'분류 전체보기' 카테고리의 글 목록 (6 Page)

forensic

Misplaced

hmm what is it?

I changed file extension. (file.what to file.zip)

oh i got something but i don't know the password.

when i opened it using 7 zip file manager, i could get a password!

i thought it is ppt file.

then i found a flag.

Nice Duck!

mp4 file

Splitted flag

split file

fix file

done.

web

Flag Script

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I googled 'BZh9' and i found it is bzip2 file

s3cr3t

interesting

export!

bitlocker.

same with 'little tricks' at starctf mandu-mandu.tistory.com/416

starCTF 2021 write-up

*CTF 2021 Crypto 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 from random import randint import o..

mandu-mandu.tistory.com

try harder

802.11 wpa

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2020?

Reversing

That's not crypto

decompile it

bruteforce!

저작자표시 비영리 동일조건

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starCTF 2021 write-up (0)	2021.01.18

blockchain

sanity check

remix.ethereum.org/

Remix - Ethereum IDE

remix.ethereum.org

code copy&paste

compile

connect metamask rinkeby wallet, load contract address and call welcome func. done.

secure enclave

0x9B0780E30442df1A00C6de19237a43d4404C5237

It can store text and return text

I thought constroctor of this contract store the flag in this contract.

When you call set_secret to store some string, transaction occurs.

I could find a transaction at etherscan

crackme.sol

0xDb2F21c03Efb692b65feE7c4B5D7614531DC45BE

"arg3 is a overflow" is a hint in line 6.

uint MaxValue + 1 = 0

so, i can pass line 7 and 11 condition.

uint is short for uint256, so maximum value is 2**256-1

arg3 = 115792089237316195423570985008687907853269984665640564039457584007913129639935

arg1 = 20 ^ 0x70 = 100

decrypt(arg2) = "offshift ftw"

ah just brute force a-z :)

arg2 = evvixyvj vjm

"100","evvixyvj vjm","115792089237316195423570985008687907853269984665640564039457584007913129639935"

compile error

To find correct uint array size, brute force all value.

if size is over 26 :

if size is equal 26:

got a flag.

crypto casino

0x186d5d064545f6211dD1B5286aB2Bc755dfF2F59

Edit code to get ouput value correctly.

To get a flag, need guessing and correct two times.

seed = b37c910f4e0df0efafb35a55489604369808b6de642ff1dbab5062680afaddcd

the block.number is the number of the mined block containing the transaction.

deploy a contract that returning result of uint(keccak256(abi.encodePacked(seed, block.number+3))) ^ 0x539

get block.number and hash

hash <= block.number <= 7956519 + 3

Then pend transaction that call bet(60284626633715439770582715312106605696128214001910194886489589200393495573074) before #7956522 block mined.

hmm out of gas

increase gas price and gas limit.

done.

RICH CLUB

0xC7bEc01281648D3A7F9BB86B811A2de5B1E0cc61

hmm

for line 21, swap eth to uniswap app.uniswap.org/#/swap

Uniswap Interface

app.uniswap.org

hmm

grant_membership() function에 pubkey를 넘겨주면 encoded flag를 넘겨준다고 한다.
그럼 그걸 decrypt 하면 된다고 하는데

그럼 solidity 코드는 어떻게 수정해서 써먹어야되는거야 :(

crypto

factorize

p와 q의 상위 512비트는 base로 동일함.

따라서 sqrt(n)의 상위 512비트와 동일.

hmm

엥 그냥 factorize 돌리면 됐었던 문제였넹

optimizer

별거 없다. simple.

주어진 배열 정렬할 때 이동 횟수 구해서 입력해주면 된다.

pyjail

hmm

0x414141

go to www.offshift.io/

You are being redirected...

www.offshift.io

go to github

github.com/offshift-dev

offshift-dev - Overview

Cryptographically private offshore storage & DeFi protocol on Ethereum - offshift-dev

github.com

There are two commits in january 2021

interesting

click

download pyc file.

decompile pyc using uncompyle6

decrypt

a link!

download a smashing.pdf via mega link.

hmm is it xor?

correct!

bitcoin paper..?

compare with smashing.pdf and original bitcoin paper pdf.

smashing.pdf is a little bigger.

a zip file is here :)

is locked.

dictionary attack

dictionary file i used : rockyou.txt

file reader

it filters "/flag.txt"

but we can get all .txt files using glob.glob("*.txt").

