# CryptoI don't know why, but crypto challenges are always fun for me.## 50 points: WelcomeCrypto```~C8 =39 A?2V8 73J:C 8FG7 AF?JJC2ECP
DF?LHb=r_>b0%_0520<c8bPN
Author: Loavso```I'm pretty embarassed this took me as long as it did. I didn't recognize the cipher right away, but its just a ROT47. Use any <a href='https://www.dcode.fr/rot-47-cipher'>online decoder you want. The one I used outputted:```Org lbh pna'g fbyir guvf punyyratr!
sun{w3lC0m3_T0_da_k4g3!}```Literally no clue what the first line was supposed to be, but the flag was valid.
flag: `sun{w3lC0m3_T0_da_k4g3!}`
*Edit*: The first line is a ROT13 after the ROT47: `Bet you can't solve this challenge!` Why they would make the flavor text another step, I have no idea.## 50 points: CB1```We picked up a new numbers station that's been active in the shortwave bands. We need to figure out how to crack his code.
Here's an example file, crack the cipher and send us the plaintext message.
```CB1.wav```
NOTE: NON-Standard Flag Format
Author: leviathan```When I saw this, I thought it would be an audio steganography challenge. But let's give it a listen! I transcribed:```Code Number: 6Begin MessageHotel Kilo Charlie Golf Xray Kilo Zulu November Kilo Oscar Juliette Kilo Yankee Uniform Lima Sierra Golf Xray India November```It just repeated after that. The words immediately stood out to me as the NATO Phonetic Alphabet; it's just the first letter of each word.```6HKCGXKZNKOJKYULSGXIN```6 seems to suggest a Caesar Cipher. I used this site to decode it with a shift of 6, and got the flag. The flag was in a nonstandard format, as was said by the description.
flag: `bewaretheidesofmarch`## 100 points: CB2```That numbers station is still active, they've just switched codes. We need you to crack it before they switch again.
Here's an example file, crack the cipher and send us the plaintext message.```CB2.wav```NOTE: NON-Standard Flag Format
Author: leviathan```Another audio file. Transcription (already using the NATO Phonetic Alphabet to replace to letters):```Codeword: ClarinetDBDAABEDDDDCDEACADBBDDADDEABBB```I got stuck on this one for a while. The key point to notice is that there are only 5 characters, A-E. So, searching 'Cipher 5 unique letters' this website popped up. It suggested that it was a polybius square cipher. I used dcode once again.
![](/Images/2019/SunshineCTF/CB2.PNG)
I entered in the key of CLARINET, chose the no 'j' alphabet for the rest of the deranged alphabet, and changed 1-5 to A-E. The website has an explanation for how Polybius works and its pretty simple. The website did its work and outputted the flag, albeit in uppercase.
flag: `polysquaresrule`
## 100 points: 16-bit-AES```Why so small?
nc aes.sunshinectf.org 4200
Author: ps_iclimbthings```This challenge was released at around 13:00 EST on the last day of the competition that would end at 21:00 EST. Luckily it didn't take too long. It's an AES challenge, obviously, but the twist was that it used a 16 bit key for a AES-128 encryption. Running the netcat command gets you:
![](/Images/2019/SunshineCTF/AESnc.PNG)
Okay, so it seems that you get to give it some text, and it outputs the resulting encrypted version. Pretty simple. Interestingly enough, running it again confirmed that the same key was being used each time. In other words, at this point there are two ways to go about doing this. I was dumb and overthought it, so I did it the "legit way": creating a script in python.```from Crypto.Cipher import AESimport itertoolsimport stringgoal = 'f312cf9c53af89447e652e73b9754a0c'                       //asdfasdfasdfasdf encoded using their keyfor combo in itertools.product(string.letters, repeat = 2):     //bash all combinations of two letters (16 bit)	key = ''.join(combo) * 8                                //AES-128 requires a 16 byte key, so hopefully the key is just 8 of the 16 bit key.	cipher = AES.new(key, AES.MODE_ECB)	msg = cipher.encrypt('asdfasdfasdfasdf')	if msg.encode('hex') == goal:		print key		break```![](/Images/2019/SunshineCTF/aeskey.PNG)
Nice. Run the netcat command again, use an online encoder or a python script to encode their string, and send it back to get the flag.
