DoubleS1405 CTF - Write-ups

Informations

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By Version Comment
noraj 1.0 Creation

CTF

  • Name : DoubleS1405 CTF
  • Website : doubles1405.com
  • Type : Online
  • Format : Jeopardy
  • CTF Time : link

1 - Mic Check - Misc

flag{W3lc0me_T0_DoubleS1405_CTF!}

50 - Easy_Crypt - Crypto

hui awavmiis hqklse mx o gvslfr tn zrqecuhcef ecdmiwihvg ysrk ac lgnvb e grvnsm fe mehjzrsjrr hoyjzv twupzvg oexsx fm xys qmoxsew tt u bdcncwl. dx wf e kcld nj gcqgvpduegsnzb wlpxbdxigmtb. zczk zg mmgpc_imlshvqi

It's Vigenere cipher.

Key is: ONEFOURZEROFIVE

Cleartext is:

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the vigenere cipher is a method of encrypting alphabetic text by using a series of interwoven caesar ciphers based on the letters of a keyword. it is a form of polyalphabetic substitution. flag is hello_vigenere

You can try it by youself.

50 - Crypto? - Crypto

https://drive.google.com/open?id=0B4SaQn817BNcUVR2OUIzRTVFNUU

#lowercase

Source text

We can see some ASCII decimals in the text:

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$ cat Crypto.txt| egrep -oh '\[[0-9]{1,3}\]' | tr -d '\n'
[123][82][48][43][86][71][80][73][79][95][78][48][74][95][51][68][65][75][84][95][49][73][125][25][88][38][28][29][30][31][32][33][34][35][36][37][38][39][40]

So a little ruby script to convert this into a string:

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open('Crypto.txt', 'r') do |f|
text = f.read()
chars = text.scan(/\[([0-9]{1,3})\]/).flatten
print chars.map{|c| c.to_i.chr}.join
end

And the output is: {R0+VGPIO_N0J_3DAKT_1I} plus some garbage. SO we keep only what's between curly braces. And now let's apply a Caesar bruteforce.

The key was +11: {C0+GRATZ_Y0U_3OLVE_1T}.

To submit we need to lowercase (see challenge's description).

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irb(main):003:0> "{C0+GRATZ_Y0U_3OLVE_1T}".downcase
=> "{c0+gratz_y0u_3olve_1t}"

But that was not good (no flag format...), I needed to remove curly braces and change + into n: c0ngratz_y0u_3olve_1t.

150 - RSA - Crypto

RSA 쉽던데..?

RSA is easy, huh?

#Change Code due to logic error

#Change server address due to server overload

nc 203.251.182.94 4000

https://drive.google.com/open?id=0B4SaQn817BNcck83M3lyS1dHc1E

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> if True:
> enc_data = SECRET_LOGIC()
> enc = powmod(enc_data, e, n)
> conn.send("enc : %d\ne : %d\np : %d\nq : %d\n" % (enc, e, p, q))
>
> recv = conn.recv(1024)
>
> if end - start > 1:
> conn.send("time out")
> conn.close()
>
> if recv == enc_data:
> conn.send(FLAG)
> conn.close()
>
> conn.send("wrong")
> conn.close()
>
> else:
> conn.close()
> print "error"
>

I wrote a ruby script to solve this RSA:

