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signature=a7ab3f52fd3143e911ffec68c5ce32d7,2019年强网杯crypto部分题解

發(fā)布時(shí)間：2024/9/19 编程问答 39 豆豆

生活随笔收集整理的這篇文章主要介紹了 signature=a7ab3f52fd3143e911ffec68c5ce32d7,2019年强网杯crypto部分题解小編覺得挺不錯(cuò)的,現(xiàn)在分享給大家,幫大家做個(gè)參考.

一、random study

這個(gè)題目中共給出了三個(gè)challenge。

1. challenge 1

服務(wù)器將python中的random模塊的種子設(shè)置為int(time())，然后生成隨機(jī)數(shù)讓我們猜，只要我們猜對一次就可以通關(guān)了。

題目中給了200次機(jī)會，應(yīng)該是考慮到服務(wù)器與我們機(jī)器的時(shí)間不同步的問題(可能相差幾秒)。我這里假定服務(wù)器時(shí)間與本地時(shí)間相差不超過10秒，然后對這20種情況進(jìn)行枚舉就好了。相應(yīng)代碼如下：

2.challenge 2

服務(wù)器自己調(diào)用了一個(gè)Java程序生成了三個(gè)隨機(jī)數(shù)，然后將前兩個(gè)告訴我們，讓我們猜第三個(gè)。

用Java Decompiler對提供的Java程序進(jìn)行反編譯得到源碼如下：

public class Main {

public static void main(String[] paramArrayOfString) {

Random random = new Random();

System.out.println(random.nextInt());

}

可以看出這個(gè)Java程序是使用了Random類的nextInt函數(shù)來生成隨機(jī)數(shù)的。百度一下這個(gè)類，可以查到官方文檔。從官方文檔中可以知道：

Random的nextInt函數(shù)就是直接調(diào)用next(32)并將得到的值返回。

next(int n)會通過以下方式更新內(nèi)部存儲的seed值，并返回下一個(gè)隨機(jī)數(shù)。

//seed乘一個(gè)數(shù)，然后取低48位。

seed = (seed * 0x5DEECE66DL + 0xBL) & ((1L << 48) - 1);

// 48位種子的前n位。

return seed>>(48-n);

setSeed(long n)會將n做一些變化，然后存儲到內(nèi)部的seed中。

seed = (n^ 0x5DEECE66DL + 0xBL) & ((1L << 48) - 1);

根據(jù)以上知識，我們可以發(fā)現(xiàn)，如果我們用nextInt獲得了一個(gè)隨機(jī)數(shù)

，那么生成這個(gè)隨機(jī)數(shù)時(shí)的種子就是

，其中x是一個(gè)16位的整數(shù)。

因此我們根據(jù)題目中給出的第一個(gè)隨機(jī)數(shù)可以推斷出生成第二個(gè)隨機(jī)數(shù)的種子的前32位，然后我們通過對低16位進(jìn)行枚舉就可以得到生成第二個(gè)隨機(jī)數(shù)的種子是多少了。

3. challenge3

這一題就沒什么了，只要challenge1通過了，這里直接調(diào)函數(shù)得到隨機(jī)數(shù)，然后發(fā)過去就好了。

代碼如下:

from pwn import *

import time

from aux import *

import tqdm

import random

import subprocess

conn = remote('119.3.245.36', 23456)

def solveChaleneg1(conn):

conn.recvuntil('[-]')

cur_time = int(time.time())

for i,j in (enumerate(range(cur_time-10,cur_time+10)),initial=0,total=20):

random.seed(j)

for k in range(i):

random.randint(0,2**64)

conn.sendline(str(random.randint(0,2**64)))

if 'fail' not in conn.recvline():

break

def solveChalenge2():

conn.recvuntil('[-]')

x1 = conn.recvline()

print x1

x1 = int(x1)

print(x1)

conn.recvuntil('[-]')

x2 = int(conn.recvline(),10)

o = subprocess.check_output(["java", "MyMain",str(x1),str(x2)])

x3 = int(o.split('\n')[0],10)

print "x1=%d,x2=%d,x3=%d"%(x1,x2,x3)

conn.recvuntil('[-]')

conn.sendline(str(x3))

if 'fail' in conn.recvline()[:-1]:

print "challenge2 fails"

def solveChalenge3():

target = random.getrandbits(32)

conn.recvuntil('[-]')

conn.sendline(str(target))

conn.recvuntil("[++++++++++++++++]challenge 3 completed[++++++++++++++++]")

solveSHA(conn)

solveChaleneg1(conn)

solveChalenge2()

solveChalenge3()

conn.interactive()

import java.util.Random;

public class MyMain {

public static void main(String[] paramArrayOfString) {

long x1 = (long)(Long.parseLong(paramArrayOfString[0])&((1L<<32)-1));

long x2 = (long)(Long.parseLong(paramArrayOfString[1])&((1L<<32)-1));