![](/Images/2019/SunshineCTF/aesflag.PNG)
flag: `sun{Who_kn3w_A3$_cou1d_be_s0_vulner8ble?}`
What was the easier way you ask? Well, since the same key is used each time, you can just open two different netcat clients, and send the requested string from one into the input in the other, and let the netcat client do it for you. :P I like to think mine was more sophisticated.
![](/Images/2019/SunshineCTF/aesrip.PNG)## 150 points: CB3```The number station has switched codes yet again. This one seems similar to the last cipher used, but we still haven't been able to crack it.
Here's an example file, crack the cipher and send us the plaintext message.```CB3.wav```NOTE: NON-Standard Flag Format
Author: leviathan```Last one in the CB series. Transcription (already using the NATO Phonetic Alphabet to replace to letters)::```Codeword: Pridefulxdxgfvvvxxafvffvadgddxagaafdffff```Six unique letters: a, d, f, g, v, and x. Searching up 'Cipher six unique letters' gives you the dumbest cipher I've ever heard of:
![](/Images/2019/SunshineCTF/whytho.PNG)
Simple enough from there. dcode once again has a decoder for this. Nice. I used the default alphabet and put in 'Prideful' as the permutation key. 
![](/Images/2019/SunshineCTF/cb3.PNG)
flag: `g3rm4n3ncrypt10n`
## 200 points: ArbCrypt```It's pretty ARB-itrary. France0110.```ciphertext.txt```Author: Mesaj2000```Apparently this one was hard enough that they added a hint for it near the end of the competition, so I'm pretty proud of myself for figuring it out relatively easily and before the hint. The ciphertext:```BBcEDAJCDBMIAxUHA3gQBxEXCwdCAwQPDhxCGRMNawYHBRgDDQcNBRAWGxZCABoUAHgQBREEDwIQDgtCCx8DFR4RaxwLGFIWERQVBxdCGQgPBBARChgBaxYSFRQRAgIGDQRCFxwYAgAEGXgHDhgRChQMChgDAlITDQAWFQgGQRcNFAAQQR0LGRsWAghCAhQaa3gGExwMQRMEFRkIEhQNQRUEERYRGxdCChsQBwQFGHgEFwQRGBkLCBcKQREUDgYUElIVGAJoEhUIBQQRGFIWERwWFxAYDhdoBhAOF1IPCAYOCBkYTAEFDBsJQQEHBQIQCxRoFwgXABYID1IQBgYUElIEBwEBF3gADRsEAAcYQRwUFQYUBFIKDRwDBFIaCBcEa1IRDR0ZABsBPgsEFy0KEwMSES0XDi0IBxc9BRQ9UEJSUENRUUdTWA8=```The '`=`' at the end immediately said 'base64', so I converted it. There weren't really any printable characters, so from then on, I used cryptii for this entire challenge, knowing that it might be multiple encryption methods. From there, I looked at the description for hints. ARB is repeated a lot, so that might be significant, though its not an encryption method. The next important thing is '`France0110`'. 
'`0110`' is the truth table for XOR, which is a pretty common encryption in CTFs. It would result in the unprintables as well. XOR requires a key though, so I used 'ARB' (in hex as '61 72 62' for cryptii). And now it has printables, and the end even has the flag format! Nice.
There's only one part left: 'France'. Searching up 'French Cryptography' brings up the Vigenère cipher. Oh duh! Using 'arb' as the key and the standard alphabet order, I decoded the message. Nice challenge, pretty fun to figure out.
![](/Images/2019/SunshineCTF/arb.PNG)
flag :`sun{arb_you_happy_to_see_me_1001130519}`
