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Root-me (Cryptanalysis) - Writeup

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By crmulent 8 min read

A writeup for Root-me’s cryptanalysis challenges.

Encoding - ASCII

Flag: 2ac376481ae546cd689d5b91275d324e

Decode the string.

Input

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4C6520666C6167206465206365206368616C6C656E6765206573743A203261633337363438316165353436636436383964356239313237356433323465

CyberChef’s Recipe

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From_Hex('None')

Output

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Le flag de ce challenge est: 2ac376481ae546cd689d5b91275d324e

Encoding - UU

Flag: ULTRASIMPLE

Very used by the HTTP protocol

Get the validation password.

Input

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B5F5R>2!S:6UP;&4@.RD*4$%34R`](%5,5%)!4TE-4$Q%"@``

Output

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Very simple ;)
PASS = ULTRASIMPLE

Hash - Message Digest 5

Flag: weak

Bob found that Ronald Rivest is a terrific cryptanalyst.

Crack the given hash.

Input

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7ecc19e1a0be36ba2c6f05d06b5d3058

Output

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7ecc19e1a0be36ba2c6f05d06b5d3058 : weak

Hash - SHA-2

Flag: a7c9d5a37201c08c5b7b156173bea5ec2063edf9

This hash was stolen during a session interception on a critical application, errors may have occurred during transmission. No crack attempt has resulted so far; hash format seems unknown. Find the corresponding plaintext.

The answer is the SHA-1 of this password.

Hashes must have a-f and 0-9 characters only. Following this rule we must exclude the “k” from the hash.

So, 96719db60d8e3f498c98d94155e1296aac105ck4923290c89eeeb3ba26d3eef92 becomes 96719db60d8e3f498c98d94155e1296aac105c4923290c89eeeb3ba26d3eef92

Input

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96719db60d8e3f498c98d94155e1296aac105c4923290c89eeeb3ba26d3eef92

Output

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96719db60d8e3f498c98d94155e1296aac105c4923290c89eeeb3ba26d3eef92 : 4dM1n

Input

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4dM1n

CyberChef’s Recipe

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SHA1(80)

Output

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a7c9d5a37201c08c5b7b156173bea5ec2063edf9

Shift cipher

Flag: Yolaihu

Recover the password

Index : keep on turning.

Input

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L|ky+*^*zo*kvsno|*kom*vo*zk}}*cyvksr

Output

  • Shift = 10
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    Brao! Tu peu alider aec le pass Yolaihu

Monoalphabetic substitution - Caesar

Flag: ujqcsddessxsffes

Emperor regresses

We just caught the messenger of the Emperor. He transmitted a coded message to his son. This could be an important message. You’ve to decrypt it ! To validate, you must enter the concatenation of the first letters of each line followed by the concatenation of the last letters of each line (for example : tfhqdlhfpkmeokgq).

Input

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tm bcsv qolfp
f'dmvd xuhm exl tgak
hlrkiv sydg hxm
qiswzzwf qrf oqdueqe
dpae resd wndo
liva bu vgtokx sjzk
hmb rqch fqwbg
fmmft seront sntsdr pmsecq

The title hints that this challenge is a caesar cipher. After several trials, I found out that the plaintext is in French, making it harder to decode. We can decode the whole text by incrementing the shift of caesar by 1 for each word.

  • key=1
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    un cdtw rpmgq
g'enwe yvin fym uhbl
<...>
  • key=2
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    vo deux sqnhr
h'foxf zwjo gzn vicm
<...>
  • key=3
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    wp efvy trois
i'gpyg axkp hao wjdn
<...>

Retaining each plaintext, we have the following:

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un deux trois

which is 1, 2, 3 in French.

Plaintext

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un deux trois
j'irai dans les bois
quatre cinq six
cueillir des cerises
sept huit neuf
dans un panier neuf
dix onze douze
elles seront toutes rouges

Pixel Madness

Flag: SOLUTION

Decrypt the code.

Clue : 0 = #FFFFFF 1 = #000000

Submit password in CAPITAL LETTERS.

Use the following code to solve the challenge.