// = (seed ^ 0x5DEECE66DL) & ((1L << 48) - 1);

//seed = (seed * 0x5DEECE66DL + 0xBL) & ((1L << 48) - 1);

for(int i= 0;i

Random random = new Random(((x1<<16)+i)^0x5DEECE66DL);

if(random.nextInt() == x2){

System.out.println(random.nextInt());

}

CopperStudy

里面包含6個(gè)challenge.

1. challenge1

[+]teamtoken:[++++++++++++++++]proof completed[++++++++++++++++]

[+]Generating challenge 1

[+]n=0x931a21e45b8c30eb30f1b9767c724f57dd770432df414a5969d770e5cf41d86aa8c198baba3709d602b6de17003406b3ba9190e99c4c9aaa470afe052fa2aec57a7c448bf1a887c5c4776a789b51bf4f76b334fc27a59469d6b4fb76af90ec2ec56a4c83f0cabfcb3669c2e24243c9dfe32cc817fd2196452df47e92eeb06eb5L

[+]e=3

[+]m=random.getrandbits(512)

[+]c=pow(m,e,n)=0x842683283571aeb5d8716f2a4667ff7e61fc63bb5a4cd89858c02881e0d536e344167e06e44ef5e276253563486c3b05971acfb96f8ed1a9e8c3143c026889ab11d5f72973bf6907f773ad0d570436cc64daf16e35ffff1fb48c66b36d9a8bfc568de1407b19f621ead2fa572fce24a921d742128a7b321f381e81bae0cb9f70L

[+]((m>>72)<<72)=0xb74706a05587a42246ed7ff2f03c4edf180f5248cfb7a8b86757ea456ba134c8a81f146b3172d1b59437efeb2e59b41e8dbb9cf2ac3254000000000000000000L

給定了明文的高位，低72位未知。思考了一會，發(fā)現(xiàn)不會，于是直接百度copper，發(fā)現(xiàn)了這個(gè)。上面給出了Coppersmith 算法的三種適用情況，并給出了相應(yīng)的腳本。抄來改改就好了_。

3 Challenge 2

[+]Generating challenge 2

[+]n=0xaf10597fa3ac1e95f9eeeb0747827fbfd6a6b74e9e7824e5e58c9423e4eea8f898ef5ad96900adc54721aa922af14987fbeecd1663e639f85d4e0fc2f5cc50fc1dfdf3e0b04d8395fdb8372447f0b398a7bff92cf7745c03a613e71a7017648db905ac867d69aab2c8c9ff0fc65c3313f4080cbb46d1a13a3b6c8e263915a413L

[+]e=65537

[+]m=random.getrandbits(512)

[+]c=pow(m,e,n)=0x7be9e49497bebb936018a628148e89dc2f8cb919ac899b1ac3fa050ba7987e051e78adc51677d3ab3d2529d1f98be08459699ab381e6bd51c96cb6ae327112d0bfcdd68691f55a0b9e92516506939cbbd1612a4760d428df38dab6353b838f7bfdbfcfd3fad489258a8a289527c30a9a38bdbec751c3c3dacdb9f67a222fc8c0L

[+]((p>>128)<<128)=0xbb1c7149e21b37fb4480326fa4ce2f825ff449b22fbc55d5b9d51dea23d64d6c36d5e45a25d4a46f2559d26625dc6bd900000000000000000000000000000000L

[-]long_to_bytes(m).encode('hex')=

解法同上改一下腳本就好了。

3. Challenge 3

[++++++++++++++++]challenge 2 completed[++++++++++++++++]

[+]Generating challenge 3

[+]n=0x9dd9df9c335f720b2a01628f4433daeec452e55ed505ae66df8567c6ca66ea0796a242a7bcc41b973d0392ccd0a6e11c829890d3243cf86c90a03a5729240cb9eae191f5cff6294d83a04cee995a8d01f72e4d2166c60b7368b8061cf23e5b89d5057b7026ff7673c658f478cd0e3a6e0a21a97e21c9b2e5d8e94cbb50210d99L