Shjail

eval eval py{r..u}hon\\ -m\\ S{h..j}mpleHTTPServer\\;

hmm

perl flag.[a-z][a-z][a-z]

web

graphed 2.0

there is a input form

There is a code send graphql query.

gist.github.com/craigbeck/b90915d49fda19d5b2b17ead14dcd6da

Introspection query for GraphQL

Introspection query for GraphQL. GitHub Gist: instantly share code, notes, and snippets.

gist.github.com

i found a fake flag :(

no flag in users

no flag in coolnotes

hmm

아 역시 getNote를 사용하는 것이었다.
하나 남은게 getNote였는데, q에 뭐가 들어가는지 몰라서 포기했는데..
쿼리문을 넣어버리는 거였넹.

maze

./robots.txt

./sup3r_secr37_@p1

good.

hmm

아 username이 pop_eax가 아니라 XFT였다.

hackme

?cmd=help

?cmd=123456

hmm

nl /*
nl *

>cat
* /f*

저작자표시 비영리 동일조건

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Forensics

Peace

"the most popular social media" == twitter

holloway.nz/steg/

Twitter Secret Messages - Steganography

Hide secret messages in your tweets (or any text) with steg-of-the-dump.js. Tweet i had a great day at the beach! #sunshine Hidden Message kidnapped by pirates Tweet to post Warning: Not all of Hidden Message stored in tweet. Add more text (e.g. spaces) to

holloway.nz

Linux Help

given a png file.

Brutus Killer

caesar cipher

Suspicious Riddle

use volatility

Win7SP1x64

pid = 2816

Hash Riddle

dump it

extract s3cr3t.mp3 file

Tewm yzc bs, _n3q0rq_a4ok3w}

hmm what is it?

Source Riddle

iexplore.exe

Anarchy Riddle

procdump -p 2816 -D ./

Memory Detective

filescan | findstr "flag"

dumpfiles -Q 0x00000000???(i forgot) -D ./

pstree

procdump -p 2108 -D ./

hmm..

i couldn't find file which named volatilesecret.

Cryptography

valhalla

runes

Rick and morty on adventure

image search

I found this

Monk

dec to ascii

Hand-Work

mp3 file.

it says "dot" and "dash"

listen and write . and _

then decode it.

... .... ._ _.. ___ .__ _._. _.._. . ...._ ... _.__ _._. ._. _.__ .__. _ _____

What are Semaphores processes?

i found this.

Colored....P

www.boxentriq.com/code-breaking/hexahue

Hexahue Alphabet - decoder, translator | Boxentriq

Tool to decrypt Hexahue code. Hexahue, and variants of it, are used occasionally in geocaching mystery caches, CTFs and logic puzzles. Look specifically for the color combinations to recognize it.

www.boxentriq.com

DORAEMON

OSINT

Eased OSINT

wav file

Intel Expert

from north korea.

city : Pyongyang

저작자표시 비영리 동일조건

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System32.kr

RSA108 풀이

rsa107문제랑 거의 똑같다.

n값을 base-16 hex로 보면 rsa107문제처럼 중간에 0이 많이 있다.

000...000을 기준으로 앞 뒤 잘라서 factor을 찾으면

이렇게 나온다.

저 공통인수를 p로 두고 n/p 로 q를 구할 수 있다.

p랑 q 구했으니 d구해서 c 복호화하면 끝.

저작자표시 비영리 동일조건

'WAR GAME > System32.kr' 카테고리의 다른 글

System32.kr [RSA107] 풀이 (0)	2021.01.18
System32.kr [EZB64] 풀이 (0)	2019.12.28
System32.kr [RSA106] 풀이 (0)	2019.05.19
System32.kr [RSA104] 풀이 (0)	2019.05.19
System32.kr [RSA105] 풀이 (0)	2019.05.19

System32.kr

RSA107 풀이

간만에 system32.kr 문제를 풀었다.

보면 모듈러 n 중간에 많은 0이 특징이다.

rsa 문제 풀이들을 찾아보다가

ctftime.org/writeup/22977

CTFtime.org / Crypto CTF 2020 / Decent RSA / Writeup

Tags: polynomials rsa Rating: 4.5

ctftime.org

이런 풀이인가 싶었는데, 여기 풀이는 n값이 대부분 0으로 구성되었을 때였다.

