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Sunshine CTF 2019 All Crypto Challenges Writeups

Welcome Crypto [50]

Here's the challenge

~C8 =39 A?2V8 73J:C 8FG7 AF?JJC2ECP

DF?LHb=r_>b0%_0520<c8bPN

Its ROT 47

Org lbh pna'g fbyir guvf punyyratr!

sun{w3lC0m3_T0_da_k4g3!}

CB1 [50]

Here's the challenge

Given in CB1.wav

Its Amateur Radio Phonetic Alphabet

When we listen and convert cipher is :

hkcgxkznkojkyulsgxin

And try Caesar Cipher

Here's flag: BEWARETHEIDESOFMARCH its maybe lower case I dont remember exactly.

CB2 [100]

Here's the challenge

Given in CB2.wav

Its same thing again Phonetic Alphabet

But this time be careful in begining part he said :

Codeword : clarinet

And Cipher is : DBDAABEDDDDCDEACADBBDDADDEABBB

And as you can see cipher became only "A B C D E"

First idea come up Polybius Cipher, I divide and convert number version:

DB DA AB ED DD DC DE AC AD BB DD AD DE AB BB

42 41 12 54 44 43 45 13 14 22 44 14 45 12 22

and new cipher here is.

When you try Decode

Cipher : 42 41 12 54 44 43 45 13 14 22 44 14 45 12 22
Key: clarinet

You can use your code, by hand or quickly try online tool:

https://md5decrypt.net/en/Polybius-square/

flag is : polysquaresrule

CB3 [150]

Here's the challenge

Given in CB3.wav

Its same thing again Phonetic Alphabet

Begining part he said:

Codeword: prideful
Cipher  : XDXGFVVVXXAFVFFVADGDDXAGAAFDFFFF

Idk cipher what this is, but I just a bit google search like a identify cipher unique letters and order alphabetically

ADFGVX cipher

And when u quickly try to decode

https://www.dcode.fr/adfgvx-cipher

Cipher  : XDXGFVVVXXAFVFFVADGDDXAGAAFDFFFF
Key     : prideful 

Flag is g3rm4n3ncrypt10n

ArbCrypt [200]

Here's the challenge

This challenge made me crazy

They gave hint France0110 has a meaning, it relates to what was used to encode the message.

France Part is obviously = Vigenere Cipher

0110 = is obviously Xor operation Truth Table

but we got a base64 ciphertext, its must be xor part but there is no key.

Here's cipher

BBcEDAJCDBMIAxUHA3gQBxEXCwdCAwQPDhxCGRMNawYHBRgDDQcNBRAWGxZCABoUAHgQBREEDwIQDgtCCx8DFR4RaxwLGFIWERQVBxdCGQgPBBARChgBaxYSFRQRAgIGDQRCFxwYAgAEGXgHDhgRChQMChgDAlITDQAWFQgGQRcNFAAQQR0LGRsWAghCAhQaa3gGExwMQRMEFRkIEhQNQRUEERYRGxdCChsQBwQFGHgEFwQRGBkLCBcKQREUDgYUElIVGAJoEhUIBQQRGFIWERwWFxAYDhdoBhAOF1IPCAYOCBkYTAEFDBsJQQEHBQIQCxRoFwgXABYID1IQBgYUElIEBwEBF3gADRsEAAcYQRwUFQYUBFIKDRwDBFIaCBcEa1IRDR0ZABsBPgsEFy0KEwMSES0XDi0IBxc9BRQ9UEJSUENRUUdTWA8=

After read many times challenge description I recognized this part

It's pretty ARB-itrary. France0110.

I thought "ARB" may be key but its no luck a bit corrupted results.

After hours I try lower case xor key as "arb". Thanks to lowercase. It worked.

Here's cipher hex values

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

XOR

6172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172626172

(arb) iterate same length with cipher but be careful last part you need to delete 1 byte like a ...arbarbar[deleted]

eefmp majbgeb
rfcuju bvmon xao
tedjaluodbtzd ahva
rdcfnproy jmatls
niy tpfwfe xzmebskjc
dptfscpdlv vnzcrfx
eojskfnkjac qlrttzd eourr oixitcz cfx

drnn aftkjsfo gfpdsze kirfvgy
fvvsykiieh cvotvs wyp
sgjdvsy tpntvbzoe
gblv mitlikz-sgmik sedprjf
vzuadjn rgtvs ffscv
blifauz nvttve hlnae xief
 slo{aic_yfv_hrqpp_uo_jfe_df_1001130519}

And now we can try vigenere with same key

key: arb

enemy lasagna
robust below wax
semiautomatic aqua
accompany slacks
why coffee gymnastic
motorcycle unibrow
existential plastic extra nightly cow

damn jettison goodbye through
everything center who
spidery concubine
pale lickity-split remorse
vitamin after force
already nested human wine
 sun{arb_you_happy_to_see_me_1001130519}

16-BIT-AES [100]

Here's the challenge

This answer describe as well why is not 16 bit key aes exist. https://crypto.stackexchange.com/questions/35117/are-there-attacks-on-aes-with-16-bit-keys

So if they use AES-128-ECB and key is 16 bit, key must be iterate itself so which means that we can find that key quickly bruteforce.

And I recognized everyconnection use same key that much easy. I create basic input and took encoded version then bruteforce this script.

from Crypto.Cipher import AES

a="dd15d5edb5a240a7c8f760ac86f0fca9"
b=[]
for x in range(0,len(a),2):
    b.append(int("0x"+a[x:x+2],16))
b=str(bytearray(b))
for key_b1 in range(256):
    for key_b2 in range(256):
        decipher = AES.new(str(bytearray([key_b1,key_b2])*8), AES.MODE_ECB)
        if decipher.decrypt(b)=="a"*16:
            print(bytearray([key_b1,key_b2]))
            break
#decipher = AES.new("fL"*8, AES.MODE_ECB)
#m=decipher.encrypt("BnMb7wu7tD3wYAYv")
#print(m.encode("hex"))

And key is became fL

Then we can find quickly flag.

Encrypt given text with that key

fLfLfLfLfLfLfLfL

and get flag

Flag is : Correct! The flag is sun{Who_kn3w_A3$_cou1d_be_s0_vulner8ble?}

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