[+]e=3

[+]m=random.getrandbits(512)

[+]c=pow(m,e,n)=0x169f8a6015b955747b211adb5e6a573eb03ea164170a35582dbc45d2a5d6ac4a01e0fae44b919d9afa419d73c1c02cb8e1095738b84236d53997862e0ac4dd108a47db0fb3515f3d8f29f620ec209bde20ebc93e64e9301751470f80212072e0981db41e0b307dbac81e3e126dd0e58cc96046b993cd2d48460731bb0a8c437cL

[+]d=invmod(e,(p-1)*(q-1))

[+]d&((1<<512)-1)=0x3a3d77ea2d0e834a43efa2accc7e463a7790f5f597abface5d56f3ed88aad629299a32ae6e21f0f9b43b7b402ba9d2e4d548f99298dce0ca92bcd26dd0d7c323L

[-]long_to_bytes(m).encode('hex')=

解法同上

4. challenge 4

[+]Generating challenge 4

[+]e=3

[+]m=random.getrandbits(512)

[+]n1=0x3cfc26953d034786a32dd22249d823ec272b8a46216efae5e57f9a8b8c7fed3b9b4a3c9361447159d714a614d7501e5b628a3da24b11a4ecd8c9c3f3cafb52674985752081ddff6eea6fc6c069d27b96ae039a53d05791c5207236890229abd337f2cad946ab550f8f4c414537fd3c0135f3b4f27e6919c15f2aa81a1ffe15afL

[+]c1=pow(m,e,n1)=0x374c6727e44c56e0de065f16bce1314b857792862f5d848a88d442605d7a54214806a0ad2c72090c39770e802aaa045deecaecc112c80a022ea3c4540823a175d11ba4fa347727931776854265c62228324ee77a8f19360f80f0a0152bfab2214474e3f744f8a0a577c26cdc2ab325326b40bc7d5d6fb9a005f9dce9418b6deaL

[+]n2=0x41e446d32fd2f4143558b8abf4f9cfd1feb827f49a3e61e115157ac948a183c196531acdd66d7749ede2f5913b01cdc49f98abb77a7bf6c9f1246ab1e74288d90885a0359cb244cc71f94b19fb552c2b6ace51ddf48893e3bae0561ae78cec67e71e7864da05fcfc13b8d3d741ef988043e944f8a53bb84def356170e2a9e5a3L

[+]c2=pow(m,e,n2)=0x2d9dbdaa0b89a0d12624a34146871f0378843199c516015e25770baefb00771713a705ff48eea7d9cad2a50c7b31d73612ac4c68ff29ba5a230546ff364fe6d2dc807b5eab1b3633baffe89c421bd683da0e4c828cd0709d08b2fb6941b94e3b73f87f1db368ef77d0597d1478192722b7ccfdab10dd4c9f6321676f5397e0ffL

[+]n3=0x1f7f8868ee329980957490a294da7507071ac5b90c029dddfef65a62a9c6010fa8cb735a00a7524e5633d3f807cc639d8956486bd29f46d16e731db107c8f1d58f995ec39c5582155ee06d4db32317fa1b3b9148052c18245c8b48f87844cb68d81f1f995044fa22eb18c197e437ba3badc12bc749572a49de66828e391b1dbL

[+]c3=pow(m,e,n3)=0x1e1e3ed0522464ccde39974aa9664f2fe9dad909ba4d42e9c2266dafccb5dc7e2074a9d4ac180653985e1b89def78d06143800d678365041abe2278dbcac81fce8f1122a117bd77b455ff7585a46d7e439265baa9c493d3f2fb2172a9bc07ba1483e1ed58a0eb5ea8efbd1d5f33cb62e01ab9ccecf7f2d396d4034b73cbf77cL

首先嘗試發(fā)現(xiàn)

、

沒有公約數(shù)(如果有的話可以直接分解)。然后，利用中國剩余定理得到：

可以發(fā)現(xiàn)m是512位的，因此

大概是1536位的，少于

的3072位。因此，我們可以知道

直接求立方根就可以得到m了。

5. chalenge 5

[++++++++++++++++]challenge 4 completed[++++++++++++++++]

[+]Generating challenge 5

[+]n=0x3298d508dc5255e5f53c2890326aceacf9439b9f2f06e2abf3e19ece5a503315852cbaf9ab9f730f9bde1082331f0697491dade2c57463d610b540697506ce3441ced4262ca150b1c28535f266e954967895f6d4b3fc86281cdfaa21fa8aa1b440de1bb0b0b4332969f24bff01d574f94235de3bc3da43401915adf253589b63L

