[question] what would this openbullet function be in terms of python

[617]

K1= 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

K2 =010001

PASS = 'password123'

#RSA FUNCTION RSAPKCS1PAD2 "<K1>" "<K2>" "<PASS>" -> VAR "RSA" 

how would this translate to python?

[917]

all thats doing is

settings 3 variables:with the data then it compined those 3 variables with a rsa encryption

[617]

all thats doing is

settings 3 variables:with the data then it compined those 3 variables with a rsa encryption

what would I have to do to replicate this on an online tool like https://www.devglan.com/online-tools/rsa...decryption

https://cracked.io/Thread-⭐Editify-The-1...s-OP-Edits

[1.262]

Code:

import base64
from Crypto.PublicKey import RSA
from Crypto.Cipher import PKCS1_v1_5
publickey = "d0c3ff22c9a9c9fb8a92e162a38f10309c9e94b9a212ddb7b22b32eb4fc903b2a272eed35e4916728bef782eec7ba6e2c61a653b7e6ce191deffe6c38742018bf827e1de22505a5e5ee924f51eca3b8dea5265b87b9eb0876f58e33d2661d17c8b6c4cd125720f38dae1a40fa921e96939e2ca22e908fe674e8bbfa2b4f293c5a6064fbb53cc019f2d26656f9ff1b018ecbc2dd8b6b5c09d9477399566ce80861c4b0e030fda2b3e16705e3b36657d7b1cc339fd5741ab8aa2082e835295e145ac6447bd1e420bb07fea2e97861275365f450d09192da988a401fbb58d53c3ad4265e0b8b01deb50f11a2ac3980d007eda83377983b93eb29a9624ccc2867af9"

key = RSA.construct((int(publickey,16),int("010001",16)))
cipher = PKCS1_v1_5.new(key)
password = b"password"

rsa = base64.b64encode(cipher.encrypt(password))
print(rsa)

[617]

Code:

import base64
from Crypto.PublicKey import RSA
from Crypto.Cipher import PKCS1_v1_5
publickey = "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"

key = RSA.construct((int(publickey,16),int("010001",16)))
cipher = PKCS1_v1_5.new(key)
password = b"password"

rsa = base64.b64encode(cipher.encrypt(password))
print(rsa)

thx aoppreciate this !! now all i have to do is install pycrypto when i find fix for this 50000 errors

https://cracked.io/Thread-⭐Editify-The-1...s-OP-Edits