27 November 2022

Pentesting Fun Stuff

following the cyber security path…

c4ptur3-th3-fl4g

This is a challenge from TryHackMe and the first task is:

Task 1 ~ Translation & Shifting

Translate, shift and decode the following:

Answers are all case sensitive.

#1

c4n y0u c4p7u23 7h3 f149?

Leetspeak…..so a substitution of the numbers will be enough.

can you capture the flag?

#2

01101100 01100101 01110100 01110011 00100000 01110100 01110010 01111001 00100000 01110011 01101111 01101101 01100101 00100000 01100010 01101001 01101110 01100001 01110010 01111001 00100000 01101111 01110101 01110100 00100001

Binary. To convert binary to text, you need to convert binary to decimal and then look up the decimal value in a ASCII table. How does this work?
Let’s take the first string: 01101100. To convert this binary string to decimal you calculate the value:

0 1 1 0 1 1 0 0
128 64 32 16 8 4 2 1

In this case: 64 + 32 + 8 + 4 = 108

In the ASCII table 108 equals the character lowercase l. And you can do this for the entire string (LoL) or you let a program like CyberChef do all the hard work.

lets try some binary out!

#3

MJQXGZJTGIQGS4ZAON2XAZLSEBRW63LNN5XCA2LOEBBVIRRHOM======

Looks like base32. For this you can use a program in Kali.

root@lab:~/tryhackme/ctf# echo 'MJQXGZJTGIQGS4ZAON2XAZLSEBRW63LNN5XCA2LOEBBVIRRHOM======' | base32 -d
base32 is super common in CTF's

#4

RWFjaCBCYXNlNjQgZGlnaXQgcmVwcmVzZW50cyBleGFjdGx5IDYgYml0cyBvZiBkYXRhLg==

Looks the similar as number 3, but this is base64. Again….in Kali you can use a program.

root@lab:~/tryhackme/ctf# echo 'RWFjaCBCYXNlNjQgZGlnaXQgcmVwcmVzZW50cyBleGFjdGx5IDYgYml0cyBvZiBkYXRhLg==' | base64 -d
Each Base64 digit represents exactly 6 bits of data.

#5

68 65 78 61 64 65 63 69 6d 61 6c 20 6f 72 20 62 61 73 65 31 36 3f

This is a Hex string. Hex is base16 and you can do this like all the others by hand. Again like with the binary string you can convert a hex value into a decimal value and look up the ASCII character that corresponds with it.

hexadecimal or base16?

#6

Ebgngr zr 13 cynprf!

This looks like an rotation cypher. When rotating the letters according the alphabet a fixed number of times, you can encrypt your plain text. For example the word cat encrypted with ROT3 becomes fdw. In this case it’s ROT13

Rotate me 13 places!

#7

*@F DA:? >6 C:89E C@F?5 323J C:89E C@F?5 Wcf E:>6DX

This one is a bit more obfuscated, but it’s still a rotation cypher. In this case it’s ROT47. ROT47 comes from the ROT13 algorithm and differs only by additional numbers and characters that it uses in its alphabet. Just like ROT13, ROT47 shifts each text symbol by 47 positions.

You spin me right round baby right round (47 times)

#8

- . .-.. . -.-. --- -- -- ..- -. .. -.-. .- - .. --- -.

. -. -.-. --- -.. .. -. --.

If you’ve watched a lot of survival shows or have been to the army, this will look familiar. It’s morse and every dot or dash combination is a letter from the alphabet. This code can be transmitted in text, light or even sound.

TELECOMMUNICATIONENCODING

#9

85 110 112 97 99 107 32 116 104 105 115 32 66 67 68

This one is already a bit explained with number 5. In encoding standards like ASCII and Unicode each character can be represented by a numeric code point. Again, with the ASCII table you can look up the corresponding character.

Unpack this BCD

#10

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This one is a little bit more tricky, because it combines multiple encoding. I’ve mentioned it before, but a really good tool is CyberChef. You can create a decoding recipe one decoding block after another.

Let's make this a bit trickier...

Task 2 ~ Hashes

For this I use a different approach. On Github there is a nifty tool, called hash-buster. It will identify a hash and try to find it’s plaintext counterpart on the internet on different sources.

First I put all the strings in a file.

root@lab:~/tryhackme/ctf# cat -n hashes.list 
     1	39d4a2ba07e44421c9bedd54dc4e1182
     2	e0418e7c6c2f630c71b2acabbcf8a2fb
     3	efbd448a935421a54dda43da43a701e1
     4	11FE61CE0639AC2A1E815D62D7DEEC53
     5	a361f05487b879f25cc4d7d7fae3c7442e7849ed15c94010b389faafaf8763f0dd022e52364027283d55dcb10974b09e7937f901584c092da65a14d1aa8dc4d8
     6	d48a2f790f7294a4ecbac10b99a1a4271cdc67fff7246a314297f2bca2aaa71f
     7	a34e50c78f67d3ec5d0479cde1406c6f82ff6cd0

Then I run hash-buster to see what it will find.

root@lab:~/tryhackme/ctf# hashbuster -f hashes.list -t 10
_  _ ____ ____ _  _    ___  _  _ ____ ___ ____ ____
|__| |__| [__  |__|    |__] |  | [__   |  |___ |__/
|  | |  | ___] |  |    |__] |__| ___]  |  |___ |  \  v3.0

[!] Hashes found: 6
d48a2f790f7294a4ecbac10b99a1a4271cdc67fff7246a314297f2bca2aaa71f : Commonly used in Blockchain
a34e50c78f67d3ec5d0479cde1406c6f82ff6cd0 : The OG
a361f05487b879f25cc4d7d7fae3c7442e7849ed15c94010b389faafaf8763f0dd022e52364027283d55dcb10974b09e7937f901584c092da65a14d1aa8dc4d8 : 1024 bit blocks!
[!] Results saved in cracked-hashes.list

It identified 6 hashes (so it missed 1) and it found 3 known hashes form the sources. I’m still missing 4. For this task I’m switching to online crackers, because cracking them myself is a hell of a task. There are a few online cracking sites that helped out a lot: https://md5decrypt.net and https://md5hashing.net

Task 3 ~ Spectrograms

For this part there is an audio file to download. To render the hidden message there are several tools which can help. One of them is sonic-visualiser. I’m going for Audacity this time.

Task 4 ~ Steganography

Again there is a file to download. This time an image with a hidden message. A tool which can be used is called steghide.

root@lab:~/tryhackme/ctf# steghide extract -sf stegosteg.jpg 
Enter passphrase: 
wrote extracted data to "steganopayload2248.txt".
root@lab:~/tryhackme/ctf# cat steganopayload2248.txt 
SpaghettiSteg

Task 5 ~ Security through obscurity

For this final part there is yet another file to download. The mission states that you need to ‘get in’. So best thing is to use the command strings or something like hexdump -C.

root@lab:~/tryhackme/ctf# strings meme.jpg | tail
S~j@6h
vA}=
*s&__
@9Xs
{@84
2$Es
i2Mc
IEND
"AHH_YOU_FOUND_ME!" 
hackerchat.png

This way you get both answers.

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