[+]e=3

[+]m=random.getrandbits(512)

[+]c=pow(m,e,n)=0x1b124c087004fd532a26f2cb6cc5c9ee2176c2070d87c159bd0e74b8811ab3af84c96e05dd7bf6bb6b5a251c8558c357036aff678525c8734570e142e38e77419993892a29d9a9f54bd3d4c048895482466d749eb0cf5b12b427c1753f6ebe3eba2ad9dcdc87b7c2040bfab6a5c124eb21b821fa58e167d8cb57e9c551ab9dbfL

[+]x=pow(m+1,e,n)=0x2d6fcf825d2fcb98f37f951fc000520f4aab11222f92a1af01e0988943fdde016664b706f4197bd085e79f57daedbecf4ee59cc9f010998c22f1c2df6223df57e86b9efd8bcca779f16e6da8a54778459b4d956a255d8ecc3d3a9e9a72a7c70cb2bec8bbaa2bb5c2f2d3f95769cee6fdd7fd04bbd6d84070e50739b5b059c6e6L

[-]long_to_bytes(m).encode('hex')=

題目中給出了m和m+1加密后的結(jié)果。即我們知道：

二者相減得到：

兩邊減1并乘以3模n的逆得到：

我們知道m(xù)是512位的，由此可以推出

不超過1024位。進(jìn)而可以得到：

k = 0或1(不確定自己分析的對不對，所以寫代碼的時(shí)候把k的范圍稍微擴(kuò)大了一點(diǎn)點(diǎn))

然后，對每一個(gè)k值，嘗試求解上面的方程就可以得到結(jié)果了。因?yàn)?/p>

在

單調(diào)遞增，因此求解方法用二分法即可。

6. challenge 6

[++++++++++++++++]challenge 5 completed[++++++++++++++++]

[+]Generating challenge 6

[+]n=0xbadd260d14ea665b62e7d2e634f20a6382ac369cd44017305b69cf3a2694667ee651acded7085e0757d169b090f29f3f86fec255746674ffa8a6a3e1c9e1861003eb39f82cf74d84cc18e345f60865f998b33fc182a1a4ffa71f5ae48a1b5cb4c5f154b0997dc9b001e441815ce59c6c825f064fdca678858758dc2cebbc4d27L

[+]d=random.getrandbits(1024*0.270)

[+]e=invmod(d,phin)

[+]hex(e)=0x11722b54dd6f3ad9ce81da6f6ecb0acaf2cbc3885841d08b32abc0672d1a7293f9856db8f9407dc05f6f373a2d9246752a7cc7b1b6923f1827adfaeefc811e6e5989cce9f00897cfc1fc57987cce4862b5343bc8e91ddf2bd9e23aea9316a69f28f407cfe324d546a7dde13eb0bd052f694aefe8ec0f5298800277dbab4a33bbL

[+]m=random.getrandbits(512)

[+]c=pow(m,e,n)=0xe3505f41ec936cf6bd8ae344bfec85746dc7d87a5943b3a7136482dd7b980f68f52c887585d1c7ca099310c4da2f70d4d5345d3641428797030177da6cc0d41e7b28d0abce694157c611697df8d0add3d900c00f778ac3428f341f47ecc4d868c6c5de0724b0c3403296d84f26736aa66f7905d498fa1862ca59e97f8f866cL

注意到題目中提示

的位數(shù)為0.27*1024，相當(dāng)于

。不知道怎么辦，后來在mhx的提示下發(fā)現(xiàn)Boneh and Durfee attack在

時(shí)分解n。然后又是直接抄個(gè)腳本過來分解掉了。

代碼如下(不包含Sage腳本，因?yàn)樯厦嫣峁┑倪B接中都有)：

from Crypto.Util import number

from pwn import *

import hashlib

import libnum

from aux import *

import tqdm

import gmpy2

conn = remote('119.3.245.36', 12345)

x0 = 3031202121516888756254+0xb74706a05587a42246ed7ff2f03c4edf180f5248cfb7a8b86757ea456ba134c8a81f146b3172d1b59437efeb2e59b41e8dbb9cf2ac3254000000000000000000

p = 157455850951539875413707605384569848509+0xbb1c7149e21b37fb4480326fa4ce2f825ff449b22fbc55d5b9d51dea23d64d6c36d5e45a25d4a46f2559d26625dc6bd900000000000000000000000000000000L

