crackme-py is a Reverse Engineering challenge worth 30 points.

This puzzle did not include a description. A small python script crackme.py is provided:

 # Hiding this really important number in an obscure piece of code is brilliant ! # AND it's encrypted! # We want our biggest client to know his information is safe with us. bezos_cc_secret = "A:4@r%uL`M-^M0c0AbcM-MFE067d3eh2bN"  # Reference alphabet alphabet = "!\"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ"+ \             "[\\]^_`abcdefghijklmnopqrstuvwxyz{|}~"   def decode_secret(secret):     """ROT47 decode      NOTE: encode and decode are the same operation in the ROT cipher family.     """      # Encryption key     rotate_const = 47      # Storage for decoded secret     decoded = ""      # decode loop     for c in secret:         index = alphabet.find(c)         original_index = (index + rotate_const) % len(alphabet)         decoded = decoded + alphabet[original_index]      print(decoded)   def choose_greatest():     """Echo the largest of the two numbers given by the user to the program      Warning: this function was written quickly and needs proper error handling     """      user_value_1 = input("What's your first number? ")     user_value_2 = input("What's your second number? ")     greatest_value = user_value_1 # need a value to return if 1 & 2 are equal      if user_value_1 > user_value_2:         greatest_value = user_value_1     elif user_value_1 < user_value_2:         greatest_value = user_value_2      print( "The number with largest positive magnitude is "         + str(greatest_value) )   choose_greatest()

The decode_secret function provides a big clue with ROT47. I placed bezos_cc_secret into CyberChef and used the ROT47 recipe to decode the flag: