BUUCTF - rsa - Code World

Test file: https://buuoj.cn/files/ed10ec009d5aab0050022aee131a7293/41c4e672-98c5-43e5-adf4-49d75db307e4.zip?token=eyJ0ZWFtX2lkIjpudWxsLCJ1c2VyX2lkIjoxOTAzLCJmaWxlX2lkIjoyMDV9.XYl7ag.BZcdhjGojXT726Y0wj4m2X_OOyw

1. File Analysis

The first time this topic, see the topic but know that this is a problem rsa decryption.

Public key n = p * q, where p and q are two large prime numbers

e is a randomly selected number, as the public key

d is a number associated with the e, satisfy the conditional formula: ed = 1 (mod phi (n))

phi (n) is the Euler function, phi (n) = (p-1) (q-1)

Transfer: https://blog.csdn.net/kevin66654/article/details/54087647

pub.key is the public key, flag.enc the rsa encrypted file, so long as we parsed n, e, p, q, d public key file, and then using a script file to decrypt the encrypted rsa.

2. The public process

2.1 extract the public key

IDA open pub.key, which converts a string of hexadecimal extracted public key.

-----BEGIN PUBLIC KEY-----
MDwwDQYJKoZIhvcNAQEBBQADKwAwKAIhAMAzLFxkrkcYL2wch21CM2kQVFpY9+7+
/AvKr1rzQczdAgMBAAE=
-----END PUBLIC KEY-----

2.2 Analytical public key (extract e, n)

In http://tool.chacuo.net/cryptrsakeyparse the public key is resolved

Public exponent and module information:

key length:	256
Model number:	C0332C5C64AE47182F6C1C876D42336910545A58F7EEFEFC0BCAAF5AF341CCDD
index:	65537 (0x10001)

Get to

e = 65537

n = 86934482296048119190666062003494800588905656017203025617216654058378322103517 (ADC decimal)

2.3 Analytical public key (extraction p, q)

Use yahu or http://www.factordb.com/index.php?query=86934482296048119190666062003494800588905656017203025617216654058378322103517 by parsing n p, q

p = 285960468890451637935629440372639283459

q = 304008741604601924494328155975272418463

3. decryption script

import gmpy2
import rsa

e = 65537
n = 86934482296048119190666062003494800588905656017203025617216654058378322103517
p = 285960468890451637935629440372639283459
q = 304008741604601924494328155975272418463

phin = (q-1)*(p-1)
d = gmpy2.invert(e, phin)

key = rsa.PrivateKey(n, e, int(d), p, q)

with open("C:\\Users\\10245\\Desktop\\output\\flag.enc", "rb+") as f:
    f = f.read()
    print(rsa.decrypt(f, key))

4.get flag！

flag{decrypt_256}

Want to learn more about the CTF's rsa, you can take a look at https://err0rzz.github.io/2017/11/14/CTF%E4%B8%ADRSA%E5%A5%97%E8%B7%AF/

BUUCTF - rsa