BUUCTF--rsa

admin 2024年4月25日19:39:33BUUCTF--rsa已关闭评论1 views字数 891阅读2分58秒阅读模式
模数: C0332C5C64AE47182F6C1C876D42336910545A58F7EEFEFC0BCAAF5AF341CCDD
指数: 65537 (0x10001)

获取到

e = 65537

n=86934482296048119190666062003494800588905656017203025617216654058378322103517(模数转换为十进制)

公钥解析(提取p,q)

使用yahu或者http://www.factordb.com/index.php?query=86934482296048119190666062003494800588905656017203025617216654058378322103517 通过n解析p,q

p = 285960468890451637935629440372639283459

q = 304008741604601924494328155975272418463

脚本解密

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))

get flag!

flag{decrypt_256}

想要详细了解CTF中的rsa,可以看看https://err0rzz.github.io/2017/11/14/CTF%E4%B8%ADRSA%E5%A5%97%E8%B7%AF/

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  • 本文由 发表于 2024年4月25日19:39:33
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