n = 0xaf10597fa3ac1e95f9eeeb0747827fbfd6a6b74e9e7824e5e58c9423e4eea8f898ef5ad96900adc54721aa922af14987fbeecd1663e639f85d4e0fc2f5cc50fc1dfdf3e0b04d8395fdb8372447f0b398a7bff92cf7745c03a613e71a7017648db905ac867d69aab2c8c9ff0fc65c3313f4080cbb46d1a13a3b6c8e263915a413

q = n/p

c = 0x7be9e49497bebb936018a628148e89dc2f8cb919ac899b1ac3fa050ba7987e051e78adc51677d3ab3d2529d1f98be08459699ab381e6bd51c96cb6ae327112d0bfcdd68691f55a0b9e92516506939cbbd1612a4760d428df38dab6353b838f7bfdbfcfd3fad489258a8a289527c30a9a38bdbec751c3c3dacdb9f67a222fc8c0

x1 = pow(c,gmpy2.invert(65537,(p-1)*(q-1)),n)

d = 73897859833360735083833018291216138538391694441421128347796113898176347610218862647497820864455327205623491646417194479391497879610944659647766341364110862866967065387388720221408859326315301232384151863747183910278699184043306720913086501616478917318589938736632770485961912898351468620093203923268728767267

c = 0x169f8a6015b955747b211adb5e6a573eb03ea164170a35582dbc45d2a5d6ac4a01e0fae44b919d9afa419d73c1c02cb8e1095738b84236d53997862e0ac4dd108a47db0fb3515f3d8f29f620ec209bde20ebc93e64e9301751470f80212072e0981db41e0b307dbac81e3e126dd0e58cc96046b993cd2d48460731bb0a8c437c

n = 0x9dd9df9c335f720b2a01628f4433daeec452e55ed505ae66df8567c6ca66ea0796a242a7bcc41b973d0392ccd0a6e11c829890d3243cf86c90a03a5729240cb9eae191f5cff6294d83a04cee995a8d01f72e4d2166c60b7368b8061cf23e5b89d5057b7026ff7673c658f478cd0e3a6e0a21a97e21c9b2e5d8e94cbb50210d99L

x2 = pow(c,d,n)

n1 = 0x3cfc26953d034786a32dd22249d823ec272b8a46216efae5e57f9a8b8c7fed3b9b4a3c9361447159d714a614d7501e5b628a3da24b11a4ecd8c9c3f3cafb52674985752081ddff6eea6fc6c069d27b96ae039a53d05791c5207236890229abd337f2cad946ab550f8f4c414537fd3c0135f3b4f27e6919c15f2aa81a1ffe15af

n2 = 0x41e446d32fd2f4143558b8abf4f9cfd1feb827f49a3e61e115157ac948a183c196531acdd66d7749ede2f5913b01cdc49f98abb77a7bf6c9f1246ab1e74288d90885a0359cb244cc71f94b19fb552c2b6ace51ddf48893e3bae0561ae78cec67e71e7864da05fcfc13b8d3d741ef988043e944f8a53bb84def356170e2a9e5a3

n3 = 0x1f7f8868ee329980957490a294da7507071ac5b90c029dddfef65a62a9c6010fa8cb735a00a7524e5633d3f807cc639d8956486bd29f46d16e731db107c8f1d58f995ec39c5582155ee06d4db32317fa1b3b9148052c18245c8b48f87844cb68d81f1f995044fa22eb18c197e437ba3badc12bc749572a49de66828e391b1db

c1 = 0x374c6727e44c56e0de065f16bce1314b857792862f5d848a88d442605d7a54214806a0ad2c72090c39770e802aaa045deecaecc112c80a022ea3c4540823a175d11ba4fa347727931776854265c62228324ee77a8f19360f80f0a0152bfab2214474e3f744f8a0a577c26cdc2ab325326b40bc7d5d6fb9a005f9dce9418b6dea

c2 = 0x2d9dbdaa0b89a0d12624a34146871f0378843199c516015e25770baefb00771713a705ff48eea7d9cad2a50c7b31d73612ac4c68ff29ba5a230546ff364fe6d2dc807b5eab1b3633baffe89c421bd683da0e4c828cd0709d08b2fb6941b94e3b73f87f1db368ef77d0597d1478192722b7ccfdab10dd4c9f6321676f5397e0ff