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#!/usr/bin/ruby
require 'socket'
require 'openssl'
require 'base64'
# Source of int2Text: http://stackoverflow.com/questions/42993763/how-to-convert-bytes-in-number-into-a-string-of-characters-character-represent#42999986
def int2Text(int)
a = []
while int>0
a << (int & 0xFF)
int >>= 8
end
return a.reverse.pack('C*')
end
# Source of egcd: https://gist.github.com/jsanders/6735046
def egcd(a, b)
u_a, v_a, u_b, v_b = [ 1, 0, 0, 1 ]
while a != 0
q = b / a
a, b = [ b - q*a, a ]
u_a, v_a, u_b, v_b = [ u_b - q*u_a, v_b - q*v_a, u_a, v_a ]
# Each time, `u_a*a' + v_a*b' = a` and `u_b*a' + v_b*b' = b`
end
[ b, u_b, v_b ]
end
def modinv(a, m)
g, x, y = egcd(a, m)
if g != 1
raise 'modular inverse does not exist'
else
return x % m
end
end
# Server connection
hostname = '203.251.182.94'
port = 4000
s = TCPSocket.open(hostname, port)
raw = ''
answer_sent = false
while line = s.gets # Read lines from the socket
puts line.chop # And print with platform line terminator
raw += line
if raw.match(/q :/) && answer_sent == false
c = raw.match(/^enc : ([0-9]*)$/).captures[0].to_i
e = raw.match(/^e : ([0-9]*)$/).captures[0].to_i
p_int = raw.match(/^p : ([0-9]*)$/).captures[0].to_i
q = raw.match(/^q : ([0-9]*)$/).captures[0].to_i
phi = (p_int - 1) * (q - 1)
d = modinv(e, phi)
n = p_int*q
# more efficient than m_int = (c ** d) % n
m_int = c.to_bn.mod_exp(d, n).to_i
# cleartext is base64 containing an integer
m_b64 = int2Text(m_int)
puts "base64 : #{m_b64}"
m = Base64.decode64(m_b64)
puts "int : #{m}"
# integer must be transform to text again
m_text = int2Text(m.to_i)
puts "text : #{m_text}"
s.puts m_text
answer_sent = true
end
end
s.close # Close the socket when done

An example of output from my script:

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$ ruby rsa.rb
enc : 491122313213059047386587355703000827178576243139630115107824179230229619638230359264090916176374104821742487423477640046091956570072907849863979916992012277080517371737063659607380884337353809683315958180812970513119961236305718314856508806575062351521772465118704267216578937782652026891140564554699280499155421872621380063738498801480672462553217846124110631365349374404141214553794903345385240728734320886304119126434949216541748009316097394118225456235324726150812532266854157628444800299612332540488915092939585065747993258534210496110027307168741103227954668422763393074609550128115559442563807710071873205546623595056058012799686847872624717521357873024268621752914303954208374908780786312254128829726625119613542022988078172368143379162734941970190142937076819213707077010206986159149431729867187504847837987139017023926853873484450256772068033378979496285329969789758899711272261594435793646971822229226412599644872158251842719957278621048558842287038138868659481907505634015392299324362317019730935313919966703702616226085783306686686638481716975435704933248720402377816087988961312653435421407390516891418321288636502551122036111983361547523717812270804583522804832003079689675943949319910428221021037993090021553176551473
e : 65537
p : 32317006071311007300714876688669951960444102669715484032130345427524655138867890893197201411522913463688717960921898019494119559150490921095088152386448283120630877367300996091750197750389652106796057638384067568276792218642619756161838094338476170470581645852036305042887575891541065808607552399123930385521954285833276606292740174507176908054077273016103644389803261062635470374515595892199454891155463898488297024308700957247533881208055894474582694028535079545281620566442541400114261729854235365927395115457109476960042332821732358509197923144094801013581965651112146928918286923938064987973879624251895591220179
q : 32317006071311007300714876688669951960444102669715484032130345427524655138867890893197201411522913463688717960921898019494119559150490921095088152386448283120630877367300996091750197750389652106796057638384067568276792218642619756161838094338476170470581645852036305042887575891541065808607552399123930385521990685772174514834944123086486002362345153147580453526134037171595087108668773961917317502849945855689432886442889958513294157709640362363734479327004391952407569596153273880472331909250263593691635107321048666489395316204775782962517724272901158130972610802371589601746375325078943967095960733617174538141999
base64 : NTEzMjU2MzI3MTA4MDgzNzE3Mjg4MzE2OTU4MTc4MTEwNTI1NzE4OTEyMTM0MDMzNTY3Mjc3Njc2NTI1MjYxMjI5MjExMzE0MDE1ODk=
int : 51325632710808371728831695817811052571891213403356727767652526122921131401589
text : qyGUcwNSzMF8uCOxByuBpTZ53Bn2IJau
flag{Y34hh_RSA_1S_S0_E4sy}

Flag is: Y34hh_RSA_1S_S0_E4sy.

