- This topic has 186 replies, 15 voices, and was last updated 4 years, 2 months ago by
.
-
Posts
-
@Mattyrat2027, thanks!
@Mattyrat2027 Thanks for the challenge!
MD5 of PLAINTEXTPANDARRAN is 9e3002b8ac3ff11dac0f7bd880bcac51
@Kford-academy 11 1101 11 01 011 0 1111 10 0 1111 0 0 1001 111 000111 01101 01101 100 100 100 101010 00 011 11111 1111 10 111 1111 001 1 01 1 101 10 0 000 101 101010 0101 000 00 01101 100 000 101 010 111 100 1 1011 1011
@TLW creating a hash never seems to work, but I can trust that you found it. You could send me a ciphertext with this code, just to confirm, but I am pretty confidet you cracked it!
@Mattyrat and @Kford-academy the ! is not a standard in the EU’s International Telecommunication Union (ITU)
(really best avoided though it can be used).@The-letter-wriggler ah so you worked it out.
By the way that tells me that you definitely solved it!
@Pandarran that’s a match
Here is a quick cipher that I created for anyone if they are bored that should not be too challenging. It relies on a special type of key..
333 9 666 55 4 11 5 7 99 11 5 8 9 666 22 9 666 333 9 777 0 99 3 5 8 666 55 777 6 88 3 6 555 9 11 4 33 333 66 11 99 99 3 666 55 3 999 8 44 6 9 7 333 11 666 4 3 11 33 5 66 8 22 888 5 66 3 666 0 9 22 5 11 777 33 44 8 444 3 66 11 22 66 9 44 5 66 3 0 99 11 8 666 5 3 222 5 44 9 99 99 9 2 3 33 555 6 6 9 7 4 666 11 777 3 11 99 99 333 11 0 22
Note: Just in case there are people from other countries (or if you happen to a have a non-standard version of the device I used to code this) I have used the British Standard Version of this cipher.
Hmmmmm. I have researched what I could about the cipher I think you are using, and the logical way to do it was not my thinking. I used one tool and it gave me a plaintext. PLAINTEXTMATTYRAT2027 – d07ba3d430e2c864b6e8f6786428c124.
To make sure, a message for you: 444 0 2 7777 7777 88 6 33 0 999 666 88 0 3 444 3 0 88 7777 33 0 8 44 444 7777 0 2 66 3 0 444 8 0 9 666 88 555 3 0 6 2 55 33 0 7777 33 66 7777 33 0 44 666 9 33 888 33 777 0 444 0 2 6 0 66 666 8 0 7777 88 777 33 0 444 333 0 444 0 2 6 0 777 444 4 44 8 0 7777 2 999 0 44 6 6 6 0 888 33 777 999 0 444 66 8 33 777 33 7777 8 444 66 4 0 444 66 0 8 44 33 0 333 666 777 88 6
@Mattyrat2027 Not quite. My cipher is a modification of the one you have written which follows a similar rule, just using a different device, and while being the inspiration for my code, it is not the correct method of solving it – keep thinking
@Everyone – For the MD5, I forgot to say take all the punctutation out of the plaintext and make it all capital letters before appending your name
Replying to @Mattyrat2027
Beware the code!
01 0000 0000 1 0000 0 010 0 1011 111 001 01 010 0 100001 1011 0 000 1 0000 0 100 00 110 00 1 000 011 0 010 0 00 10 0001 0 010 1 0 100 010101 1000 1011 1 0000 0 011 01 1011 00 110 01 0001 0 1011 111 001 1 0000 0 011 010 111 10 110 0000 01 000 0000 1010 111 100 0 100001 00 0000 01 100 0100 0 0010 1 111 001 1 1 0000 0 100 00 110 00 1 00000 011 0000 0 10 11 1011 1 0 1001 1 0 100 00 1 111 010 000 1 010 00 0110 0110 0 100 01 0100 0100 1000 001 1 1 0000 0 0100 0 1 1 0 010 000 010101 00 011 00 0100 0100 10 111 1 1000 111 1 0000 0 010 1011 111 001 011 00 1 0000 1 0000 0 1010 111 010 010 0 1010 1 11 100 00000 010101
Hint for my cipher:
If you’re board of trying to find the key to crack this cipher, perhaps a google search may help (not that I think you will find the answer on the internet).
All letters only PLAINTEXT+TLW = MD5: d5cddb67047d587f85e211bb0c3415b5
I don’t know what device you used – I simply solved it as a Monoalphabetic Substitution 😉
@Madness No such markings on mine!
- You must be logged in to reply to this topic.