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Il contenuto di Opencaching.de è rilasciato sotto licenza Creative Commons BY-NC-ND 3.0 DE.

Richtig Sichere Angelegenheit

Bob needs your help! Please help him decrypt the secret message!

di BafTux Austria > Wien > Wien

N 48° 15.689' E 016° 23.811' (WGS84)
Conversione coordinate Dimensioni: micro Stato: disponibile Nascosta il: 15.06.2022 Pubblicata il: 07.04.2026 Ultimo aggiornamento: 07.04.2026 Listing: https://opencaching.de/OC18C6F Pubblicata anche su: geocaching.com	1 trovata 0 non trovata 0 Note 1 Osservatore 0 Ignorata 19 Pagina visite 0 Log immagini Storia Geoketry

Mappa grande

Stagionale

Necessaria preparazione

Descrizione Deutsch · English (Inglese)

Changelog

2022-06-15 - Initial release (gc only)
2026-04-07 - Publish on oc; No content changes.

Description

Bob needs your help! He and Alice are communicating via encrypted texts and Bob just received a new message from Alice:

WUaPSMWHPjsZ/t0MPYksZsfxH2en56TDyu+9CUirg/l4c5yqQxeUG8hmIYYsB7nU6XUaM1yzMp3gVco9w+olvw==

Unfortunately, Bob has forgotten the private key to decrypt the message. However, he still has the self-implemented source code of the decryption tool. Can you help Bob to recover the original message from Alice?

import base64

def main():

print('Richtig Sicherer Austausch (v1.0 - 2022-06-05)')

cipher = read_cipher("WUaPSMWHPjsZ/t0MPYksZsfxH2en56TDyu+9CUirg/l4c5yqQxeUG8hmIYYsB7nU6XUaM1yzMp3gVco9w+olvw==")

# This line asks for the key on the command line

# key = read_key()

# You can also directly pass in the key:

key = read_key(17)

decrypt(cipher, key)

def decrypt(cipher, key):

cipher_num = int.from_bytes(cipher, byteorder='big')

# Remember, e = 0x10001, Hopefully no one knows the prime factors of this number

modulus = 10941738641570527421809707322040357612003732945449205990913842131476349984288934784717997257891267332497625752899781833797076537244027146743531593354333897

plain_num = pow(cipher_num, key, modulus)

plain_bytes = plain_num.to_bytes( (plain_num.bit_length() + 7) // 8, byteorder='big')

try:

plain_text = plain_bytes.decode('utf8')

print(f'plaintext = {plain_text}')

except UnicodeDecodeError:

print('Failed to decrypt message. Wrong key!?')

def read_cipher(cipher=None):

"""

Reads the base64 encoded cipher and converts it into a byte array

"""

if cipher is None:

cipher = input('Enter the ciphertext: ')

decoded = base64.b64decode(cipher)

return [bb for bb in decoded]

def read_key(key=None):

if key is None:

key = input('Enter the key (in base 10): ')

return int(key)

if __name__ == '__main__':

main()

Further information

Brute-Force is not necessary. The result should be unambiguous. You dont have to modify the given decryption code, you only have to provide the correct key.

Acknowledgements

Thanks to the beta testers for testing & their feedback.

AimyBits
Dementophobia

Suggerimenti addizionali Decripta

[puzzle] Eba, Nqv, naq Yrbaneq pna uryc lbh. Urezna naq uvf grnz cebonoyl nf jryy.
[cache] Frr zrffntr sebz Nyvpr

A|B|C|D|E|F|G|H|I|J|K|L|M
N|O|P|Q|R|S|T|U|V|W|X|Y|Z

Utilità

Cerca geocache vicine: tutte - ricercabile - stesso tipo
Scarica come file: GPX - LOC - KML - OV2 - OVL - TXT - QR-Code
Scaricando questo file accetti i nostri termini di utilizzo e la Licenza dati.

Log per Richtig Sichere Angelegenheit 1x 0x 0x

08.08.2022, 06:46 Tankred der Zweite ha trovato la geocache

Tja, was soll ich sagen?
Das ist ganz sicher nicht mein Spezialgebiet 😔
Aber man muss ja nicht alles wissen, sondern nur wissen, wo man nachschauen/-fragen kann 😁
Und selbst dann ist es fast unmöglich zu verstehen 🤪
Aber der Fund ging rasch und ungestört über die Bühne 👍🏻
Danke für dieses knackige Rätsel!
TFTC