c3 = 0x1e1e3ed0522464ccde39974aa9664f2fe9dad909ba4d42e9c2266dafccb5dc7e2074a9d4ac180653985e1b89def78d06143800d678365041abe2278dbcac81fce8f1122a117bd77b455ff7585a46d7e439265baa9c493d3f2fb2172a9bc07ba1483e1ed58a0eb5ea8efbd1d5f33cb62e01ab9ccecf7f2d396d4034b73cbf77c

x = c1*gmpy2.invert(n2*n3,n1)*n3*n2 + c2*gmpy2.invert(n1*n3,n2)*n1*n3 + c3 * gmpy2.invert(n1*n2,n3)*n1*n2

x = int(x%(n1*n2*n3))

def cubeRoot(x):

l,r = 0,x

while l

mid = (l+r)//2

if mid**3 == x:

return mid

elif mid**3 < x:

l = mid+1

else:

r = mid-1

return l

x3 = cubeRoot(x)

assert x3**3 %n1 == c1 and x3**3%n2 == c2 and x3**3%n3 == c3

n = 0x3298d508dc5255e5f53c2890326aceacf9439b9f2f06e2abf3e19ece5a503315852cbaf9ab9f730f9bde1082331f0697491dade2c57463d610b540697506ce3441ced4262ca150b1c28535f266e954967895f6d4b3fc86281cdfaa21fa8aa1b440de1bb0b0b4332969f24bff01d574f94235de3bc3da43401915adf253589b63

e = 3

c = 0x1b124c087004fd532a26f2cb6cc5c9ee2176c2070d87c159bd0e74b8811ab3af84c96e05dd7bf6bb6b5a251c8558c357036aff678525c8734570e142e38e77419993892a29d9a9f54bd3d4c048895482466d749eb0cf5b12b427c1753f6ebe3eba2ad9dcdc87b7c2040bfab6a5c124eb21b821fa58e167d8cb57e9c551ab9dbf

x = 0x2d6fcf825d2fcb98f37f951fc000520f4aab11222f92a1af01e0988943fdde016664b706f4197bd085e79f57daedbecf4ee59cc9f010998c22f1c2df6223df57e86b9efd8bcca779f16e6da8a54778459b4d956a255d8ecc3d3a9e9a72a7c70cb2bec8bbaa2bb5c2f2d3f95769cee6fdd7fd04bbd6d84070e50739b5b059c6e6

tmp = (x-c-1)*gmpy2.invert(3,n)%n

def findm(x,n):

l,r = 0,n

while l

mid = (l+r)//2

t = mid**2+mid

if t==x:

return mid

elif t

l = mid +1

else:

r = mid -1

return l

for i in range(5):

m = findm(tmp+i*n,n)

if (m**2+m) == tmp:

break

assert (m**3)%n == c

x4 = m

n = 0xbadd260d14ea665b62e7d2e634f20a6382ac369cd44017305b69cf3a2694667ee651acded7085e0757d169b090f29f3f86fec255746674ffa8a6a3e1c9e1861003eb39f82cf74d84cc18e345f60865f998b33fc182a1a4ffa71f5ae48a1b5cb4c5f154b0997dc9b001e441815ce59c6c825f064fdca678858758dc2cebbc4d27

e = 0x11722b54dd6f3ad9ce81da6f6ecb0acaf2cbc3885841d08b32abc0672d1a7293f9856db8f9407dc05f6f373a2d9246752a7cc7b1b6923f1827adfaeefc811e6e5989cce9f00897cfc1fc57987cce4862b5343bc8e91ddf2bd9e23aea9316a69f28f407cfe324d546a7dde13eb0bd052f694aefe8ec0f5298800277dbab4a33bb

c = 0xe3505f41ec936cf6bd8ae344bfec85746dc7d87a5943b3a7136482dd7b980f68f52c887585d1c7ca099310c4da2f70d4d5345d3641428797030177da6cc0d41e7b28d0abce694157c611697df8d0add3d900c00f778ac3428f341f47ecc4d868c6c5de0724b0c3403296d84f26736aa66f7905d498fa1862ca59e97f8f866c

d = 776765455081795377117377680209510234887230129318575063382634593357724998207571

assert c == pow(c,e*d,n)

x5 = pow(c,d,n)

solveSHA(conn)

conn.recvuntil('long_to_bytes(m).encode(\'hex\')=')

conn.sendline(number.long_to_bytes(x0).encode('hex'))