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from PIL import Image

data = [
            "0x3+1x1+0x1+0x1+0x7+1x2+0x15+1x1+0x8+1x1+0x8+1x1+0x1+1x1+0x1+1x1+0x1+1x1+0x1+1x1+0x3+1x1+0x1+1x1+0x3+1x1+0x1+1x4+0x2+1x1+0x25",
            "0x2+1x1+0x4+1x1+0x4+1x3+0x1+1x2+0x2+1x8+0x11+1x4+0x1+1x3+0x6+1x2+0x4+1x1+0x4+1x2+0x7+1x4+0x4+1x2+0x7+1x2+0x3+1x2+0x3",
            "0x3+1x1+0x2+1x1+0x2+1x1+0x11+1x2+0x2+1x3+0x7+1x1+0x4+1x2+0x2+1x2+0x7+1x1+0x6+1x1+0x2+1x1+0x4+1x3+0x1+1x1+0x4+1x1+0x2+1x1+0x2+1x1+0x3+1x1+0x2+1x3+0x2+1x2+0x3",
            "1x1+0x2+1x1+0x4+1x1+0x2+1x1+0x1+1x1+0x2+1x1+0x2+1x1+0x1+1x2+0x2+1x2+0x1+1x2+0x3+1x1+0x3+1x1+0x2+1x2+0x1+1x3+0x3+1x1+0x2+1x1+0x4+1x2+0x1+1x1+0x4+1x1+0x3+1x2+0x12+1x2+0x1+1x1+0x3+1x7+0x3",
            "0x3+1x1+0x7+1x1+0x1+1x1+0x4+1x1+0x2+1x2+0x2+1x2+0x4+1x1+0x2+1x1+0x1+1x2+0x1+1x8+0x1+1x1+0x4+1x1+0x5+1x1+0x3+1x2+0x2+1x1+0x1+1x2+0x2+1x1+0x3+1x2+0x9+1x1+0x1+1x2+0x2+1x3+0x2+1x1",
            "0x7+1x1+0x4+1x1+0x4+1x1+0x1+1x1+0x1+1x7+0x3+1x1+0x1+1x2+0x3+1x1+0x1+1x6+0x1+1x1+0x3+1x1+0x2+1x1+0x14+1x2+0x8+1x1+0x10+1x2+0x3+1x2+0x1+1x1+0x1",
            "0x6+1x5+0x4+1x1+0x7+1x1+0x2+1x1+0x3+1x2+0x4+1x1+0x8+1x1+0x3+1x2+0x1+1x2+0x3+1x1+0x8+1x1+0x2+1x2+0x1+1x1+0x3+1x7+0x5+1x2+0x2+1x1+0x2+1x2+0x3",
            "0x1+1x1+0x2+1x1+0x1+1x2+0x5+1x1+0x6+1x2+0x3+1x1+0x2+1x1+0x1+1x2+0x20+1x8+0x1+1x1+0x1+1x1+0x4+1x2+0x3+1x1+0x2+1x2+0x3+1x2+0x7+1x2+0x3+1x2+0x4",
            "0x2+1x1+0x3+1x5+0x5+1x2+0x7+1x1+0x4+1x2+0x2+1x1+0x2+1x2+0x1+1x1+0x3+1x1+0x6+1x2+0x2+1x2+0x3+1x2+0x2+1x3+0x1+1x1+0x6+1x3+0x3+1x5+0x3+1x1+0x4+1x1+0x5",
            "0x4+1x2+0x3+1x2+0x3+1x1+0x5+1x2+0x2+1x1+0x1+1x1+0x1+1x1+0x1+1x2+0x9+1x1+0x3+1x1+0x2+1x1+0x1+1x1+0x2+1x1+0x1+1x2+0x2+1x1+0x2+1x1+0x1+1x1+0x4+1x3+0x1+1x1+0x2+1x2+0x3+1x2+0x3+1x1+0x5+1x1+0x4+1x1+0x2",
            "0x6+1x5+0x4+1x1+0x1+1x1+0x2+1x2+0x6+1x1+0x1+1x7+0x4+1x3+0x3+1x1+0x4+1x1+0x2+1x2+0x4+1x1+0x6+1x1+0x6+1x8+0x3+1x1+0x5+1x1+0x7",
            "0x2+1x1+0x3+1x6+0x4+1x1+0x1+1x3+0x4+1x1+0x2+1x2+0x4+1x1+0x5+1x1+0x2+1x1+0x3+1x2+0x3+1x1+0x2+1x3+0x1+1x1+0x2+1x2+0x3+1x3+0x2+1x3+0x9+1x1+0x4+1x2+0x7+1x2",
]