그러다가 포스트 하나를 찾았다.

party4bread.github.io/a-year-of-ctf-rsa/#pq%EC%97%90-0%EC%9D%B4-%EB%A7%8E%EC%9D%84-%EA%B2%BD%EC%9A%B0

A year of CTF RSA | Haven 4 BREAD

1년정도 CTF 뉴비로 있으면서 (아직도 뉴비지만) 겪었던 RSA 문제들의 유형을 대략 정리했습니다.

party4bread.github.io

오 출제자님 블로그

1234..00000000...1234 이런 식으로 구성되어 있으면 p나 q중에 하나는 100000...00 + z 형태를 취하고 있을 가능성이 높다.

따라서 n값을 0000...000을 기준으로 큰 부분을 a로 두고

작은 부분을 b로 두어 b/a를 계산하였더니 나누어 떨어졌다.

p = 1000...0000 + 961

이렇게 p와 q를 구할 수 있다.

저작자표시 비영리 동일조건

'WAR GAME > System32.kr' 카테고리의 다른 글

System32.kr [RSA108] 풀이 (0)	2021.01.19
System32.kr [EZB64] 풀이 (0)	2019.12.28
System32.kr [RSA106] 풀이 (0)	2019.05.19
System32.kr [RSA104] 풀이 (0)	2019.05.19
System32.kr [RSA105] 풀이 (0)	2019.05.19

*CTF 2021
Crypto

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from random import randint
import os
 
flag = "flag"
N=64
key=randint(0,2**N)
print key
key=bin(key)[2:].rjust(N,'0')
count=0
while True:
    p=0
    q=0
    new_key=''
    zeros=[0]
    for j in range(len(key)):
        if key[j]=='0':
            zeros.append(j)
    p=zeros[randint(0,len(zeros))-1]
    q=zeros[randint(0,len(zeros))-1]
    try:
        mask=int(raw_input("mask:"))
    except:
        exit(0)
    mask=bin(mask)[2:]
    if p>q:
        tmp=q
        q=p
        p=tmp
    cnt=0
    for j in range(0,N):
        if j in range(p,q+1):
            new_key+=str(int(mask[cnt])^int(key[j]))
        else:
            new_key+=key[j]
        cnt+=1
        cnt%=len(mask)
    key=new_key
    try:
        guess=int(raw_input("guess:"))
    except:
        exit(0)
    if guess==int(key,2):
        count+=1
        print 'Nice.'
    else:
        count=0
        print 'Oops.'
    if count>2:
        print flag
 
Colored by Color Scripter

cs

랜덤 key를 생성하고

generate random key

mask 값을 입력받아 여러번 xor을 해서

xor mask value and key

만들어진 key값을 3번 맞추면 되는데

need guessing key 3 times

mask에 0을 넣으면 xor을 몇 번을 하던지 그대로다.

however, when you put 0 in mask, xor calculation is useless.

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from random import randint
import os
from flag import flag
N=64
key=randint(0,2**N)
# print key
key=bin(key)[2:].rjust(N,'0')
count=0
while True:
    p=0
    q=0
    new_key=''
    zeros=[0]
    for j in range(len(key)):
        if key[j]=='0':
            zeros.append(j)
    p=zeros[randint(0,len(zeros))-1]
    q=zeros[randint(0,len(zeros))-1]
    try:
        mask=int(raw_input("mask:"))
    except:
        exit(0)
    mask=bin(mask)[2:]
    if p>q:
        tmp=q
        q=p
        p=tmp
    cnt=0
    for j in range(0,N):
        if j in range(p,q+1):
            new_key+=str(int(mask[cnt])^int(key[j]))
        else:
            new_key+=key[j]
        cnt+=1
        cnt%=len(mask)
    key=new_key
    try:
        guess=int(raw_input("guess:"))
    except:
        exit(0)
    if guess==int(key,2):
        count+=1
        print 'Nice.'
    else:
        count=0
        print 'Oops.'
    if count>2:
        print flag
Colored by Color Scripter

cs

guesskey 문제 잘못냈나보다.

처음 print key가 주석처리되었다.

no print key :p

key를 2진법으로 나타냈을 때 0의 개수를 구해서

p와 q를 생성하고

p와 q번째 사이번째 비트는 mask와 xor을 하고

그 외 비트는 그대로 가져온다.

mask = 1 로 두고 계속 돌리면

하위 비트 쪽은 1로 되고

0의 개수가 작아지면서 p와 q도 작아지고

그럼 더더욱 하위 비트는 유지되고

just put 1 in mask.

