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hack the web note
hack the web
challenge 0 start
The Answer to the Great Question… Of Life, the Universe and
Everything… Is… Forty-two.
42
challenge 5 lemon juice
ctrl a, the word with revel
invisible
challenge 15 username
reverse of your username
vkkkv
challenge 24 dont blink
if you cant read the word
marvelous
challenge 110 calculator
2x2x2x2x2x2x2x2
challenge 111 calculator II
2x5x2x5...
challenge 116 calculator III
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➤ calc C-style arbitrary precision calculator (version 2.15.1.1) Calc is open software. For license details type: help copyright [Type "exit" to exit, or "help" for help.]
translate from International
maritime signal flags
HOHESEE
challenge 8 hard work
1 2 3 4 5 6 7
✗ calc C-style arbitrary precision calculator (version 2.15.1.1) Calc is open software. For license details type: help copyright [Type "exit" to exit, or "help" for help.]
; 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 55
or
(1+10)*5
Arithmetic
progression
55
challenge 9 hard work II
1 2 3 4 5 6
➤ calc C-style arbitrary precision calculator (version 2.15.1.1) Calc is open software. For license details type: help copyright [Type "exit" to exit, or "help" for help.]
; 1 + 2 + 3 + 4 + 5 + 95 + 96 + 97 + 98 + 99
500
challenge 10 hard work III
Arithmetic
progression
1 2 3 4 5 6
➤ calc C-style arbitrary precision calculator (version 2.15.1.1) Calc is open software. For license details type: help copyright [Type "exit" to exit, or "help" for help.]
; 101*50
5050
challenge 18 rot13
terng lbh unir fhpprffshyyl qrpbqrq gur grkg nf n erjneq lbh abj trg
gur nafjre naq vg vf fcvrtryovyq (gur trezna jbeq sbe zveebe vzntr)
drag circle
or
i use vim btw, so i copy the string to vim and press g?? to tr
rot13
great you have successfully decoded the text as a reward you now get
the answer and it is spiegelbild (the german word for mirror image)
<p> <!-- noinspection SpellCheckingInspection --> <em> <strong>T</strong>he hours stretch before the glowing screen.<br> <strong>H</strong>uman language is so imprecise.<br> <strong>E</strong>verything depends on clarity.<br> <strong>A</strong>nd clarity emerges through code's decree.<br> <br> <strong>N</strong>ow is the moment, a canvas to create,<br> <strong>s</strong>culpting worlds in the digital state.<br> <strong>W</strong>ith every line, a universe unfurls.<br> <strong>E</strong>ager minds contemplate the code's weight.<br> <br> <strong>R</strong>evealing secrets in each algorithm's dance,<br> <strong>i</strong>n the binary, we find our cosmic trance. <br> <strong>S</strong>ynchronizing bytes in a digital romance: <br> <br> <strong>S</strong>eeking clarity, as we advance.<br> <strong>T</strong>apping keys, we're bound to transcend,<br> <strong>e</strong>levating thought, in this realm we intend,<br> <strong>n</strong>urturing ideas, our journey won't end,<br> <strong>o</strong>ffline is good, but online is better. </em> </p>
THEA NSWE RIS STENO
STENO
challenge 28 ads
use ublock
Litfaßsäule
challenge 79 suspicious
secure-bank-login.com
challenge 64 smiley
run in console
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const d = [68, 105, 101, 32, 65, 110, 116, 119, 111, 114, 116, 32, 108, 97, 117, 116, 101, 116, 32, 75, 111, 114, 111, 115, 101, 110, 115 , 101, 105 ,46 ]; const e = d .map( (x )=>String.fromCharCode(x)) .join ('' ) /* x x x xxxxx xxx*/ window.alert(e)
korosensei
challenge 47 progressbar
in console
for(let i = 0; i < 999; i++) work(true)
challenge 27 tactility
learn about Braille
licht
challenge 87 scratch
play game in scratch
miau
challenge 52 binary II
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➤ python Python 3.13.5 (main, Jun 21 2025, 09:35:00) [GCC 15.1.1 20250425] on linux Type "help", "copyright", "credits" or "license" for more information. >>> bin(45) '0b101101'
challenge 60 scan me
scan qrcode, i use qrazybox btw
Final Decoded string : I start with the letter Z. I have the same
colors as a QR code. Who am I?
i expect thats a letter about color in german, but not
zebra
challenge 45 characters
サンドイッチ a カタカナ
sandwich
challenge 59 secret text
its easy to manual decrypt
julian
challenge
press ctrl p in browser
or
search source code
1
<pclass="print-only">The answer is ????????????.</p>
recursive_unzip() { for file in "$1"/*; do if [ -f "$file" ]; then case "$file" in *.7z|*.zip|*.rar|*.tar|*.gz|*.bz2) dir="${file%.*}" mkdir -p "$dir" 7z x -o"$dir" "$file" > /dev/null recursive_unzip "$dir" ;; esac fi done }
recursive_unzip "."