# Function to know the max width
x = lambda x: [int(i[2]) for i in x.split('+')]
max_width = max([sum(i) for i in list(map(x, data))])
print(([sum(i) for i in list(map(x, data))]))

im = []

for row in data:
    k = 0
    for column in row.split('+'):
        column = list(map(int,column.split('x')))
        if column[0] == 0:
            for i in range(column[1]):
                if k == 100: break
                im.append((255,255,255))
                k+=1
        elif column[0] == 1:
            for i in range(column[1]):
                im.append((0,0,0))
                k+=1

image = Image.new("RGB", (max_width, len(data)))
image.putdata(im)
image.show()

# SOLUTION

File - PKZIP

Flag: 14535

A protected ZIP file, you have to find what’s inside.

Use zip2john and john to crack the password

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$ zip2john ch5.zip > hash.txt
$ john hash.txt 
<...>
14535            (ch5.zip/readme.txt)     
<...>

Polyalphabetic substitution - Vigenère

Flag: Loyd Blankenship

We need your expert opinion on this document. This is an old letter and it appears that it is important for the pirates that we are searching for. Your mission is to decipher the text and give us the full name of the author (example : “John Doe”).

Input

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Moi Tepdsi Fhrujrlhf

Nu egxex g'vla jmmg ifvgkvq ehcclkk'lgm, p'xgk ihvfshm rrgz pqw whiighyj. "Wptbutsi: Gr nwccxzgqrg tfixai bshk qibti urshfdtamcyr,
"Tfixzxmxvhb u'nu 'lmgxxf' riyie pr iwitaesi q'nbv uhrcyr"...

Loktuie kblgvl, asgw yxg dxtie.
<...>

Output

  • Key = THEMENTOR
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    THEHACKERMANIFESTOUNAUTRESESTFAITPRENDREAUJOURDHUICESTPARTOUTDANSLESJOURNAUXSCANDALEUNADOLESCENTARRETEPOURCRIMEINFORMATIQUEARRESTATIONDUNHACKERAPRESLEPIRATAGEDUNEBANQUESATANESGOSSESTOUSLESMEMESMAISAVEZVOUSDANSVOTREPSYCHOLOGIEENTROISPIECEETVOTREPROFILTECHNOCRATIQUEDEUNJOURPENSEAREGARDERLEMONDEDERRIERELESYEUXDUNHACKERNEVOUSETESVOUSJAMAISDEMANDECEQUILAVAITFAITAGIRQUELLESFORCESLAVAIENTMODELEJESUISUNHACKERENTREZDANSMONMONDELEMIENESTUNMONDEQUICOMMENCEAVECLECOLEJESUISPLUSASTUCIEUXQUELAPLUPARTDESAUTRESENFANTSLESCONNERIESQUILSMAPPRENNENTMELASSENTJESUISAUCOLLEGEOUAULYCEEJAIECOUTELESPROFESSEURSEXPLIQUERPOURLAQUINZIEMEFOISCOMMENTREDUIREUNEFRACTIONJELAICOMPRISNONMMEDUBOISJENEPEUXPASMONTRERMONTRAVAILJELAIFAITDANSMATETESATANEGOSSESILLACERTAINEMENTCOPIETOUSLESMEMESJAIFAITUNEDECOUVERTEAUJOURDHUIJAITROUVEUNORDINATEURATTENDSUNEMINUTECESTCOOLCAFAITCEQUEJEVEUXSIOXDMLLCEYJOPQXJKVMYMYDFWYLYOBKQVMQJJQXLFHHIJJFOAQTMQCHJJYUPWXRMSFNMUUMHONIQQVWVKGJKYOHHIIGTFLUSSZTYVRGXSWVJYVRCVHKCZMUBRUWXQCIZULUSSZTYVRGXQSQDYUXQ

To make it more readable, let’s decode it using CyberChef.