뭐 이런 식으로 비트들이 1로 되어간다.

it's gonna be 1s

한 150~200번 돌리면

try 150-200 times

값은 두 개중 하나가 된다.

이제 mask = 0 으로 두고 두 개중 하나를 때려박으면 된다.

choice one between 92~ and 18~

Misc

vhdxfile

encrypted by BitLocker

bitlocker2john

hashcat

password 12345678

open with tool like ftk imager

extract two pdf files

pdfcandy.com/extract-images.html

extract images from pdf files

flag image. done.

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2020 Christmas CTF [show me the pcap] Write-up (2)	2020.12.27

my write ups during the ctf & after the ctf

공부겸 오피셜 롸업 보고 풀어서 내용들을 추가했습니다.

Forensic

gchq.github.io/CyberChef/

CyberChef

gchq.github.io

SAM은 있는데 SYSTEM이 없네...;;

다른 방법을 찾아야 한다.

이 내용이 힌트였다.

저 내용만 가지고 temp 에 있는 lsass.DMP를 찾아내야 했다..

crackstation.net/

CrackStation - Online Password Hash Cracking - MD5, SHA1, Linux, Rainbow Tables, etc.

Free Password Hash Cracker Enter up to 20 non-salted hashes, one per line: Supports: LM, NTLM, md2, md4, md5, md5(md5_hex), md5-half, sha1, sha224, sha256, sha384, sha512, ripeMD160, whirlpool, MySQL 4.1+ (sha1(sha1_bin)), QubesV3.1BackupDefaults How Crack

crackstation.net

아 왜 이거 아님

여기 있는 이 stargazer.jpg를 사용해야되는 문제였다.

여기서 이 문자열을 패스워드로 해서

steghide를 돌리면 플래그가 나오는 문제였다... 아니 이걸 어케 알지..

Crypto

CipherText : @$$@@@$$@$$$$@@$@$$@@@$@@$$@@$@$@$$$@@$@@$$@@$$$@$$$@@$@@$$@@@@$@$$@@@$@@$$$@@$$@$$$$@$$@$$$@@@$@$$$@$@$@$$@$@@$@$$$@$@@@@$$@@$$@$@$$$$$@$$@@$$$@@$$@@@@@@$$@@@@@$$@@$@@@$@$$$$$@$$$@$$$@@$$@$@@@$$$@@$@@$$@$$@$@$$$@$@$@$$$@@@@@$@$$$$$@$$@$@@$@$$$@@$$@$$@$$$@@@$@@$$$@$$$@$@@@$@$$$$$@$$@$@@$@$$$@$@@@$$$$$@$

@ -> 0

$ -> 1

bin to ascii

morsecode.world/international/decoder/audio-decoder-adaptive.html

Morse Code Adaptive Audio Decoder | Morse Code World

Notes The decoder will analyse sound coming from the microphone or from an audio file. The spectrogram of the sound is shown in the main graph along with a pink region showing the frequency being analysed. If the volume in the chosen frequency is louder th

morsecode.world

Misc

holloway.nz/steg/

Twitter Secret Messages - Steganography

Hide secret messages in your tweets (or any text) with steg-of-the-dump.js. Tweet i had a great day at the beach! #sunshine Hidden Message kidnapped by pirates Tweet to post Warning: Not all of Hidden Message stored in tweet. Add more text (e.g. spaces) to

holloway.nz

이런게 있넹

www.boxentriq.com/code-breaking/hexahue

Hexahue Alphabet - decoder, translator | Boxentriq

Tool to decrypt Hexahue code. Hexahue, and variants of it, are used occasionally in geocaching mystery caches, CTFs and logic puzzles. Look specifically for the color combinations to recognize it.

www.boxentriq.com

color code cryptography

이런게 있네

저작자표시 비영리 동일조건

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2020 Christmas CTF [show me the pcap] Write-up (2)	2020.12.27
CyberTalents Digital Forensics CTF write up (0)	2020.11.29

Forenscis

3Cd7MMS7GmZMwxmRTfQwoXvxcDkGt8o4cFzwPUrg

~~then what?~~

아 base58이었넹

ah it was base58 encoded strings.

ascii 85 -> pastebin.pl/view

use "d02a7960"

pastebin.pl/view/d02a7960

morse code

S0METIM35SH4RK1SFR13ND1Y

png file.

