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I am trying to figure out how to decrypt and also re-encrypt potentially the following, meaning that i could write a random text string and it would generate that encryption and also decrypting it produces the same text string.

Railusa.100 = PW_2AE07A4495B46110F5793D

Railusa.200 = PW_2AE07A4495B46110F6793D

Railusa.300 = PW_2AE07A4495B46110F7793D

Railusa.400 = PW_2AE07A4495B46110F0793D

Railusa.500 = PW_2AE07A4495B46110F1793D

Railusa.600 = PW_2AE07A4495B46110F2793D

Railusa.700 = PW_2AE07A4495B46110F3793D

Railusa.800 = PW_2AE07A4495B46110FC793D

Railusa.900 = PW_2AE07A4495B46110FD793D

Railusa.910 = PW_2AE07A4495B46110FD783D

Railusa.920 = PW_2AE07A4495B46110FD7B3D

Railusa.930 = PW_2AE07A4495B46110FD7A3D

Railusa.940 = PW_2AE07A4495B46110FD7D3D

Railusa.950 = PW_2AE07A4495B46110FD7C3D

Any help in determining the type of encryption used will help me tremendously.

You are my last resort in helping me figure out what this is.

Thanks!

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  • $\begingroup$ How do you know it is a cipher? What is its source? $\endgroup$
    – Weather Vane
    Commented Feb 25, 2019 at 15:51
  • $\begingroup$ @WeatherVane I believe it is a XOR Cipher, i just dont know how to confirm it. Also my grammer may be off since i speak 3 languages, i ask for forgiveness there. $\endgroup$
    – DeerSpotter
    Commented Feb 25, 2019 at 16:20
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    $\begingroup$ Is there a typo in the 4th one (Railusa.500)? Should C3 be 93 as in the others? $\endgroup$
    – Weather Vane
    Commented Feb 25, 2019 at 16:50
  • $\begingroup$ @DeerSpotter due to the shortness of the sample text, I am not optimistic about this being solvable. $\endgroup$
    – Brandon_J
    Commented Feb 25, 2019 at 16:58
  • $\begingroup$ @Brandon_J i will add more strings for us to see, i would really like to solve this... $\endgroup$
    – DeerSpotter
    Commented Feb 25, 2019 at 17:03

1 Answer 1

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Possibly this is

An XOR Cipher with a key (at least the first 11 characters) of
78 81 13 28 E0 C7 00 3E C4 49 0D

What points to this is as a possibility is

If you assume two hex characters per character, then the numbers 2, 3, 4, 5, 6 correspond to F6, F7, F0, F1, F2. XOR-ing the values together gives C4.

With the new information

The XOR with the key given below is consistent

But

There is no way to confirm with the information given. However, if it is an XOR Cipher, then the key (at least the first 11 characters) are 78 81 13 28 E0 C7 00 3E C4 49 0D. This key is XOR'd with the ASCII values of the original text and a "PW_" is pre-pended.

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  • $\begingroup$ i will generate a little bit more of these strings, i am trying to figure out how my tool creates these ciphers so i could recreate them manually. The 37 could potentially just follow a pattern as something unique this software uses, this is why i think i need at least 10-15 strings so we could see it. $\endgroup$
    – DeerSpotter
    Commented Feb 25, 2019 at 17:00
  • $\begingroup$ So it turns out the 37 was completely random, i tried a different PC to generate the strings the same way and the 37 dissapeared. I updated the above OP to include more data. $\endgroup$
    – DeerSpotter
    Commented Feb 25, 2019 at 17:24

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