739053365339856393e = 65537c = 126721104148692049427127809839057445790
p = 12546190522253739887q = 18207136478875858439
phi = (p-1)*(q-1)
d = inverse(e, phi)
flag = pow(c,d,n)
print(long_to_bytes(flag).decode())``` 
## Flag
```flag{68ab82df34}```