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Challenge "Factorization Cipher — Part 3"  

  By: admin on Jan. 6, 2012, 1:11 p.m.

In the third part of this challenge the cipher has been used to encrypt a plaintext but the spaces have not been removed before.
Read more...

 Last edited by: admin on Oct. 31, 2021, 2:54 a.m., edited 1 time in total.

Re: Challenge  

  By: DarkFibre on Jan. 7, 2012, 2:17 a.m.

45507615745573443767693145918845818417826462286915787032566954236226173685937788898568409536616579162837280388561142148740532201770389710966667592858299105619076041530050393198090089723725207320021206407956023430700647320029061021217193836480456744080948969842256330766311919254008043235604471

;)

Re: Challenge  

  By: Veselovský on Jan. 7, 2012, 3 a.m.

Nice!!!
"DARKFIBRE WAS HERE" …and many others who do not write in ciphers ;-)

I see some of you have already prepared their algorithms. Then I guess after uploading the ciphertext all you have to do is one click and the challenge will be solved in a second…
I am curious how they want to deal the points for this in such a way that it would be fair for everybody.

I do not agree with reseting the time to zero after uploading the ciphertext. My suggestion is to leave it as it is now… i.e. to count time from uploading the PDF file, because anybody who read the PDF file could prepare an algorithm.

Re: Challenge "Factorization Cipher - Part 3"  

  By: admin on Jan. 9, 2012, 3:49 p.m.

The ciphertext was now uploaded in the correct way and is provided next to the challenge. I split everything away that did reference the missing ciphertext but I wanted to keep the other posts, since I didn't want to ruin your fun :)

If there are any issues, please let me know. Sorry that my update took some time, I was occupied during the weekend.

Re: Challenge "Factorization Cipher - Part 3"  

  By: Annanasss on May 18, 2012, 4:29 p.m.

What does mean the word "longest"? Is it longest in letters or in figures?

Re: Challenge "Factorization Cipher - Part 3"  

  By: Veselovský on May 18, 2012, 10 p.m.

Quote from PDF:

P are two words from the plaintext that results from the second question: The first word is the word before the longest word, the second is the longest word itself…

The plaintext consists of letters not figures, so the longest word of plaintext is the one that has the most letters.

Re: Challenge "Factorization Cipher - Part 3"  

  By: Nikita on Nov. 12, 2012, 12:27 a.m.

Hey! Is there any other chance to handle greater numbers ( longer than 45 signs ) without using Sage? I tried it with Sage, but I don't get along with it. I'm to stupid for Sage. :(

Re: Challenge  

  By: Samacian on Aug. 8, 2013, 7:18 a.m.

Is there somebody that is able to check my working on this? I am (fairly) sure I have got this right.

I have tried multiple ways and got the same answer. Which, unfortunately, is not the required answer.

The format I have submitted the answer in is:
##,TTTTTTTT TTTTTTTTTT

:= digit, T := Uppercase character

Re: Challenge  

  By: Veselovský on Aug. 8, 2013, 1:57 p.m.

I have tried multiple ways and got the same answer. Which, unfortunately, is not the required answer.

There is a "little trap" in the first question. So I guess this is where you might have made a mistake. Check whether each of your possible plaintexts really encrypts to the given ciphertext. If that is not the case feel free to send me your solution in PM (personal message).

Re: Challenge  

  By: fmoraes on Nov. 17, 2013, 5:29 p.m.

Hmm, I thought I had it all correct but it is not taking my answer. Any help on what I am doing wrong?

Francisco

Re: Challenge  

  By: sir5 on April 16, 2021, 11:13 a.m.

There is a "little trap" in the first question. So I guess this is where you might have made a mistake. Check whether each of your possible plaintexts really encrypts to the given ciphertext. If that is not the case feel free to send me your solution in PM (personal message).

before starting the digits-shuffle: any hint about the trap? [HTML_REMOVED]

Re: Challenge  

  By: Veselovský on April 16, 2021, 12:33 p.m.

There is a "little trap" in the first question. So I guess this is where you might have made a mistake. Check whether each of your possible plaintexts really encrypts to the given ciphertext. If that is not the case feel free to send me your solution in PM (personal message).

before starting the digits-shuffle: any hint about the trap? [HTML_REMOVED]

The trap is that not all purported plaintexts encrypts to the given ciphertext. And the hint is already given - Check whether each of your possible plaintexts really encrypts to the given ciphertext.


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