信息收集

下载，解压 (密码都是hackthebox)

enc后缀的文件是一种加密文件，打开为乱码，key.pub内容如下，是 RSA 加密算法的公钥

-----BEGIN PUBLIC KEY-----
MIIBHzANBgkqhkiG9w0BAQEFAAOCAQwAMIIBBwKBgQMwO3kPsUnaNAbUlaubn7ip
4pNEXjvUOxjvLwUhtybr6Ng4undLtSQPCPf7ygoUKh1KYeqXMpTmhKjRos3xioTy
23CZuOl3WIsLiRKSVYyqBc9d8rxjNMXuUIOiNO38ealcR4p44zfHI66INPuKmTG3
RQP/6p5hv1PYcWmErEeDewKBgGEXxgRIsTlFGrW2C2JXoSvakMCWD60eAH0W2PpD
qlqqOFD8JA5UFK0roQkOjhLWSVu8c6DLpWJQQlXHPqP702qIg/gx2o0bm4EzrCEJ
4gYo6Ax+U7q6TOWhQpiBHnC0ojE8kUoqMhfALpUaruTJ6zmj8IA1e1M6bMqVF8sr
lb/N
-----END PUBLIC KEY-----

RSA 算法

计算机的数据都是以 01 形式存储的，一个字节 8 位可以用两位 16 进制数表示，如 1111 1111 = FF，很多算法都是对数字进行运算与求逆的过程

RSA 算法的步骤：

选择一对不相等的大质数 p、q

• 质数要足够大，不容易破解，举例使用小数字

• 质数：除了 1 和它本身无法被其他自然数整除，例如：3 的因数是 1和3，11 的因数是 1和11
  $$p=3,q=11$$

计算 p、q 的乘积 n

$$n=p\times q=3\times 11=33$$

计算 n 的欧拉函数 φ(n)

• 欧拉函数：小于或等于n 的正整数中与 n 互质的数的数目，例如 φ(3)={1,2} = 2；φ(11)={1,2,3,4,5,6,7,8,9,10}=10，即质数除了它本身都与其互质，所以 φ(n) = n-1

• 互质：两个数的公因数只有1 时称为互质数，例如 5和6互质，5和9互质，6和9不互质，因为 3 也是它们的公因数

$$\phi (n)=(p-1)(q-1)=(3-1)(11-1)=2\times 10=20$$

选择一个与 φ(n)互质的整数 e，随便选一个 3 吧，方便计算

1 < e < φ ( n ) , e ∈ { 3 , 7 , 9 , 11 , 13 , 17 , 19 } 1<e<φ(n)\quad,\quad e∈\{3,7,9,11,13,17,19\} 1<e<φ(n),e∈{3,7,9,11,13,17,19}

计算 e对于φ(n)的模反元素 d

• 模反元素：正整数 a、n互质，可以找到整数 b，满足： a·b mod n=1

• mod：取余，a mod b即 a除b的余数，例如：10 mod 3 =1，19 mod 4 = 3，3 mod 5 = 5，36 mod -10 = - 4
  $$d\cdot emod\phi (n)=1$$

• k 的取值为整数集
  $$d=[k\phi (n)+1]/e,k\in \mathbb{Z}$$

• 还是取最简单的 k=1，d=7
  $$d=[1\times 20+1]/3=7$$

计算公钥 KU

$$KU=(e,n)=(3,33)$$

计算私钥 KR

$$KR=(d,n)=(7,33)$$

加密：M 是明文，C是密文，例如 M = 20

$$C=M^{e}modn=20^{3}mod33=8000mod33=14$$

解密

$$M=C^{d}modn=14^{7}mod33=105413504mod33=20$$

题目解密

openssl：一个常用的加解密的库，执行如下命令

openssl rsa -pubin -text -modulus -in warmup -in key.pub

Modulus：代表 n ；Exponent：代表 e ；

上面是 16 进制，下面是去除冒号的，在线工具获取e和n：http://www.hiencode.com/pub_asys.html

拿到了 n 首先要转换为 10进制再计算，使用 python 转换，在线转换工具：https://c.runoob.com/unit-conversion/7999/

