rsa题目

Posted 2021-10-07 Nullan

easyrsa

还记得veryeasy RSA吗？是不是不难？那继续来看看这题吧，这题也不难。
已知一段RSA加密的信息为：0xdc2eeeb2782c且已知加密所用的公钥：
(N=322831561921859 e = 23)
请解密出明文，提交时请将数字转化为ascii码提交
比如你解出的明文是0x6162，那么请提交字符串ab
提交格式:PCTF{明文字符串}

factorord分解n
http://factordb.com/
分解链接

#!/usr/bin/python  
#coding:utf-8  
 
import libnum  
from Crypto.Util.number import*
c = 0xdc2eeeb2782c  
n = 322831561921859  
e = 23  
q = 13574881  
p = 23781539  
 
d = libnum.invmod(e, (p - 1) *(q - 1))  
m = pow(c, d, n)  
print "m的值为:"  
print long_to_bytes(m)

meduimRSA

#!/usr/bin/python  
#coding:utf-8  

import libnum  
from Crypto.Util.number import* 
 
n = 87924348264132406875276140514499937145050893665602592992418171647042491658461  
p = 275127860351348928173285174381581152299  
q = 319576316814478949870590164193048041239  
e = 65537  
phi = (p-1)*(q-1)  
d = libnum.invmod(e,phi)  
 
with open('./flag.enc') as f:  
    c = f.read().encode('hex')  
    c = int(c,16)  
m = pow(c,d,n)  
print long_to_bytes(m)

meduimRSa

利用命令 openssl rsa -pubin -in pubkey.pem -text -modulus
EXponent那一处是e
Modulus那一处是n
利用网站分解n
得到p和q
脚本

#!/usr/bin/python  
#coding:utf-8  

import libnum  
from Crypto.Util.number import* 
 
n = 87924348264132406875276140514499937145050893665602592992418171647042491658461  
p = 275127860351348928173285174381581152299  
q = 319576316814478949870590164193048041239  
e = 65537  
phi = (p-1)*(q-1)  
d = libnum.invmod(e,phi)  
 
with open('./flag.enc') as f:  
    c = f.read().encode('hex')  
    c = int(c,16)  
m = pow(c,d,n)  
print long_to_bytes(m)