CyberChef’s Recipe

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Vigenère_Decode('THEMENTOR')

Output

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The Hacker Manifesto

Un autre s'est fait prendre aujourd'hui, c'est partout dans les journaux. "Scandale: Un adolescent arrete pour crime informatique,
"Arrestation d'un 'hacker' apres le piratage d'une banque"...

Satanes gosses, tous les memes.
<...>

By searching the web about The Hacker Manifesto Author, I found out that the author was Loyd Blankenship.

File - Insecure storage 1

Flag: F1rstP4sSw0rD

Mozilla Firefox 14

Retrieve the user’s password.

To retrieve the user’s password, I used the tool firepwd.

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$ python3 firepwd.py -d ../.mozilla/firefox/o0s0xxhl.default/
 SEQUENCE {
   SEQUENCE {
     OBJECTIDENTIFIER 1.2.840.113549.1.12.5.1.3 pbeWithSha1AndTripleDES-CBC
     SEQUENCE {
       OCTETSTRING b'cdc0d5a7ce257959a27755b057ba3764afa6ecc6'
       INTEGER b'01'
     }
   }
   OCTETSTRING b'842aac4fc88d6c99993695d323dc4440576318024446f6fbf3131de5ca5cf524b2661f1efab6ce736ef94b67c8d18c0bbcd8741b4a363ad4adb8c228bb5e6e644fdfdcb599a8f6fd00f0d6fb45f347d43154ce129b917efc180205190539e4ca'
 }
decrypting privKeyData
 SEQUENCE {
   INTEGER b'00'
   SEQUENCE {
     OBJECTIDENTIFIER 1.2.840.113549.1.1.1 pkcs-1
     NULL 0
   }
   OCTETSTRING b'3043020100021100f8000000000000000000000000000001020100021900fecb1985f7f1522089086b0894e6a8ab684507321af2f11f020100020100020100020100020115'
 }
decoding b'3043020100021100f8000000000000000000000000000001020100021900fecb1985f7f1522089086b0894e6a8ab684507321af2f11f020100020100020100020100020115'
 SEQUENCE {
   INTEGER b'00'
   INTEGER b'00f8000000000000000000000000000001'
   INTEGER b'00'
   INTEGER b'00fecb1985f7f1522089086b0894e6a8ab684507321af2f11f'
   INTEGER b'00'
   INTEGER b'00'
   INTEGER b'00'
   INTEGER b'00'
   INTEGER b'15'
 }
sqlite
decrypting login/password pairs
http://www.root-me.org:b'shell1cracked',b'F1rstP4sSw0rD'

Known plaintext - XOR

Flag: ICONOCLASTE

For this challenge you will need to decypher a simple XORed picture.

This BMP picture was mistakenly encrypted. Can you recover it ?

I used this tool to know the possible keys used to encrypt the image. After getting the possible key, I used CyberChef to decrypt the image.

CyberChef’s Recipe

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XOR({'option':'UTF8','string':'fallen'},'Standard',false)
Render_Image('Raw')

ELF64 - PID encryption

Flag: ``

Bad idea to use predictable stuff.

Source Code

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/*
 * gcc ch21.c -lcrypt -o ch21
 */
 
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <crypt.h>
#include <sys/types.h>
#include <unistd.h>
 
int main (int argc, char *argv[]) {
    char pid[16];
    char *args[] = { "/bin/bash", "-p", 0 };
 
    snprintf(pid, sizeof(pid), "%i", getpid());
    if (argc != 2)
        return 0;
 
    printf("%s=%s",argv[1], crypt(pid, "$1$awesome"));
 
    if (strcmp(argv[1], crypt(pid, "$1$awesome")) == 0) {
        printf("WIN!\n");
        execve(args[0], &args[0], NULL);
 
    } else {
        printf("Fail... :/\n");
    }
    return 0;
}

Hash - LM

Flag: admin!!

