basic-mod2 - PicoCTF2022
100 points
AUTHOR: WILL HONG
Description
A new modular challenge! Download the message here. Take each number mod 41 and find the modular inverse for the result. Then map to the following character set: 1-26 are the alphabet, 27-36 are the decimal digits, and 37 is an underscore.Wrap your decrypted message in the picoCTF flag format (i.e. picoCTF{decrypted_message}
)
message
104 290 356 313 262 337 354 229 146 297 118 373 221 359 338 321 288 79 214 277 131 190 377
Answer
import string
alphabet=' '+string.ascii_lowercase+string.digits+'_'
#print(alphabet[37])
text='104 290 356 313 262 337 354 229 146 297 118 373 221 359 338 321 288 79 214 277 131 190 377'
tab_text=text.split()
#print(tab_text)
for digit in tab_text:
digit=int(digit)%41
for i in range(41):
if (digit*i)%41==1: print(alphabet[i], end="")
#print(alphabet)
Po zastosowaniu takiego kodu otrzymałem 1nv3r53ly_h4rd_8a05d939
. Co było poprawną odpowiedzią :)
picoCTF{1nv3r53ly_h4rd_8a05d939}