4F -> 4E

43 -> 44

46 -> 4e

nothing here :(

move to 5527125

start and end...?

xor 6

Crypto

{ ^),!&),!!$,^$,!^%,!$&,!%),!&#,!!#,^#,!&!,!$@,^),!$!,!^@,!$$,!#&,^!,!^#,!#&,!!),^#,!!$,!@),!)^,!@%,^!,^!,!&% }

! -> 1

@ -> 2

# -> 3

...

60 170 114 64 165 147 150 173 113 63 171 142 60 141 162 144 137 61 163 137 110 63 114 120 106 125 61 61 175

~~hmm..~~

~~Isn't it this?~~

아 8진수였네

ah octal..

n = 5028492424316659784848610571868499830635784588253436599431884204425304126574506051458282629520844349077718907065343861952658055912723193332988900049704385076586516440137002407618568563003151764276775720948938528351773075093802636408325577864234115127871390168096496816499360494036227508350983216047669122408034583867561383118909895952974973292619495653073541886055538702432092425858482003930575665792421982301721054750712657799039327522613062264704797422340254020326514065801221180376851065029216809710795296030568379075073865984532498070572310229403940699763425130520414160563102491810814915288755251220179858773367510455580835421154668619370583787024315600566549750956030977653030065606416521363336014610142446739352985652335981500656145027999377047563266566792989553932335258615049158885853966867137798471757467768769820421797075336546511982769835420524203920252434351263053140580327108189404503020910499228438500946012560331269890809392427093030932508389051070445428793625564099729529982492671019322403728879286539821165627370580739998221464217677185178817064155665872550466352067822943073454133105879256544996546945106521271564937390984619840428052621074566596529317714264401833493628083147272364024196348602285804117877
e = 65537
c = 4690057718147075505522680135959473215321622692923721213835300886402444910436674094980456964526719786485709929645871497583481786451712108343985733309427211434750949557522557087475715799166136616546091244246093209194216096205011115055709130990240778725741521267153888212132276867942685123502211572949952162376597662509054070693025973089923370015547373862589488928782901235791144433788299046705518327561160954291094820233386528023713184029738780555483600166071578613803010858511582163397706626459433456365568227181855121476317779040965290548179086133039864725660837003894485377993939038122515590380127757353399577646033195886942935498851291625325622687406058565345707842924577200871090281931390828399034387159796711570518912284855782049322766568438776035673997640836043767460584670094065481165095303859142188605921710309909549354356478577687136627040919972987279885429990570784611705563443122226291405511409355924588407638851356402686178076614729462505897314633054448103933929160379080620408454649164684464952565103672481604538187885457480005006907884784443460386864548916037417492123123957243478299871616131317172973941585334012558762947082226744473068190488648000780008598569174088053018903156614111943478152720349210983651343

search n at www.factordb.com/

factordb.com

www.factordb.com

good.

0xL4ugh{I_Th1nk_Br1ll3_W45_$m4rt}

Web

make money~

I found base32 strings in cookie.

Reverse Engineering

HxD

WA! Flag!

rand() % 4 값을 반복 생성하여 파일 크기와 동일한 크기의 배열에 저장한다.

파일의 j번째 바이트와 랜덤값이 들어간 배열의 j번째 바이트를 가져와서

mystery함수 호출

Repeat generation of values (rand() %4) and push them on an array of the same size as the file size.
Gets the jth byte of the file and the jth byte of the array containing random values.

call mystery function

랜덤값이 2또는 0인경우 파일의 값과 1을 xor연산하여 리턴.

If the random value is 2 or 0, xor computes the byte and 1 of the file and returns it.

main 함수에서 srand(0)을 사용했으므로 시드가 고정.

랜덤값을 구할 수 있음.

The seed is fixed because it used sand(0) in main func.
I can get random values.

Programming

Misc

Sonic Visualiser

spectrogram

FLAG : 0xL4ugh{SP3c7ro_1s_Gr347}

github.com/jzyra/DecodeAndroidGesture

jzyra/DecodeAndroidGesture

This tool decode Android's pattern lock screen with gesture.key file in Android device. - jzyra/DecodeAndroidGesture

github.com

github.com/ribt/dtmf-decoder

ribt/dtmf-decoder

Extract phone numbers from an audio recording of the dial tones. - ribt/dtmf-decoder

github.com

66#666#8#33#888#33#777#999#8#44#444#66#4#666#66#7777#2#6#33#9#2#999#

->
noteverythingonsameway

저작자표시 비영리 동일조건

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