1 2
➤ find . -type f -not -regex ".*.zip" | xargs grep -i antwort Die Antwort auf diese Aufgabe lautet Matrjoschka.
➤ calc C-style arbitrary precision calculator (version 2.15.1.1) Calc is open software. For license details type: help copyright [Type "exit" to exit, or "help" for help.]
deffind_n(): n = 1 target = "000000" whileTrue: s = "hacktheweb" + str(n) hash_value = hashlib.md5(s.encode()).hexdigest() if hash_value.startswith(target): print(hash_value) print(n) return n n += 1
if __name__ == "__main__": result_n = find_n()
1688157
challenge 95 handwriting
download fontfile and check
i use fontforge btw
kalligraph
challenge 40 terminal
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# to GOP/053/vjer > cat vjer Hmm, have a look in GDA/644/sdvd > cat sdvd Oops, I meant GDA/644/dnei > cat dnei It is worth looking in the directory ZFD, in it the directory with the highest number, and in it the alphabetically first file. > cat fgst The answer to this task is the content of the file SHY/666/pfpz > cat pfpz ????????????????????
Sonne Garten ist wir
challenge 38 metadata
download img and check
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➤ exiftool chal38-en.jpg ... Image Description : The answer is: ??????????????? (exposure time in german) ...
belichtungszeit
challenge 100 nostalgia
i use hackbar btw, change User Agent and execute
1
Mozilla/5.0 (Windows NT 10.0; WOW64; Trident/7.0; rv:11.0) like Gecko
quirky
challenge 76 timeframe
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➤ TZ='Europe/Berlin' date -d "???????????????????" +%s time1
➤ TZ='Europe/Berlin' date +%s time2
time1-time2 /60
challenge 94 original
i use google image btw, oldest link to unsplash is target
there are different levels and variants of leet from easily readable
to tota...
notice its a german word
ultimativ
challenge 92 constitution
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➤ cat tmp|sed "s/ /\n/g" |sed "s/\,//g" | sed "s/\.//g"| sort |uniq -c | sort -nr |head 32 und 28 der 22 18 die 13 zu 10 Recht 10 in 10 Die 10 den 9 werden
➤ calc C-style arbitrary precision calculator (version 2.15.1.1) Calc is open software. For license details type: help copyright [Type "exit" to exit, or "help" for help.]
; 1500+1286 2786 ; 1000+225 1225
touhou hijack
challenge 119 password
i dont know Taylor, so i make a dict and fuzz with zap
but it seem easy to guess password
lovestory
challenge 120 password II
just google swedish chess youtube
challenge 117 treasure
chamber
lrllr
back rlrrl
challenge 73 phone number
learn about DTMF
you can use your golden ear find the key
or
i use audacity btw, open audio file select a field, click analyze
-> plot spectrum
contrast two peaks with wiki
1337
challenge 103 background
search source code in map, theres a element under this challenge
challenge 83 freedom
i use zap fuzz 0-999, interval 10000ms
stupid but simple
274
challenge 43 post it
modify post date
1
answer=Klamauk
challenge 93 cipher
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a = "6d45454e0a4045480b0a7e424f0a4b44595d4f580a43590a6b6f7904" b = bytes.fromhex(a) c = [byte ^42for byte in b] print(bytes(c).decode())
a = "4c314c205f37542d4c3819345c274f2c01781d79072a44205d38502f0266183556344d600c631d7f1e661f7716624f3159374f30587b56137b046c46381561037b0429452658224a35022f70146a0e680a29" b = bytes.fromhex(a)
for key inrange(0,255): previousOutput = 0 output_bytes = [] for d in b: current_key = previousOutput ^ key decrypted_byte = d ^ current_key previousOutput = d output_bytes.append(decrypted_byte) s = bytes(output_bytes).decode() if"answer"in s.lower(): print(f"{key}, {s}") break
<!-- the byte in b is equal to xor_byte of source code--> <!-- so previousOutput = byte -->
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➤ python tmp.py 13, Apparently, even this cipher is not absolutely secure. Here's your answer: ??????.
defcipher_iii(key1,key2,input_bytes): previousOutput = 0 output_bytes = [] for i, d inenumerate(input_bytes): k = key1 if i % 2 == 0else key2 current_key = previousOutput ^ k xor_byte = d ^ current_key output_bytes.append(xor_byte) previousOutput = d s = bytes(output_bytes).decode(errors='ignore') if"antwort"in s.lower(): print(s.lower())
for key1 inrange(0,255): for key2 inrange(0,255): cipher_iii(key1,key2,b)
rosen sind rot, veilchen sind blau, zucker ist süß, und ich gebe auf.
die antwort lautet ?????????????
@>--->--->---
challenge username III
i have a server btw
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from http.server import HTTPServer, BaseHTTPRequestHandler as Handler USERNAME = "your_username_here" Handler.do_GET = lambdaself: [self.send_response(200), self.end_headers(), self.wfile.write(USERNAME.encode())] HTTPServer(('', 8000), Handler).serve_forever()