# 16进制转10进制
>>> s='3303B790FB149DA3406D495AB9B9FB8A9E293445E3BD43B18EF2F0521B726EBE8D838BA774BB5240F08F7FBCA0A142A1D4A61EA973294E684A8D1A2CDF18A84F2DB7099B8E977588B0B891292558CAA05CF5DF2BC6334C5EE5083A234EDFC79A95C478A78E337C723AE8834FB8A9931B74503FFEA9E61BF53D8716984AC47837B'
>>> int(s,16)
573177824579630911668469272712547865443556654086190104722795509756891670023259031275433509121481030331598569379383505928315495462888788593695945321417676298471525243254143375622365552296949413920679290535717172319562064308937342567483690486592868352763021360051776130919666984258847567032959931761686072492923

分解为 p 和 q，使用在线工具分解：http://www.factordb.com/

p=20423438101489158688419303567277343858734758547418158024698288475832952556286241362315755217906372987360487170945062468605428809604025093949866146482515539

q=28064707897434668850640509471577294090270496538072109622258544167653888581330848582140666982973481448008792075646342219560082338772652988896389532152684857

e的处理，去除冒号，转为10 进制

echo '61:17:c6:04:48:b1:39:45:1a:b5:b6:0b:62:57:a1:
    2b:da:90:c0:96:0f:ad:1e:00:7d:16:d8:fa:43:aa:
    5a:aa:38:50:fc:24:0e:54:14:ad:2b:a1:09:0e:8e:
    12:d6:49:5b:bc:73:a0:cb:a5:62:50:42:55:c7:3e:
    a3:fb:d3:6a:88:83:f8:31:da:8d:1b:9b:81:33:ac:
    21:09:e2:06:28:e8:0c:7e:53:ba:ba:4c:e5:a1:42:
    98:81:1e:70:b4:a2:31:3c:91:4a:2a:32:17:c0:2e:
    95:1a:ae:e4:c9:eb:39:a3:f0:80:35:7b:53:3a:6c:
    ca:95:17:cb:2b:95:bf:cd'|tr -d ':'

e=68180928631284147212820507192605734632035524131139938618069575375591806315288775310503696874509130847529572462608728019290710149661300246138036579342079580434777344111245495187927881132138357958744974243365962204835089753987667395511682829391276714359582055290140617797814443530797154040685978229936907206605

求出 d

import gmpy2
p=20423438101489158688419303567277343858734758547418158024698288475832952556286241362315755217906372987360487170945062468605428809604025093949866146482515539
q=28064707897434668850640509471577294090270496538072109622258544167653888581330848582140666982973481448008792075646342219560082338772652988896389532152684857
e=68180928631284147212820507192605734632035524131139938618069575375591806315288775310503696874509130847529572462608728019290710149661300246138036579342079580434777344111245495187927881132138357958744974243365962204835089753987667395511682829391276714359582055290140617797814443530797154040685978229936907206605
phi = (p-1) * (q-1)
d = gmpy2.invert(e, phi)
print(d)

d=44217944188473654528518593968293401521897205851340809945591908757815783834933

python2脚本生成私钥文件private.pem

#coding=utf-8
import math
import sys
from Crypto.PublicKey import RSA
arsa=RSA.generate(1024)
arsa.p=20423438101489158688419303567277343858734758547418158024698288475832952556286241362315755217906372987360487170945062468605428809604025093949866146482515539
arsa.q=28064707897434668850640509471577294090270496538072109622258544167653888581330848582140666982973481448008792075646342219560082338772652988896389532152684857
arsa.e=68180928631284147212820507192605734632035524131139938618069575375591806315288775310503696874509130847529572462608728019290710149661300246138036579342079580434777344111245495187927881132138357958744974243365962204835089753987667395511682829391276714359582055290140617797814443530797154040685978229936907206605
arsa.n=arsa.p*arsa.q
arsa.d=44217944188473654528518593968293401521897205851340809945591908757815783834933
Fn=long((arsa.p-1)*(arsa.q-1))
private=open('private.pem','w')
private.write(arsa.exportKey())
private.close()

ps：kali 安装crypto报错

apt-get install python2-dev
pip install pycrypto

最后使用 openssl解密

OpenSSL> rsautl -decrypt -in flag.enc -inkey private.pem