50 - Instead of 5 billion won, I give you a cute handcuffs - Misc

In the fall of 2016, an e-mail arrived to detective Kim, who is tracking up economic fugitives who have shed large amounts of money from a large corporation.

I know the location of the perpetrator, but I can not tell you, and if you pay me 5 billion, I was attached with a photograph that would tell you the exact location.

The attached photo was taken nearest to his refuge.

Can Kim find out where the shooter escaped?

(The correct answer is the address in Korea. Example: 52-3, Yeonsan-dong, Yeonje-gu, Busan, Republic of Korea)

https://drive.google.com/open?id=0B4SaQn817BNceWIxTlE5RVlIbkU

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$ exiftool 1.JPG | grep -i gps
GPS Latitude Ref : North
GPS Longitude Ref : East
GPS Altitude Ref : Above Sea Level
GPS Time Stamp : 04:50:51
GPS Speed Ref : km/h
GPS Speed : 0
GPS Img Direction Ref : True North
GPS Img Direction : 115.244898
GPS Dest Bearing Ref : True North
GPS Dest Bearing : 115.244898
GPS Date Stamp : 2016:11:06
GPS Horizontal Positioning Error: 10 m
GPS Altitude : 95.3 m Above Sea Level
GPS Date/Time : 2016:11:06 04:50:51Z
GPS Latitude : 37 deg 40' 11.91" N
GPS Longitude : 127 deg 0' 29.88" E
GPS Position : 37 deg 40' 11.91" N, 127 deg 0' 29.88" E

Google Maps sucks: DEG40, 94551, Deggendorf, Niederbayern, Bayern, Allemagne

findlatitudeandlongitude.com works better: 238-1 Ui-dong, Gangbuk-gu, Seoul, South Korea but we need to respect description format: 238-1, Ui-dong, Gangbuk-gu, Seoul, Republic of Korea.

50 - Cute - Forensic

Very cute..

https://drive.google.com/open?id=0B4SaQn817BNcOFJmel91OFA4azA

The file is a mkv (Matroska):

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$ file file
file: Matroska data
$ mv file file.mkv

There is a subtitle track. We can see with mkvtoolnix that the subtitle track id for mkextract is 2.

So I extracted it:

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$ mkvextract tracks "file.mkv" 2:out.srt

And we get someting like:

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00:00:01,000 --> 00:00:01,021
e
2
00:00:01,022 --> 00:00:01,042
eh
3
00:00:01,043 --> 00:00:01,064
eh~
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00:00:01,065 --> 00:00:01,086
eh~!
5
00:00:01,087 --> 00:00:01,108
eh~!@
[...]
81
00:00:10,539 --> 00:00:10,539
f
82
00:00:10,539 --> 00:00:10,539
l
83
00:00:10,539 --> 00:00:10,539
a
84
00:00:10,539 --> 00:00:10,539
g
[...]

So we can see flag from the entry 81. Flag is written between entries 81 and 110: flag{v3r_cut3_p4Rr0t_12nt_!t?}.

Flag is v3r_cut3_p4Rr0t_12nt_!t?.

100 - iterator - Web

Warning : we didn't flag this challenge!

The server is executing:

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<?php
$directory = $_GET['dir'];
$cond = explode("=", urldecode($_SERVER['QUERY_STRING']));
if (!is_dir($cond[1])) echo 'folder does not exist.';
else {
$directory = new DirectoryIterator($directory);
foreach($directory as $file) {
echo "\n<!--";
if (preg_match("/txt$/i", $file->getFilename())) echo $file->getFilename();
else;
echo "-->";
}
}
?>

Ok so let's try /var/www/html:

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<!--534dddm3.txt-->
<!---->
<!---->
<!---->
<!---->
<!---->

Now try to reach http://175.214.81.36/534dddm3.txt:

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Hello! If you are reading this article you have succeeded in reading the file.
what is left is finding a hidden diretory!

We didn't manage to find the hidden directory.

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