Retrieve the password of the Administrator user from the information output by the secretsdump tool of the Impacket suite.

Note: the flag is in lowercase

Input

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d3bf255c530633b9aad3b435b51404ee:31d6cfe0d16ae931b73c59d7e0c089c0

Output

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31d6cfe0d16ae931b73c59d7e0c089c0:
d3bf255c530633b9aad3b435b51404ee:ADMIN!!

Hash - DCC

Flag: ilikethat

Retrieve the password of the Administrator user from the information output by the secretsdump tool of the Impacket suite.

Input

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Administrator:15a57c279ebdfea574ad1ff91eb6ef0c

Output

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$ john --format=mscash --wordlist=rockyou.txt hash.txt
<...>
ilikethat        (Administrator)     
<...>

System - Android lock pattern

Flag: 145263780

Having doubts about the loyalty of your wife, you’ve decided to read SMS, mail, etc in her smarpthone. Unfortunately it is locked by schema. In spite you still manage to retrieve system files.

You need to find this test scheme to unlock smartphone.

NB : validation password is a number (archive sha256 is 525daa911d4dddb7f3f4b4ec24bff594c4a1994b2e9558ee10329144a6657f98)

Resources

Procedure: https://www.forensicfocus.com/articles/android-forensics-study-of-password-and-pattern-lock-protection/

SHA-1 Dictionary: https://github.com/Machiry/AndroidGestureBreaker/blob/master/AndroidGestureSHA1.txt

Tool https://github.com/bolisettynihith/android-pattern-decoder

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https://github.com/MGF15/P-Decode

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$ python3 android-pattern-decoder/androidpatterndecode.py -g ./android/data/system/gesture.key -d AndroidGestureSHA1.txt 
[+] Pattern retrieved from gesture.key file is: 256374891

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$ python3 P-Decode/P-Decode.py -f ./android/data/system/gesture.key 

        |~)  |~\ _ _ _  _| _
        |~ ~~|_/}_(_(_)(_|}_ v0.5

             [ {41}ndr0id Pa77ern Cr4ck t00l. ]
       
[*] Pattern SHA1 Hash   : 2C3422D33FB9DD9CDE87657408E48F4E635713CB

[+] Pattern Length      : 9

[+] Pattern             : 145263780

[+] Pattern SVG         : 145263780.svg

[*] Time                : 0.4 sec

Hash - DCC2

Flag: ihatepasswords

Retrieve the password of the Administrator user from the information output by the secretsdump tool of the Impacket suite.

Input

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$DCC2$10240#administrator#23d97555681813db79b2ade4b4a6ff25

Output

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$ hashcat -m 2100 -a 0 hash.txt rockyou.txt
<...>
$DCC2$10240#administrator#23d97555681813db79b2ade4b4a6ff25:ihatepasswords
<...>

Transposition - Rail Fence

Flag: Frozen chicken

Invaders Must Die

USA, American Civil War, August 3, 1862. You are on patrol around the camp when you see an enemy rider. Once you intercepted him, you discover that he carries a message but nobody at the camp manages to uncipher it. You are the only hope to find the hidden information. It could be crucial !

Input

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Wnb.r.ietoeh Fo"lKutrts"znl cc hi ee ekOtggsnkidy hini cna neea civo lh

CyberChef’s Recipe

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Rail_Fence_Cipher_Decode(8,0)

Output

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Will invade Kentucky on October the eighth. signal is "Frozen chicken".

Hash - NT

Flag: iloveyou99!

One rule to rule them all

Retrieve the password of the Administrator user from the information given by the secretsdump tool of the Impacket suite.

Input

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aad3b435b51404eeaad3b435b51404ee:b4f79698831d92b61f886438e36c0c52

Output

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b4f79698831d92b61f886438e36c0c52:iloveyou99!

RSA - Factorisation

Flag: ``

cryptography
This post is licensed under CC BY 4.0 by the author.

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